Answer:
V₀ₓ = 10.94 m/s
V₀y = 18.87 m/s
Explanation:
To find the launch velocity, we use 1st equation of motion.
Vf = Vi + at
where,
Vf = Final Velocity of Ball = Launch Speed = V₀ = ?
Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)
a = acceleration = 376 m/s²
t = time = 0.058 s
Therefore,
V₀ = 0 m/s + (376 m/s²)(0.058 s)
V₀ = 21.81 m/s
Now, for x-component:
V₀ₓ = V₀ Cos θ
where,
V₀ₓ = x-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀ₓ = (21.81 m/s)(Cos 59.9°)
V₀ₓ = 10.94 m/s
for y-component:
V₀ₓ = V₀ Sin θ
where,
V₀y = y-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀y = (21.81 m/s)(Sin 59.9°)
V₀y = 18.87 m/s
Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2.25 kg and is on an incline of 1=43.5∘ with coefficient of kinetic friction 1=0.205 . 2 has a mass of 5.45 kg and is on an incline of 2=32.5∘ with coefficient of kinetic friction 2=0.105 . The two‑block system is in
Answer:
The acceleration of [tex]M_2[/tex] is [tex]a = 0.7156 m/s^2[/tex]
Explanation:
From the question we are told that
The mass of first block is [tex]M_1 = 2.25 \ kg[/tex]
The angle of inclination of first block is [tex]\theta _1 = 43.5^o[/tex]
The coefficient of kinetic friction of the first block is [tex]\mu_1 = 0.205[/tex]
The mass of the second block is [tex]M_2 = 5.45 \ kg[/tex]
The angle of inclination of the second block is [tex]\theta _2 = 32.5^o[/tex]
The coefficient of kinetic friction of the second block is [tex]\mu _2 = 0.105[/tex]
The acceleration of [tex]M_1 \ and\ M_2[/tex] are same
The force acting on the mass [tex]M_1[/tex] is mathematically represented as
[tex]F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1[/tex]
=> [tex]M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1[/tex]
Where T is the tension on the rope
The force acting on the mass [tex]M_2[/tex] is mathematically represented as
[tex]F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]
[tex]M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]
At equilibrium
[tex]F_1 = F_2[/tex]
So
[tex]T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]
making a the subject of the formula
[tex]a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}[/tex]
substituting values [tex]a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}[/tex]
=> [tex]a = 0.7156 m/s^2[/tex]
The acceleration of the second block to the right is 2.21 m/s².
The normal force on block1 is calculated as follows;
[tex]F_n_1 = m_1g cos(\theta_1)[/tex]
The parallel force on block 1 is calculated as;
[tex]F_x_1 = m_1gsin(\theta)[/tex]
The frictional force on block 1 is calculated as;
[tex]F_k_1 = \mu_k F_n = \mu_k m_1gcos\theta_1[/tex]
The net force on block 1 is calculated as;
[tex]\Sigma F_x_1 = m_1gsin(\theta_1) - \mu_k_1m_1gcos(\theta_1)[/tex]
The normal force on block 2 is calculated as follows;
[tex]F_n_2 = m_2gcos\theta _2[/tex]
The frictional force on block 2 is calculated as;
[tex]F_k_2 = \mu k_2 m_2g cos\theta _2[/tex]
The net force on block 2 is calculated as follows;
[tex]\Sigma F_x_2 = m_2a_2\\\\ m_2g_2 sin(\theta _2) - F_k_2 - \Sigma F_x_1 = m_2a_2 \\\\m_2gsin(\theta) - F_k_2 - (m_1gsin(\theta) - \mu_k _1 m_1g cos(\theta)) = m_2a_2\\\\m_2gsin(\theta) -\mu_k_2 m_2gcos(\theta) + \mu_k _1 m_1g cos(\theta) - m_1gsin(\theta) = m_2a_2\\\\5.45( 9.8) sin(32.5) -(0.105)(5.45)(9.8)cos(32.5) + \\\\0.205( 2.25) ( 9.8)cos(43.5) - 2.25( 9.8) sin (43.5) = 5.45a_2\\\\ 12.07 = 5.45a_2\\\\a_2= \frac{12.07}{5.45} \\\\a_2 = 2.21 \ m/s^2[/tex]
Thus, the acceleration of the second block to the right is 2.21 m/s².
Learn more here: https://brainly.com/question/18839638
The energy delivered to the resistive coil is dissipated as heat at a rate equal to the power input of the circuit. However, not all of the energy in the circuit is dissipated by the coil. Because the emf source has internal resistance, energy is also dissipated by the battery as heat. Calculate the rate of dissipation of energy PbatPbatP_bat in the battery.
Answer:
P = I²r
Explanation:
ε= IR + Ir
where r is the internal resistance
A plane is flying to a city 756 km directly north of its initial location. The plane maintains a speed of 203 km/h relative to the air during its flight. (a) If the plane flies through a constant headwind blowing south at 53.5 km/h, how much time (in h) will it take to reach the city
Answer:
The answer is 5.05 hours.
Explanation:
If the plane has an airspeed of 203 km/h which only applies for air and not the ground speed, we can subtract the speed of the wind since it is a headwind in the directly opposite direction.
So the speed of the plane becomes 203 - 53.5 = 149.5 km/h which will give us the true airspeed of the plane and the ground speed as well.
From here we can calculate the time it will take to reach the city as
756 km / 149.5 km/h = 5.05 hours.
I hope this answer helps.
A child bounces a 48 g superball on the sidewalk. The velocity change of the superball is from 28 m/s downward to 17 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answer:
F = 1.2×10⁻³ N
Explanation:
From the question,
Applying newton's second law of motion,
F = m(v-u)/t................... Equation 1
Given: F = magnitude of the average force exerted on the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time of contact.
Note: let downward be negative and upward be positive.
Given: m = 48 g = 48/1000 = 0.048 kg, v = 17 m/s, u = -28 m/s (downward),
t = 1800 s
Substitute into equation 1
F = 0.048(17-[28])/1800
F = 1.2×10⁻³ N
Tired of being chased by a jaguar, you set a trap. Hoping to drop it on the jaguar, you try to push a
44.0 kg stone boulder off of the edge of a cliff that slopes down at an angle of 15.0°. Being weak with
hunger, the best you can do is push the boulder with a force of 222 N. The coefficient of kinetic friction
between the boulder and the ground is is 0.700. (Ignore static friction.)
What is the acceleration of the boulder while you push it down the incline?
Answer: acceleration = 3.27m/s^2
Explanation:
Given that the
Mass M = 44kg
Angle Ø = 15 degree
Coefficient of friction ų = 0.7
Force F = 222N
F - Fr = ma ...... (1)
Where Fr = frictional force
Fr = ųN
N = normal reaction = mg
Fr = ųmgsinØ
Fr = 0.7 × 44 × 9.81 × sin 15
Fr = 78.2N
Substitutes Fr, F and M into equation one.
222 - 78.2 = 44a
143.79 = 44a
Make a the subject of formula
a = 143.79/44
Acceleration a = 3.27 m/s^2
PLEASE HELP ME, A person has a 340 g can of organic frozen apple juice concentrate. In order to make the most dilute solution, which amount of water should he add? A. 709 mL B. 1064 mL C. 1419 mL D. 1774 mL
An ideal diatomic gas undergoes a cyclic process. In the first step, the gas undergoes an isothermal expansion from V1 to 3.00 V1. In the second step of the process the gas undergoes an isovolumetric decrease in pressure. In the third step the gas undergoes an adiabatic compression from 3.00 V1 back to V1 completing the cycle.
Required:
a. Sketch the cycle.
b. In terms of P., V. and T., determine P2, P3, T3.
c. In terms of P., V. To determine W, Q and ΔE int for each step. Take T, to be between 100K and 1000K
Answer:
Step 1
Work done = -9134.4 J
ΔQ = -9134.4 J
Step 2
ΔQ = -3570.32 J = ΔU
W = 0
Step 3
The pdV work done = 3570.32 J
The Vdp work done = 11053.37 J
Heat transferred, ΔE = 0.
Explanation:
For diatomic gases γ = 1.4
Step 1
Where:
v₂ = 3.00·v₁
On isothermal expansion of an ideal gas by Boyle's law, we have;
p₁·v₁ = p₂·v₂ which gives;
p₁·v₁ = p₂×3·v₁
Dividing both sides by v₁, we have;
p₁= 3·p₂
[tex]p_2 = \dfrac{p_1}{3}[/tex]
Hence, the pressure is reduced by a factor of 3
Work done =
[tex]n\cdot R\cdot T\cdot ln\dfrac{v_{f}}{v_{i}}[/tex]
Where:
n = 1 mole
R = 8.3145 J/(mole·K)
T = 1000 K we have
[tex]1 \times 8.3145 \times 1000 \times ln\left (\dfrac{1}{3} \right ) = -9134.4 J[/tex]
Step 2
The gas undergoes a constant volume decrease in pressure given by Charles law as follows;
[tex]\dfrac{p_2}{p3} = \dfrac{T_1}{T_3}[/tex]
Whereby p₂ > p₃, T₁ will be larger than T₃
W = 0 for constant volume process
ΔQ = m×cv×ΔT = 1 × 3.97 × -900 = -3570.32 J = ΔU
Step 3
For adiabatic compression, we have;
[tex]\dfrac{p_3}{p_1} = \left (\dfrac{V_1}{V_3} \right )^{\gamma } = \left (\dfrac{T_3}{T_1} \right )^{\frac{\gamma }{\gamma -1}}[/tex]
Where:
T₁ = 1000 K
T₃ = 100 K
We have;
[tex]\left (\dfrac{V_1}{3\cdot V_1} \right )^{\gamma } = \left (\dfrac{100}{1000} \right )^{\dfrac{\gamma}{\gamma -1}}[/tex]
[tex]\left (\dfrac{1}{3} \right ) = \left (\dfrac{1}{10} \right )^{\dfrac{1}{\gamma -1}}[/tex]
[tex]log\left (\dfrac{1}{3} \right ) = {\dfrac{1}{\gamma -1}} \times log \left (\dfrac{1}{10} \right )^[/tex]
[tex]\gamma -1 =\dfrac{log \left (\dfrac{1}{10} \right )}{ log\left (\dfrac{1}{3} \right ) } {[/tex]
∴ γ-1 = 2.096
γ = 3.096
The pdV work done =
[tex]m \times c_v \times (T_1 - T_3)[/tex]
m×R/(γ - 1)×(T₁ - T₃) =
3.97×(1000 - 100) = 3570.32 J
The Vdp work done =
[tex]m \times c_p \times (T_1 - T_3)[/tex]
[tex]c_p = k \times c_v = 3.096 \times 3.97 = 12.3 \, J/(mol\cdot K)[/tex]
12.3×(1000 - 100) = 11053.37 J
Heat transferred, ΔE = 0.
An electron moving in a wire collides again and again with atoms and travels an average distance between collisions that is called the mean free path. If the mean free path is less in some metals, what can you say about the resistance of these metals? For a given conductor, what can you do to lengthen the mean free path?
Explanation:
A substance with a short , medium, free path has improved electron flow resistance and a higher electrical resistance . Heat applications impose more molecular chaos on all materials and shorten the track further, increasing resistance of most materials. So, just refresh the material to expand the course. In certain materials, when cooled to the minimum temperature, the conductivity is substantially increased.
In a particular lab, a cube of ice (Tice = -5.5˚C) is taken and dropped into a calorimeter cup (98g) partially filled with 326 g of water (Water = 20˚C). The cup was at the same initial temperature as the water and is perfectly insulating. The final temperature of the system is 15˚C. What was the mass of ice added?
Answer:
The mass of the ice added = 16.71 g
Explanation:
The heat gained by the ice is equal to the heat lost by the calorimeter cup and the water in the cup.
But for this question, the cup is said to be perfectly insulated, hence, there is no loss of heat from the calorimeter cup.
Heat gained by the ice = Heat lost by the 326 g of water.
Let the mass of ice be m
The heat gained by the ice = (Heat gained by ice in temperature from -5.5°C to 0°C) + (Heat used by the ice to melt at 0°C) + (Heat required for the melted ice to rise in temperature from 0°C to 15°C)
Heat gained by ice in temperature from -5.5°C to 0°C = mCΔT
m = unknown mass of ice
C = Specific Heat capacity of ice = 2.108 J/g°C
ΔT = change in temperature = 0 - (-5.5) = 5.5°C
Heat gained by ice in temperature from -5.5°C to 0°C = m×2.108×5.5 = (11.594m) J
Heat used by the ice to melt at 0°C = mL
m = unknown mass of ice
L = Latent Heat of fusion of ice to water = 334 J/g
Heat used by the ice to melt at 0°C = m×334 = (334m) J
Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = mCΔT
m = unknown mass of water (which was ice)
C = Specific Heat capacity of water = 4.186 J/g°C
ΔT = change in temperature = 15 - 0 = 15°C
Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = m×4.186×15 = (62.79m) J
Total heat gained by the ice = 11.594m + 334m + 62.79m = (408.384m) J
Heat lost by the water in the calorimeter cup = MCΔT
M = mass of water in the calorimeter cup = 326 g
C = specific heat capacity of water = 4.186 J/g°C
ΔT = change in temperature = 20 - 15 = 5°C
Heat lost by the water in the calorimeter cup = 326×4.186×5 = 6,823.18 J
Heat gained by the ice = Heat lost by the 326 g of water.
408.384m = 6,823.18
m = (6,823.18/408.384)
m = 16.71 g
Hope this Helps!!!
Use Hooke's Law to determine the variable force in the spring problem. A force of 450 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 40 centimeters to 70 centimeters?
Answer:
Work Done = 67.5 J
Explanation:
First we find the value of spring constant (k) using Hooke's Law. Hooke's is formulated as:
F = kx
where,
F = Force Applied = 450 N
k = Spring Constant = ?
x = Stretched Length = 30 cm = 0.3 m
Therefore,
450 N = k(0.3 m)
k = 450 N/0.3 m
k = 1500 N/m
Now, the formula for the work done in stretching the spring is given as:
W = (1/2)kx²
Where,
W = Work done = ?
k = 1500 N/m
x = 70 cm - 40 cm = 0.3 m
Therefore,
W = (1/2)(1500 N/m)(0.3 m)²
W = 67.5 J
A small block is released from rest at the top of a frictionless incline. The block travels a distance 0.633 m in the first second after it is released. How far does it travel in the next second
Answer:1.89 m
Explanation:
Given
Block travels [tex]0.63\ m[/tex] in first second
It is released from rest i.e. initial speed is zero (u=0)
using
[tex]s=ut+\frac{1}{2}at^2[/tex]
where a=acceleration
here acceleration is the component of gravity on incline plane (say [tex]\theta [/tex])
so
[tex]s_1=\frac{1}{2}\times g\sin \theta (1)^2[/tex]
[tex]0.633\times 2=9.8\sin \theta \times 1^2[/tex]
[tex]\sin\theta =0.1291[/tex]
[tex]\theta =7.41^{\circ}[/tex]
So distance traveled in [tex]2\ sec[/tex]
[tex]s=\frac{1}{2}\times g\sin \theta (2)^2[/tex]
[tex]s=0.5\times 9.8\times \sin (7.41)\times 4[/tex]
[tex]s=2.52\ m[/tex]
So distance traveled in [tex]2^{nd}\ sec[/tex] is
[tex]s-s_1=2.52-0.633=1.89\ m[/tex]
Consider a blackbody that radiates with an intensity I1I1I_1 at a room temperature of 300K300K. At what intensity I2I2I_2 will this blackbody radiate when it is at a temperature of 400K400K
Answer:
Explanation:
We shall apply Stefan's formula
E = AσT⁴
When T = 300
I₁ = Aσ x 300⁴
When T = 400K
I₂ = Aσ x 400⁴
I₂ / I₁ = 400⁴ / 300⁴
= 256 / 81
= 3.16
I₂ = 3.16 I₁ .
You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock so it has velocity 12.0 m/s relative to the earth at an angle of 35.0∘ above the horizontal. Your mass is 72.0 kg and the rock’s mass is 3.50 kg . What is your speed after you throw the rock?
Answer:
0.4778 m/s
Explanation:
To solve this question, we will make use of law of conservation of momentum.
We are given that the rock's velocity is 12 m/s at 35°. Thus, the horizontal component of this velocity is;
V_x = (12 m/s)(cos(35°)) = 9.83 m/s.
Thus, the horizontal component of the rock's momentum is;
(3.5 kg)(9.83 m/s) = 34.405 kg·m/s.
Since the person is not pushed up off the ice or down into it, his momentum will have no vertical component and so his momentum will have the same magnitude as the horizontal component of the rock's momentum.
Thus, to get the person's speed, we know that; momentum = mass x velocity
Mass of person = 72 kg and we have momentum as 34.405 kg·m/s
Thus;
34.405 = 72 x velocity
Velocity = 34.405/72
Velocity = 0.4778 m/s
Question 1 [7]
Hydrogen gas is used in a Carnot cycle having an efficiency of 60% with a low temperature of 300K. During the heat rejection the pressure changes from 90 kPa to 120 kPa. Find the high and low temperature heat transfer and the net cycle work per unit mass of hydrogen.
Question 2 [8]
A rigid insulated container has two rooms separated by a membrane. Room A contains 1 kg of air at 200°C and Room B contains 1.5 kg of air at 20°C, both rooms are at 100 kPa. Consider two different cases
A. The Heat transfer between A and B creates a final uniform T
B. The membrane breaks and air comes to a uniform state.
For both cases find the final temperature. Are the two-process reversible and different? Explain.
Measure Your Reaction Time Here's something you can try at home-an experiment to measure your reaction time. Have a friend hold a ruler by one end, letting the other end hang down vertically. At the lower end, hold your thumb and index finger on either side of the ruler, ready to grip it. Have your friend release the ruler without warning. Catch it as quickly as you can.If you catch the ruler 5.7 cm from the lower end, what is your reaction time?
Express your answer using two significant figures.
Answer:
Explanation:
I catch the ruler 5.7 cm from lower end that means my reaction time is equal to time of fall of ruler as free fall under gravity .
h = 1/2 gt²
t = [tex]\sqrt{\frac{2h}{g} }[/tex]
= [tex]\sqrt{\frac{2\times 5.7}{9.8} }[/tex]
= 1.078 s
= 1.1 s .
A NFL linebacker runs the 100m sprint in 12s. What is his final velocity?
Answer:
Final velocity of NFL line backer is 16.67 m/s.
Explanation:
From the question, we have following data about the NFL line backer:
Initial Speed of line backer = Vi = 0 m/s (Since, he starts from rest)
Distance covered by NFL line backer = s = 100 m
Time taken by the NFL line backer to complete 100 m sprint = t = 12 s
Acceleration of NFL line backer during sprint = a
Final Velocity of NFL line backer = Vf = ?
First we need to find the acceleration of the NFL line backer. For that purpose we will use 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
using values:
100 m = (0 m/s)(12 s) + (0.5)(a)(12 s)²
100 m/72 s² = a
a = 1.39 m/s²
Now, we use 1st equation of motion to find Vf:
Vf = Vi + at
Vf = 0 m/s + (1.39 m/s²)(12 s)
Vf = 16.67 m/s
a 350g mass as attached to a spring of constant 5.2N/m and set into oscillation with amplitude of 10 cm. what is the frequency, period, maximum velocity and the maximum force in the spring?
Explanation:
It is given that,
Mass of the object, m = 350 g = 0.35 kg
Spring constant of the spring, k = 5.2 N/m
Amplitude of the oscillation, A = 10 cm = 0.1 m
Frequency of a spring mass system is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{5.2}{0.35}} \\\\f=0.613\ Hz[/tex]
Time period:
[tex]T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.613}\\\\T=1.63\ s[/tex]
Maximum velocity in the spring is given by :
[tex]v=A\omega[/tex]
[tex]v=A\sqrt{\dfrac{k}{m}} \\\\v=0.1\times \sqrt{\dfrac{5.2}{0.35}} \\\\v=0.38\ m/s[/tex]
The maximum force acting in the spring is :
[tex]F=-kx\\\\F=kA\\\\F=5.2\times 0.1\\\\F=0.52\ N[/tex]
Hence, this is the required solution.
Suppose you catch and hold a baseball, and then someone invites you to catch and hold a bowling ball with either the same momentum or the same kinetic energy as the baseball. Which would you choose
Answer:
The same momentum will be the best option.
Explanation:
Let's recall that the force will be express in terms of the momentum. We can write the force as the variation of the momentum over time.
[tex]F=\frac{dp}{dt}[/tex]
This is the force needed to stop the base ball or the bowling ball.
if we will choose the same kinetic energy it would imply an increase of momentum, because of the difference of the masses, and therefore an increase of the force. We do not want this.
Now, if we choose the same momentum the kinetic energy will increase, but the force will the same. We want the less force as possible to stop it, and we have the same at least.
Therefore the same momentum would be the best option.
I hope it helps you!
The best choice to catch and hold the bowling ball will be; with the same momentum
We know that formula for impulse is;
Impulse = Force x Time
And we know that change in momentum is equal to impulse. Thus;
Change in momentum = F × t
ΔP = F × t
F = ΔP/t
This formula represents the force required to stop the baseball or the bowling ball.
Now, momentum is proportional to the square root of kinetic energy.
Now, since momentum is directly proportional to velocity, while kinetic energy is proportional to the square of the velocity, it means that if kinetic energy is quadrupled, then the momentum will become double.
Now, the collision in the question is completely inelastic and as such, all the bowling balls kinetic energy will be in inelastic collision, the kinetic energy is lost.
Formula for the kinetic energy in terms of the momentum here is;
K = p²/2m
Looking at it overall, we can say that the best choice to catch and hold the bowling ball will be with the same momentum since it results in lesser force.
Read more at;https://brainly.com/question/13994440
A distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. A) Decide whether this galaxy is approaching or receding from the earth. Give your reasoning. B) Find the speed of the galaxy relative to the earth.
Answer:
A) receding from the earth
B) [tex]3.078x10^6m/s[/tex]
Explanation:
A) receding from the earthThe wavelength went from 434.1nm to 438.6nm, there was an increase in wavelength (also knowecn as redshift due to the doppler efft), this increase is due to the fact that the source that emits the radiation (the distant galaxy) is moving away and therefore the light waves it emits are "stretched", causing us to see a wavelength greater than the original.
B) [tex]3.078x10^6m/s[/tex]to calculate the relative speed we use the following formula:
[tex]v_{rel}=c(1-\frac{\lambda_{1}}{\lambda_{2}} )[/tex]
where [tex]c[/tex] is the speed of light: [tex]c=3x10^8m/s[/tex]
[tex]\lambda_{1}[/tex] is the wavelength emited by the source, and
[tex]\lambda_{2}[/tex] is the wavelength measured on earth.
we substitute all the values and do the calculations:
[tex]v_{rel}=(3x10^8m/s)(1-\frac{434.1nm}{438.6nm} )\\\\v_{rel}=(3x10^8m/s)(1-0.98974)\\\\v_{rel}=(3x10^8m/s)(0.01026)\\\\v_{rel}=3.078x10^6m/s[/tex]
the relative speed is: [tex]3.078x10^6m/s[/tex]
______ ______ are created when something is caused to vibrate.
Answer: the answer is MIRANDA COOPER
Explanation:
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You are designing a hydraulic lift for a machine shop. The average mass of a car it needs to lift is about 1500 kg. You wish to exert a force on a smaller piston of not more than 550N .a) What should be the specifications on the dimensions of the pistons?Asmall piston/Alarge piston = ???b) How far down will you need to push the piston in order to lift the car 50cm ?h = ???
Answer:
(a) Area(small piston)/Area(large piston) = 0.037
(b) h = 1336.36 cm = 13.36 m
Explanation:
(a)
The stress on the smaller piston is equally transmitted to the larger piston, in a hydraulic lift. Therefore,
Stress (small piston) = Stress (large piston)
Force (small piston)/Area (small piston) = Force (Large Piston)/Area (Large Piston)
Area(small piston)/Area(large piston) = Force (small piston)/Force(Large piston)
Area(small piston)/Area(large piston) = 550 N/(1500 kg)(9.8 m/s²)
Area(small piston)/Area(large piston) = 0.037
(b)
The work is also transmitted equally to the large piston. So,
Work(small piston) = Work(Large Piston)
Force(small piston).Displacement(small piston) = Force(large piston).Displacement(small piston)
(550 N)(h) = (1500 kg)(9.8 m/s²)(50 cm)
h = 735000 N.cm/550 N
h = 1336.36 cm = 13.36 m
(a) The ratio of area smaller piston to area of larger piston is 0.037.
(b) The distance the smaller piston will be pushed down to lift the car at the given height is 13.36 m.
The given parameters;
mass of the car, m = 1500 kgforce on the smaller piston, F₁ = 550 Nlet the area of the small piston = A₁
let the area of the large piston = A₂
Apply constant pressure principle as shown below;
[tex]P = \frac{F_1}{A_1} = \frac{F_2}{A_2} \\\\\frac{A_1}{A_2} = \frac{A_{small}}{A_{large}} = \frac{F_1}{F_2} = \frac{550}{mg} \\\\ \frac{A_{small}}{A_{large}} = \frac{550}{1500 \times 9.8} \\\\ \frac{A_{small}}{A_{large}} = 0.037[/tex]
The height the car was raised = 50 cm = 0.5 mThe distance the effort will be applied is calculated as follows;
[tex]550 d = mgh\\\\550d = (1500 \times 9.8 \times 0.5)\\\\550 d = 7350\\\\d = \frac{7350}{550} \\\\d = 13.36 \ m[/tex]
Thus, the distance the smaller piston will be pushed down to lift the car at the given height is 13.36 m.
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What is the mass of the heaviest book this person can hold onto vertically before it slips out of his or her fingers? The coefficient of static friction of the surface between the fingers and the book cover is 0.65.
Answer:
0.795 kg
Explanation:
Assuming the complete question:
A person with compromised pinch strength in their fingers can only exert a normal force of 6.0 N to either side of a pinch-held object. What is the mass of the heaviest book this person can hold onto vertically before it slips out of his or her fingers? The coefficient of static friction of the surface between the fingers and the book cover is 0.60
SOLUTION:
The maximum weight of the book will equal the maximal friction force that can be produced:
m g = 2 f [tex]F{normal}[/tex]
Note that there are two sides of the book, so the friction force equals 2 times the friction force on one side (hence the factor 2).
So the maximum mass of the book is
m = 2 f [tex]F_n[/tex]/ g
m = 2[tex]\times[/tex]0.65[tex]\times[/tex] 6.0N / (9.81N/kg)
m = 0.795 kg
4–72 A person puts a few apples into the freezer at 215°C to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of 20°C, and the heat transfer coefficient on the surfaces is 8 W/m2·K. Treating the apples as 9-cm-diameter spheres and taking their properties to be r 5 840 kg/m3, cp 5 3.81 kJ/kg·K, k 5 0.418 W/m·K, and a 5 1.3 3 1027 m2/s, determine the center and surface temperatures of the apples in 1 h. Also, determine the amount of heat transfer from each apple. Solve this problem using analytical one-term approximation method (not the Heisler charts).
Complete and Clear Question:
A person puts a few apples into the freezer at -15°C to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of 20°C, and the heat transfer coefficient on the surfaces is 8 W/m2·K. Treating the apples as 9-cm-diameter spheres and taking their properties to be [tex]\rho =[/tex] 840 kg/m3, [tex]c_{p} =[/tex] 3.81 kJ/kg·K, k = 0.418 W/m·K, and [tex]\alpha = 1.3 * 10^{-7} m^{2} /s[/tex], determine the center and surface temperatures of the apples in 1 h. Also, determine the amount of heat transfer from each apple. Solve this problem using analytical one-term approximation method (not the Heisler charts).
Answer:
Temperature at the center of the apple, T(t) = 11.215°C
Temperature at the surface of the apple, T(r,t) = 2.68°C
Amount of heat transfer from each apple, Q = 21.47 kJ
Explanation:
For clarity and easiness of expression, the calculations are handwritten and attached as a file. Check the attached files for the complete calculation.
which statement about the image Formed by a plane mirror is correct?
1. the image is larger then the object
2. the image is smaller then the object
3. the image is twice as far from the mirror as the object
4. the image is virtual.
Answer:
The image is virtual
number-4
A long glass rod A is initially at 22.0°C. A second rod B is identical to rod A and has the same mass and initial temperature as A. The same amount of heat is supplied to both rods and the two rods A and B reach final temperatures of 86.3°C and 190.0°C respectively. If the specific heat of glass is 0.2007 kcal/(kg· °C), what is the specific heat of the material from which rod B is made?
Answer:
[tex]c_B=0.0768\frac{kcal}{kg\cdot\°C}[/tex]
Explanation:
In order to calculate the specific heat of the material B, you use the following formula for the change in the temperature of a substance, where an amount of heat Q is given to the substance:
[tex]Q=mc(T_2-T_1)[/tex]
Q: amount oh heat
m: mass of the substance
T2: final temperature
T1: initial temperature
c: specific heat of the substance.
If QA and QB are the heat of material A and B, you have:
[tex]Q_A=m_Ac_A(T_{2A}-T_{1A})\\\\Q_B=m_Bc_B(T_{2B}-T_{1B})[/tex]
both materials have the same mass, mA = mB
cA: specific heat of A = 0.2007 kcal/(kg.°C)
cB: specific heat of B = ?
T2A: final temperature of A = 86.3°C
T1A: initial temperature of A = 22.0°C
T2B: final temperature of B = 190.0°C
T1B: initial temperature of B = 22.0°C
In this case you have that both material A and B receive the same amount of heat Q. Then, you equal QA with QB and solve for cB:
[tex]m_Ac_A(T_{2A}-T_{1A})=m_Bc_B(T_{2B}-T_{1B})\\\\c_B=\frac{c_A(T_{2A}-T_{1A})}{(T_{2B}-T_{1B})}\\\\c_B=\frac{(0.2007kcal/(kg.\°C))(86.3\°C-22.0\°C)}{190.0\°C-22.0\°C}\\\\c_B=0.0768\frac{kcal}{kg\cdot\°C}[/tex]
hence, the specific heat of the second rod B is 0.0768kcal/(kg°C)
A red ball is thrown down with an initial speed of 1.6 m/s from a height of 25 meters above the ground. Then, 0.4 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24 m/s, from a height of 1.2 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s^2. 1) What is the speed of the red ball right before it hits the ground?
2) How long does it take the red ball to reach the ground?
3) What is the height of the blue ball 2 seconds after the red ball is thrown?
4) How long after the red ball is thrown are the two balls in the air at the same height?
Answer:
1. v = 22.2 m/s
2. t = 2.25 seconds
3. h = 27.05 m
4. t = 1.16 seconds
Explanation:
The questions involve motion under the influence of gravity
1. Using the formula v² = u² + 2gh
where u = 1.6 m/s; g= 9.81 m/s²; h = 25 m; v = ?
v² = (1.6)² + 2 * 9.81 * 25
√v² = √493.06
v = 22.2 m/s
2. Using h = ut + 1/2 gt²
where h = 25 m; u = 0 (since velocity on reaching the ground is zero); g = 9.81 m/s²; t = ?
therefore, h = 1/2 gt²
making t subject of the formula, t = √ (2*h /g)
t = √ (2 * 25 / 9.81)
t = 2.25 seconds
3. Time of travel for the blue ball, t = 2 - 0.4 = 1.6s
using h = ut - gt²
u = 24 m/s; t = 1.6 s; g = 9.81 m/s²
note: since the ball is travelling against gravity, g is negative
h = 24 * 1.6 - 11/2 * 9.81 * 1.6²
h = 38.4 - 12.55 = 25.85 m
since height above the ground is 1.2 m,
total height h = 25.85 m + 1.2 m
h = 27.05 m
4. Let the time of travel of the red ball be t seconds.
So the time of travel of the blue ball = (t - 0.4) seconds.
Both the balls are at the same height :
25 - s = 1.2 + h where s & h are the displacements of the red & the blue ball respectively.
25 - (ut + 1/2 gt2) = 1.2 + (ut - 1/2 gt2)
25 - (1.6 t + 0.5 * 9.8 t²) = 1.2 + (24(t-0.4) - 0.5*9.8*(t-0.4)²)
solving the equation above for the time after which both the balls are at the same height.
25 - 1.6t - 4.9t² = 1.2 + 24t - 9.6 - 4.9t² + 3.92t - 0.784
collecting like terms
(25 - 1.2 + 9.6 + 0.784) = (24 + 3.92 + 1.6) * t
t = 34.184 / 33.44
t = 1.16 seconds
A radar antenna is rotating and makes one revolution every 24 s, as measured on earth. However, instruments on a spaceship moving with respect to the earth at a speed v measure that the antenna makes one revolution every 44 s. What is the ratio v/c of the speed v to the speed c of light in a vacuum
Answer:
0.838
Explanation:
The ratio v/c of the speed v to the speed c of light in a vacuum is shown below:
Given that
[tex]\triangle t_0 = 24\ seconds[/tex] = time interval for one revolution
[tex]\triangle t = 44\ seconds[/tex] = time interval measured with speed v
based on the given information, the ratio v/c of the speed v to the speed c of light in a vacuum is
[tex]\triangle t = \frac{\triangle t_0}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]
[tex]{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{\triangle t_0}{\triangle t}[/tex]
Now squaring both the sides
[tex]\frac{v^2}{c^2} = 1 - \frac{(\triangle t_0)^2}{(\triangle t)^2}[/tex]
Now remove the squaring root from both the sides and putting the values
[tex]\frac{v}{c} = {\sqrt{1 - \frac{(\triangle t_0)^2}{(\triangle t)^2}[/tex]
[tex]= {\sqrt{1 - \frac{(24)^2}{(44)^2}[/tex]
= 0.838
what is the application of physics
Answer:
1) We can estimate the age of the earth
2) we can calculate the speed of anything
3) we can also calculate gravity, e.t.c.
Explanation:
I could give you more just ask
Which of the following statements best describes the visible spectrum of light as seen by the human eye? The lowest frequency appears , and the highest frequency appears violet. B. The lowest frequency appears red, and the highest frequency appears yellow. C. The lowest frequency appears green, and the highest frequency appears violet. D. The lowest frequency appears green, and the highest frequency appears yellow.
Answer:
The Answer is red is the lowest and violet is the highest frequency
Explanation:
I think that means A, because the red isn't in the question. But I'm sure red is the lowest frequency and violet is the highest in the visible light spectrum
The visible spectrum as it appears to the human eye is that A. the lowest frequency appears red, and the highest frequency appears violet.
Humans can only view a portion of the electromagnetic spectrum and this portion is known as visible light.
The colors in this visible light have different frequencies which include:
Violet with a frequency range of 700 - 790 THzBlue with a frequency range of 600 - 700 THzGreen with a frequency range of 530 - 580 THz Yellow with a frequency range of 510–530 THzOrange with a frequency range of 480–510 THz and, Red with a frequency range of 400–480 THzNotice how red is the lowest frequency and violet is the highest so we can conclusively say that the lowest frequency appears red, and the highest frequency appears violet.
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On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs account for 80%. We can model a spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally. Suppose a 74.0 kg skater is 1.80 m tall, has arms that are each 70.0 cm long (including the hands), and a trunk that can be modeled as being 35.0 cm in diameter. If the skater is initially spinning at 68.0 rpm with her arms outstretched, what will her angular velocity 2 be (in rpm ) after she pulls in her arms and they are at her sides parallel to her trunk? Assume that friction between the skater and the ice is negligble.
Answer:
176.38 rpm
Explanation:
mass percentage of arms and legs = 13%
mass percentage of legs and trunk = 80%
mass percentage of head = 7%
Total mass of the skater = 74.0 kg
length of arms = 70 cm = 0.7 m
height of skater = 1.8 m
diameter of trunk = 35 cm = 0.35 m
Initial angular momentum = 68 rpm
We assume:
The spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally.friction between the skater and the ice is negligible.We split her body into two systems, the spinning hands as spinning rods
1. Each rod has moment of inertia = [tex]\frac{1}{3} mL^{2}[/tex]
mass m of the arms is 13% of 74 kg = 0.13 x 74 = 9.62 kg
mass of each side will be assumed to be 9.62/2 = 4.81 kg
L = length of each arm
therefore,
I = [tex]\frac{1}{3}[/tex] x 4.81 x [tex]0.7^{2}[/tex] = 0.79 kg-m for each arm
2. Her body as a cylinder has moment of inertia = [tex]\frac{1}{2} mr^{2}[/tex]
r = radius of her body = diameter/2 = 0.35/2 = 0.175 m
mass of body trunk = (80% + 7%) of 74 kg = 0.87 x 74 = 64.38 kg
I = [tex]\frac{1}{2}[/tex] x 64.38 x [tex]0.175^{2}[/tex] = 0.99 kg-m
We consider each case
case 1: Body spinning with arm outstretched
Total moment of inertia = sum of moments of inertia of both arms and moment of inertia of body trunk
I = (0.79 x 2) + 0.99 = 2.57 kg-m
angular momentum = Iω
where ω = angular speed = 68.0 rpm = [tex]\frac{2\pi }{60}[/tex] x 68 = 7.12 rad/s
angular momentum = 2.57 x 7.12 = 18.29 kg-rad/m-s
case 2: Arms pulled down parallel to trunk
The momentum of inertia will be due to her body trunk alone which is 0.91 kg-m
angular momentum = Iω
= 0.99 x ω = 0.91ω
according to conservation of angular momentum, both angular momentum must be equal, therefore,
18.29 = 0.99ω
ω = 18.29/0.99 = 18.47 rad/s
18.47 ÷ [tex]\frac{2\pi }{60}[/tex] = 176.38 rpm
