Answer:
t = 5.77 s
Explanation:
This exercise will use Newton's second law for rotational motion
τ = I α
α = τ / I
α = 66/175
α = 0.3771 rad/s²
now we can use the rotational kinematics relations, remember that all angles must be in radians
θ = 1 rev = 2π radians
θ = w₀ t + ½ α t²
as the wheel starts from rest w₀ = 0
t = √ (2θ/α)
let's calculate
t = √ (2 2π / 0.3771)
t = 5.77 s
Watching the World Series (only as an example of physics in action), you wonder about the ability of the catcher to throw out a base runner trying to steal second. Suppose a catcher is crouched down behind the plate when he observes the runner breaking for second. After he gets the ball from the pitcher, he throws as hard as necessary to second base without standing up. If the catcher throws the ball at an angle of 30 degrees from the horizontal so that it is caught at second base at about the same height as that catcher threw it, how much time does it take for the ball to travel the 120 feet from the catcher to second base
Answer:
The time is [tex]t_t = 3.7583 \ s [/tex]
Explanation:
From the question we are told that
The angle is [tex]\theta = 30^o[/tex]
The horizontal distance is [tex]d = 120 \ ft[/tex]
Generally when the ball is at maximum height before descending the velocity is zero and this velocity can be mathematically represented as
[tex]v = u + at[/tex]
here a = -g the negative sign is because the direction of motion is against gravity
So
[tex]v = v_i + at[/tex]
Here[tex] v_i [/tex] is the vertical component of the initial velocity of the ball which is mathematically
represented as
[tex]v_i = usin(\theta )[/tex]
So
=> [tex]0 = usin(\theta ) -9.8t[/tex]
Generally the total time taken to travel the 120 ft is mathematically represented as
[tex]t_t = \frac{120}{v_h}[/tex]
Here [tex]v_h[/tex] is the horizontal component of the initial velocity which is mathematically represented as
[tex]v_h = u cos(\theta )[/tex]
So
[tex]t_t = \frac{120}{ u cos(\theta )}[/tex]
Generally the time taken to reach the maximum height is
[tex]t = \frac{t_t}{2}[/tex]
=> [tex]t = \frac{120}{ u cos(\theta )} * \frac{1}{2} [/tex]
=> [tex]t = \frac{60}{ u cos(\theta )} [/tex]
So
[tex]0 = usin(\theta ) -9.8* [\frac{60}{ u cos(\theta )}][/tex]
[tex] usin(\theta ) = 9.8* [\frac{60}{ u cos(\theta )}][/tex]
[tex] usin(\theta ) * u cos(\theta) =60* 9.8 [/tex]
[tex] u^2 sin(\theta ) cos(\theta) =60* 9.8 [/tex]
[tex] u^2 sin(30 ) cos(30) =60* 9.8 [/tex]
[tex] u^2 * \frac{1}{2}* \frac{\sqrt{3}}{2} =588.6 [/tex]
[tex] u^2 *\sqrt{3} =2354.4 [/tex]
[tex] u^2 = 1359.31 [/tex]
[tex] u = 36.87 \ ft/s [/tex]
Substituting this value into the equation for total time
[tex]t_t = \frac{120}{36.87 cos(30 )}[/tex]
[tex]t_t = 3.7583 \ s [/tex]
I WILL MARK YOU AS BRAINLIEST IF RIGHT
A 75 kg skier travels downhill 1200 meters in 56 seconds. What is the velocity of the skier?
500m=?m² va rooooooog
Answer:
250000 [m²]
Explanation:
A unit analysis has to be performed, where the unit of length is the meter. And the unit for the area is the square meter (m²)
L = 500 [m]
Therefore if we want to convert this length to meters we must square the length.
A = 500² = 500*500 = 250000 [m²]
In an RL series circuit, an inductor of 4.74 H and a resistor of 9.33 Ω are connected to a 26.4 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor? Answer in units of J.
Answer:
The energy is [tex]U = 18.98 \ J [/tex]
Explanation:
From the question we are told that
The inductor is [tex]L = 4.74 \ H[/tex]
The resistance of the resistor is [tex]R = 9.33 \ \Omega[/tex]
The voltage of the battery is [tex]V = 26.4 \ V[/tex]
Generally the current flowing in the circuit is mathematically represented as
[tex]I = \frac{V}{R}[/tex]
=> [tex]I = \frac{26.4}{9.33 }[/tex]
=> [tex]I = 2.83 \ A[/tex]
Generally the corresponding energy stored in the circuit is
[tex]U = \frac{1}{2} * L * I^2[/tex]
[tex]U = \frac{1}{2} * 4.74 * 2.83 ^2[/tex]
[tex]U = 18.98 \ J [/tex]
A 2-m3 rigid insulated tank initially containing saturated water vapor at 1 MPa is connected through a valve to a supply line that carries steam at 400°C. Now the valve is opened, and steam is allowed to flow slowly into the tank until the pressure in the tank rises to 2 MPa. At this instant the tank temperature is measured to be 300°C. Determine the mass of the steam that has e
Answer:
5.6449
9 mpa
Explanation:
we are to determine mass of steam that has entered and also the pressure of steam.
After solving
Mass of steam = m2 - m1
= 15.925-10.2901
= 5.6449kg
Then the enthalpy of steam was calculated to be 3109.26
Using steam table, tl = 400⁰c
Hl = 3109.26
Supply line pressure = 9mpa
Please refer to attachment for all calculations
What are arrecture pill muscles
Answer:
Arrector Pili Muscle - This is a tiny muscle that attaches to the base of a hair follicle at one end and to dermal tissue on the other end. In order to generate heat when the body is cold, the arrector pili muscles contract all at once, causing the hair to "stand up straight" on the skin.
Explanation:
A small glass bead has been charged to 20 nC. What is the magnitude of acceleration in m/s^2 of an electron that is 1.0 cm from the center of the bead? (mass of an electron= 9.1x10^-31 kg)
Answer:
The acceleration is 3.16x10¹⁷ m/s².
Explanation:
First, we need to find the magnitude of the Coulombs force (F):
[tex] |F| = \frac{Kq_{1}q_{2}}{d^{2}} [/tex]
Where:
K is the Coulomb constant = 9x10⁹ Nm²/C²
q₁ is the charge = 20x10⁻⁹ C
q₂ is the electron's charge = -1.6x10⁻¹⁹ C
d is the distance = 1.0 cm = 1.0x10⁻² m
[tex]|F| = \frac{Kq_{1}q_{2}}{d^{2}} = \frac{9\cdot 10^{9}Nm^{2}/C^{2}*20 \cdot 10^{-9} C*(-1.6\cdot 10^{-19} C)}{(0.01 m)^{2}} = 2.88 \cdot 10^{-13} N[/tex]
Now, we can find the acceleration:
[tex] a = \frac{F}{m} = \frac{2.88 \cdot 10^{-13} N}{9.1 \cdot 10^{-31} kg} = 3.16 \cdot 10^{17} m/s^{2} [/tex]
Therefore, the acceleration is 3.16x10¹⁷ m/s².
I hope it helps you!
A man walking at 0.5 m/s increases his pace to 0.8 m/s in 3 seconds. What is his acceleration?
Acceleration = (change in speed) / (time for the change)
Change in speed = (New speed) - (old speed)
Change in speed = (0.8 m/s - 0.5 m/s) = 0.3 m/s
Time for the change = 3 seconds
Acceleration = (0.3 m/s) / (3 s)
Acceleration = 0.1 m/s²
(a hair more than 1 percent of 1 G.)
Listening to the radio, you can hear two stations at once. Describe this wave interaction.
A)
diffraction
B)
interference. C)
reflection
D)
refraction
Answer:
Answer is C or B
Explanation:
Consider a system of a cliff diver and the Earth. The gravitational potential energy of the system decreases by 28,000 J as the diver drops to the water from a height of 40.0 m. Determine her weight in newtons.
Answer:
700 N
Explanation:
The gravitational potential energy of the system decreases by 28,000 joules
The diver jumps from a height of 40 meters
Therefore the weight in Newton can be calculated as follows
= 28,000/40
= 700 newtons
Water flows from one reservoir to another a height, 41 m below. A turbine (η=0.77) generates power from this flow. 1 m3/s passes through the turbine. If 12 m head loss occurs between the two reservoirs, determine the actual (i.e. useable) power generated by the turbine (in kW).
Complete Question
A diagram representing this question is shown on the first uploaded image
Answer:
The value is [tex]P = 294594.3 \ W[/tex]
Explanation:
From the question we are told that
The height is [tex]h = 41 \ m[/tex]
The efficiency of the turbine is [tex]\eta = 0.77[/tex]
The flow rate is [tex]\r V = 1 m^3 / s[/tex]
The head loss is g = 2 m
Generally the head gain by the turbine is mathematically represented as
[tex]H = h - d[/tex]
=> [tex]H = 41 - 2[/tex]
=> [tex]H = 39 \ m [/tex]
Generally the actual power generated by the turbine is mathematically represented as
[tex]P = \eta *\gamma * \r V * H[/tex]
Here [tex]\gamma[/tex] is the specific density of water with value
[tex]\gamma = 9810 N/m^3 [/tex]
So
[tex]P =0.77 *9810 * 1 * 39[/tex]
[tex]P = 294594.3 \ W[/tex]
A thin, horizontal copper rod is 1.43 m long and has a mass of 236 g. What is the minimum current in the rod that can cause it to float in a horizontal magnetic field of 0.546 T
Answer:
The minimum current in the rod that can cause it to float 2.96 A.
Explanation:
Given;
length of the copper rod, L = 1.43 m
mass of the rod, m = 236 g = 0.236 kg
magnetic field, B = 0.546 T
The magnetic of the rod is given by;
Fm = BIL
The downward force on the rod is given by;
F = mg
For the rod to float, the difference in the two force will be zero;
BIL - mg = 0
BIL = mg
I = mg / BL
I = (0.236 x 9.8) / (0.546 x 1.43)
I = 2.96 A
Therefore, the minimum current in the rod that can cause it to float 2.96 A.
In the figure below the two blocks are connected by a string of negligible mass passing over a frictionless pulley. m1 = 10.0 kg and m2 = 3.80 kg and the angle of the incline is = 37.0°. Assume that the incline is smooth. (Assume the +x direction is down the incline of the plane.)
For what value of m1 (in kg) will the system be in equilibrium?
Answer:
For equilibrium, the mass m1 must be about 2.29 kg
Explanation:
The forces acting on m1 are : the weight (m1 * g) and the tension (T1) of the string.
The forces acting on m2 are:
1) along the ramp: the component of m2's weight (m2 g * sin(37)) and the tension (T2) of the string. Which by the way, must be equal in magnitude to T1 since the string is inextensible.
2) perpendicular to the ramp; the component of m2's weight (m2 g * cos(37)) and the normal from the contact with the ramp. These two compensate each other.
We therefore want the net force on each block to be zero for the system to be in equilibrium. This means:
T1 = m1 g
T2 = T1 = m2 g sin(37)
Then we have that if m2 = 3.8 kg, then:
m1 g = m2 g sin(37)
cancelling "g" on both sides:
m1 = m2 * sin(37) = 3.8 * sin(37) = 2.28689 kg
which may be rounded to about 2.29 kg.
. A cathode ray tube (CRT) is a device that produces a focused beam of electrons in a vacuum. The electrons strike a phosphor-coated glass screen at the end of the tube, which produces a bright spot of light. The position of the bright spot of light on the screen can be adjusted by deflecting the electrons with electrical fields, magnetic fields, or both. Although the CRT tube was once commonly found in televisions, computer displays, and oscilloscopes, newer appliances use a liquid crystal display (LCD) or plasma screen. You still may come across a CRT in your study of science. Consider a CRT with an electron beam average current of 25.00μA25.00μA . How many electrons strike the screen every minute?
Answer:
The value is [tex]n= 9.375 *10^{15} \ electrons [/tex]
Explanation:
From the question we are told that
The average current is [tex]I = 25.0 \mu A = 25.0 *10^{-6} \ A[/tex]
Generally the quantity of charge (electron ) is mathematically represented as
[tex]Q = ne[/tex]
Here e is the charge on a single electron with value [tex]e = 1.60 *10^{-19} \ C[/tex]
Generally current is mathematically represented as
[tex]I = \frac{Q}{t}[/tex]
=> [tex]I = \frac{ne}{t}[/tex]
Here t is time which is given as 1 minutes = 60 seconds
and n is the number of electrons
So
[tex]25.0 *10^{-6} = \frac{ n* 1.60 *10^{-19}}{60}[/tex]
=> [tex] 60 * 25.0 *10^{-6} = n* 1.60 *10^{-19} [/tex]
=> [tex]n= \frac{60 * 25.0 *10^{-6} }{ 1.60 *10^{-19} }[/tex]
=> [tex]n= 9.375 *10^{15} \ electrons [/tex]
The number of electrons that strike the screen every minute is; n = 9.375 × 10¹⁵ electrons
What is the number of electrons?
We are given;
Average Current; I = 25 μA = 25 × 10⁻⁶ A
Formula for the current is;
I = Q/t = ne/t
where;
n is number of electrons
e is electron charge = 1.6 * 10⁻¹⁹ C
t is time = 1 minute = 60 seconds
Thus making n the subject gives;
n = It/e
n = (25 × 10⁻⁶ * 60)/(1.6 * 10⁻¹⁹)
n = 9.375 × 10¹⁵ electrons
Read more about number of electrons at; https://brainly.com/question/11406294
When working with vectors, you will often see right triangles. What are the consistent properties of these triangles?
a. The hypotenuse of the triangle is always the vector itself.
b. The trangles are always equilateral.
c. The sides of the triangles always ine up with the x- and y-axes.
d. The triangles always have the same base and height
Answer:
a) and c)
Explanation:
Any vector can be expressed as a vector sum of its x- and -y components, as follows:v = vₓ* x + vy* y
where vₓ = x- component, x= unit vector in the x direction, vy = y-component, y = unit vector in the y direction.If we add the components graphically, using the head-to-tail method, it will be defined a right triangle, being the vector sum of both components (the vector itself) the hypotenuse of the triangle.As both components are perpendicular, they will be always lined up with the x- and y- axes.plz help me it is improtant
A pole-vaulter just clears the bar at 5.53 m and falls back to the ground. The change in the vaulter's potential energy during the fall is -3200 J. What is his weight?
Answer:
578.66 N
Explanation:
The first step is to calculate the mass
mgh= 3200J
3200/9.8×5.53
3200/54.194
m = 59.047 kg
Therefore the weight can be calculated as follows
Weight = m × g
= 59.047 × 9.8
= 578.66 N
Why does the brightness of a bulb not change noticeably when you use longer copper wires to connect it to the battery?
a. All the current is used up in the bulb, so the connecting wires don't matter.
b. Very little energy is dissipated in the thick connecting wires.
c. Electric field in the connecting wires is zero, so emf = E_bulb * L_bulb.
d. Current in the connecting wires is smaller than current in the bulb.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
Answer:
Options B & E are correct
Explanation:
Looking at all the options, B & E are the correct ones.
Option B is correct because the thicker the wire per unit length, the lesser resistance it will posses and the lesser the energy that will be dissipated by the wire and in return more energy will be dissipated by the bulb.
Option E is also correct because the resistance of the copper wires is low enough to ensure that there's not much drop in voltage across the copper wires. Thus, there will not be any noticeable differences in the voltage across the bulb.
Option A is not correct because the current is not used up and thus the charge is conserved, and it will circulate just through the circuit.
Option C is not correct because although the Electric field along the wire is not zero, it is very small.
Option D is not correct because the wires and the light bulb are connected in series and as such, the current in both the wires and the light bulb will be identical.
The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is :
b. Very little energy is dissipated in the thick connecting wires.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
"Energy"The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is very little energy is dissipated in the thick connecting wires and the electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
The thicker the wire per unit length, the lesser resistance it'll posses and the lesser the vitality that will be scattered by the wire and in return more vitality will be disseminated by the bulb.
The resistance of the copper wires is low sufficient to guarantee that there's not much drop in voltage over the copper wires. Hence, there will not be any noticeable contrasts within the voltage over the bulb.
Thus, the correct answer is B and E.
Learn more about "Circuit":
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Is it true or false that the displacement always equals the product of the average velocity and the time interval?
Answer:
True.
Explanation:
Applying the definition of average velocity, we know that we can always write the following expression:[tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex] (1)
By definition, Δx is just the displacement, and Δt is the time interval.So, just rearranging terms in (1), we get:[tex]\Delta x} = v_{avg}* {\Delta t}[/tex]
A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sphere about an axis through its center
Answer:
2I/7
Explanation:
Formula for moment of inertia through the centre of mass of a solid sphere is given as; I_sc = (2/5) mR²
Now, from parallel axis theorem, moment of inertia of solid sphere from tangent is given as;
I = (2/5) m R² + m R²
I = (7/5) mR²
Thus,mR² = 5I/7
Putting 5I/7 for mR² in first equation, we have;
I_sc = (2/5) × 5I/7
I_sc = 2I/7
_______ are pictures of relationships.
Which is the simplest?
Organ
Organism
Tissue
Cell
Which is most complex?
Tissue
Organ
Organ system
Whole organism
Answer:
1. Cell2. OrganHopes it helps you....Thank you ☺️
Enterprising students set an enormous slip-n-slide (a plastic sheet covered in water to reduce friction) on flat ground. If the slip-n-slide is 250 meters long.
a. How small does the average acceleration have to be for a student starting at 5 meters/second to slide to the end?
b. If the acceleration is 0.34 m/s^2, what is the minimum initial speed a student would have to run to make it to the end? Is this speed possible?
Answer:
Explanation:
a) Using the equation of motion
v² = u²+2as
v is the final velocity = 0m/s
u is the initial velocity = 5m/s
S is the distance = 250m
a is the acceleration
Substitute given values:
0² = 5²+2a(250)
-25 = 500a
a = -25/500
a = -0.05m/s²
b) If a = 0.34m/s²
We need to get the initial speed u
v² = u²+2as
0 = u²+2(0.34)(250)
-u² = 170
u² = -170
u = √-170
The square root of a negative number will give a complex number, hence the speed is not possible
QUESTION 10
An archer fires an arrow towards a tree with initial speed 65 m/s and angle 25 degrees above the horizontal. If the arrow takes 0.85
seconds to hit the tree, calculate the horizontal distance between the archer and the tree.
QUESTION 11
A monkey throws a banana from a tree into a nearby river. The banana has initial speed 7.6 m/s, is angled 40 degrees below the
horizontal, and takes 0.75 seconds to land in the river. Calculate the speed of the banana when it hits the water.
Answer:
10) The distance between the archer and the tree is 50.074 meters.
11) The speed of the banana when it hits the water is approximately 13.554 meters per second.
Explanation:
10) The arrow experiments a parabolic motion, which is the combination of horizontal motion at constant velocity and vertical uniform accelerated motion. In this case we need to find the horizontal distance between the archer and the tree, calculated by the following kinematic equation:
[tex]x = x_{o} +v_{o}\cdot t \cdot \cos \theta[/tex] (Eq. 1)
Where:
[tex]x_{o}[/tex] - Initial position of the arrow, measured in meters.
[tex]x[/tex] - Final position of the arrow, measured in meters.
[tex]v_{o}[/tex] - Initial speed of the arrow, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]\theta[/tex] - Launch angle, measured in sexagesimal degrees.
If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 65\,\frac{m}{s}[/tex], [tex]t = 0.85\,s[/tex] and [tex]\theta = 25^{\circ}[/tex], the horizontal distance between the archer and the tree is:
[tex]x = 0\,m + \left(65\,\frac{m}{s}\right)\cdot (0.85\,s)\cdot \cos 25^{\circ}[/tex]
[tex]x = 50.074\,m[/tex]
The distance between the archer and the tree is 50.074 meters.
11) The final speed of the banana ([tex]v[/tex]), measured in meters per second, just before hitting the water is determined by the Pythagorean Theorem:
[tex]v = \sqrt{v_{x}^{2}+v_{y}^{2}}[/tex] (Eq. 2)
Where:
[tex]v_{x}[/tex] - Horizontal speed of the banana, measured in meters per second.
[tex]v_{y}[/tex] - Vertical speed of the banana, measured in meters per second.
Each component of the speed are obtained by using these kinematic equations:
[tex]v_{x} = v_{o}\cdot \cos \theta[/tex] (Eq. 3)
[tex]v_{y} = v_{o}\cdot \sin \theta +g\cdot t[/tex] (Eq. 4)
Where [tex]g[/tex] is the gravitational acceleration, measured in meters per square second.
If we know that [tex]v_{o} = 7.6\,\frac{m}{s}[/tex], [tex]\theta = -40^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]t = 0.75\,s[/tex], the components of final speed are, respectively:
[tex]v_{x} = \left(7.6\,\frac{m}{s} \right)\cdot \cos (-40^{\circ})[/tex]
[tex]v_{x} = 5.822\,\frac{m}{s}[/tex]
[tex]v_{y} = \left(7.6\,\frac{m}{s}\right)\cdot \sin (-40^{\circ})+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (0.75\,s)[/tex]
[tex]v_{y} = -12.240\,\frac{m}{s}[/tex]
And the speed of the banana right before hitting the water is:
[tex]v = \sqrt{\left(5.822\,\frac{m}{s} \right)^{2}+\left(-12.240\,\frac{m}{s} \right)^{2}}[/tex]
[tex]v \approx 13.554\,\frac{m}{s}[/tex]
The speed of the banana when it hits the water is approximately 13.554 meters per second.
Part D
Next, we'll examine magnetic force. Bring the ends of your two magnets together. Explore the three
possible combinations. In two of the combinations, the two ends are the same. In one combination, the
two ends are different. Describe the force you feel in each combination
Answer:
i. The magnetic force of repulsion.
ii. The magnetic force of attraction.
Explanation:
A magnet is a material that has the attraction and repulsion capability. Magnets has two poles, north and south, thus would attract or repel another magnet in its neighborhood. It can either be a permanent or temporal magnet, and attracts ferrous metals.
i. In the case of two combinations where two ends are the same, it could be observed that the two ends (poles) repels each other. Thus since like poles repels, magnetic force of repulsion is felt.
ii. In the case of one combination in which the two ends are different, the two ends (poles) attract. Since unlike poles attracts, magnetic force of attraction is observed.
1 Which of the following is an example of a physical change?
A. Water freezing into ice.
B. A piece of wood burning.
C. A toy car rusting.
D. Zinc producing hydrogen gas when mixed with water.
Answer:
it's A trust me ok
Explanation:
i can't explain
4. According to Newton’s law of cooling, if an object at temperature T is immersed in a medium having the constant temperature M, then the rate of change of T is proportional to the difference of temperature M − T. This gives the differential equation dT dt = k(M − T). (a) Solve the differential equation for T. (b) A thermometer reading 100◦F is placed in a medium having a constant temperature of 70◦F. After 6 min, the thermometer reads 80◦F. What is the reading after 20 min?
Answer:
a) The solution of the differential equation is [tex]T(t) = M + (T_{o}-M) \cdot e^{-\frac{t}{\tau} }[/tex].
b) The reading after 20 minutes is approximately 70.770 ºF.
Explanation:
a) Newton's law of cooling is represented by the following ordinary differential equation:
[tex]\frac{dT}{dt} = -\frac{T-M}{\tau}[/tex] (Eq. 1)
Where:
[tex]\frac{dT}{dt}[/tex] - Rate of change of temperature of the object in time, measured in Fahrenheit per minute.
[tex]\tau[/tex] - Time constant, measured in minutes.
[tex]T[/tex] - Temperature of the object, measured in Fahrenheit.
[tex]M[/tex] - Medium temperature, measured in Fahrenheit.
Now we proceed to solve the differential equation:
[tex]\frac{dT}{T-M} = -\frac{t}{\tau}[/tex]
[tex]\int {\frac{dT}{T-M} } = -\frac{1}{\tau} \int \, dt[/tex]
[tex]\ln (T-M) = -\frac{t}{\tau} + C[/tex]
[tex]T(t) -M = (T_{o}-M)\cdot e^{-\frac{t}{\tau} }[/tex]
[tex]T(t) = M + (T_{o}-M) \cdot e^{-\frac{t}{\tau} }[/tex] (Eq. 2)
Where:
[tex]t[/tex] -Time, measured in minutes.
[tex]T_{o}[/tex] - Initial temperature of the object, measured in Fahrenheit.
b) From (Eq. 2) we obtain the time constant of the cooling equation for the object: ([tex]M = 70\,^{\circ}F[/tex], [tex]T_{o} = 100\,^{\circ}F[/tex], [tex]t = 6\,min[/tex], [tex]T(t) = 80\,^{\circ}F[/tex])
[tex]80\,^{\circ}F = 70\,^{\circ}F + (100\,^{\circ}F-70\,^{\circ}F)\cdot e^{-\frac{6\,min}{\tau} }[/tex]
[tex]e^{-\frac{6\,min}{\tau} } = \frac{80\,^{\circ}F-70\,^{\circ}F}{100\,^{\circ}F-70\,^{\circ}F}[/tex]
[tex]e^{-\frac{6\,min}{\tau} } = \frac{1}{3}[/tex]
[tex]-\frac{6\,min}{\tau} = \ln \frac{1}{3}[/tex]
[tex]\tau = -\frac{6\,min}{\ln \frac{1}{3} }[/tex]
[tex]\tau = 5.461\,min[/tex]
The cooling equation of the object is [tex]T(t) = 70 +30\cdot e^{-\frac{t}{5.461} }[/tex] and the temperature of the object after 20 minutes is:
[tex]T(20) = 70+30\cdot e^{-\frac{20}{5.461} }[/tex]
[tex]T(20) \approx 70.770\,^{\circ}F[/tex]
The reading after 20 minutes is approximately 70.770 ºF.
A plane travelling at 100 m/s accelerates at 5 m/s² for a distance of 125 m. What is the final velocity of the plane?
Analyzing the question:
We are given:
initial velocity (u) = 100 m/s
final velocity (v) = v m/s
distance (s) = 125 m
acceleration (a) = 5 m/s²
Solving for Final Velocity (v):
from the third equation of motion:
v² - u² = 2as
v² - (100)² = 2(5)(125)
v² - 10000 = 1250
v² = 1250 + 10000
v² = 11250
v = 106.06 m/s
Suppose the angle of incidence of a light ray is 42°.What is the angle of reflection?
Answer:
angle of reflection will be also 42°Explanation:
we know that ------------- angle of incidence=angle of reflectionSuppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m57.0 m . If the track is completely flat and the race car is traveling at a constant 24.5 m/s24.5 m/s (about 55 mph55 mph ) around the turn, what is the race car's centripetal (radial) acceleration
Answer:
10.53m/s²
Explanation:
Centripetal acceleration is the acceleration of an object about a circle. The formula for calculating centripetal acceleration is expressed by:
[tex]a = \frac{v^2}{r}[/tex]
v is the velocity of the car = 24.5m/s
r is the radius of the track = 57.0m
Substitute the given values into the formula:
[tex]a = \frac{24.5^2}{57} \\\\a = \frac{600.25}{57}\\ \\a = 10.53m/s^{2}[/tex]
Hence the centripetal acceleration of the race car is 10.53m/s²
