Answer: the work done by the force is 0
Explanation:
F (x², xy)
121 = 11²
so R = x² + y² = 11²
p = x². Q = xy
Δp/Δy = 0, ΔQ/Δx
using Green's theorem
woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA
= (x² + y² = 121)_∫∫ yΔA
now let x = rcosФ, y = rsinФ
ΔA = rΔrΔФ
so r from 0 to 11
and Ф from 0 to 2π
= 0_∫^2π 0_∫^11 rsinФ × rΔrΔФ
= 0_∫^2π SinФΔФ 0_∫^11 r²Δr
= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0
therefore the work done by the force is 0
What is the magnitude of the force exerted by the biceps FbicepsFbicepsF_biceps? What is the magnitude of the force exerted by the elbow FelbowFelbowF_elbow
Answer:
The force of forearm is 239.9 N
The force of elbow is 215.89 N
Explanation:
Suppose, When you lift an object by moving only your forearm, the main lifting muscle in your arm is the biceps. Suppose the mass of a forearm is 1.50 kg. If the biceps is connected to the forearm a distance [tex]d_{b}[/tex] 2.50 cm from the elbow, how much force [tex]F_{b}[/tex] must the biceps exert to hold a 950 g ball at the end of the forearm at distance dball J 36.0 cm from the elbow, with the forearm parallel to the floor? How much force [tex]F_{l}[/tex] must the elbow exert,
Given that,
Mass of forearm = 1.50 kg
Distance of forearm = 2.50 cm
Mass of ball = 950 g
Distance of ball = 36.0 cm
We need to calculate the force of forearm
Using balancing torque about elbow
[tex]F_{b}\times d_{b}=w_{f}\times\dfrac{d_{f}}{2}+w_{ball}\times d_{ball}[/tex]
Put the value into the formula
[tex]F_{b}\times 0.025= 1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36[/tex]
[tex]F_{b}=\dfrac{1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36}{0.025}[/tex]
[tex]F_{b}=239.9\ N[/tex]
We need to calculate the force of elbow
Using balancing force
[tex]F_{b}=F_{l}+w_{f}+w_{b}[/tex]
[tex]F_{l}=F_{b}-w_{f}-w_{b}[/tex]
Put the value into the formula
[tex]F_{l}=239.9-(1.50\times9.8)-(0.95\times9.8)[/tex]
[tex]F_{l}=215.89\ N[/tex]
Hence, The force of forearm is 239.9 N
The force of elbow is 215.89 N
A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).
a. 1.62*105 m/s2.b. 1000 m/s2.c. 0.62*105 m/s2.d. 3.62*105 m/s2.
Answer:
A. 1.62×10⁵m/s²Explanation:
Given
velocity of the bullet = 521m/s
distance = 0.840m
required
acceleration of the bullet a
Using the equation of motion:
v² = u²+2aS where:
v is the final velocity
u is the initial velocity
a is the acceleration
S is the distance
Substitute the given parameters into the equation to get a:
521² = 0²+2a(0.840)
271,441 = 0+1.68a
271,441 = 1.68a
a = 271,441/1.68
a = 161,572.02
a = 1.62×10⁵m/s²
Hence the acceleration of the bullet is 1.62×10⁵m/s²
In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 4.7 kg book is pushed from rest through a distance of 0.85 m by the horizontal 42 N force from the broom and then has a speed of 1.0 m/s, what is the coefficient of kinetic friction between the book and floor
Answer:
μ_k = 0.851
Explanation:
We are given;
Mass of book; m_book = 4.7 kg
Horizontal force; F_horiz = 42 N
Distance; d = 0.85 m
Speed; v = 1 m/s
First of all let's find the acceleration using Newton's equation of motion;
v² = u² + 2ad
u is initial velocity and it's 0 m/s in this case.
Thus;
1² = (2 × 0.85)a
1 = 1.7a
a = 1/1.7
a = 0.5882 m/s²
Now, resolving forces along the vertical direction, we have;
W - N = 0
Thus,W = N
Where W is weight = mg and N is normal force
Thus; N = mg = 4.7 × 9.81 = 46.107 N
Now, resolving forces along the horizontal direction, we have;
F_horiz - ((μ_k)N) = ma
Where μ_k is coefficient of kinetic friction.
Thus;
42 - 46.107(μ_k) = 4.7 × 0.5882
42 - 46.107(μ_k) = 2.76454
μ_k = (42 - 2.76454)/46.107
μ_k = 0.851
I need help with this answer
Sound wave A has a lower frequency than sound wave B, but both waves
have the same amplitude. What must also be true of these two waves?
O A. Wave A is softer.
O B. Wave A is louder.
C. Wave A has a lower pitch.
D. Wave A has a higher pitch.
Answer: C
Explanation: Amplitude controls loudness, and frequency controls pitch. The more frequent the higher pitch.
The true statement regarding the two sounds is wave A has a lower pitch.
What is pitch of a sound?
The pitch of a sound is sensation of the sound frequency.
A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to low frequency.
Sound of the the same amplitude, have the same loudness.
Thus, the true statement regarding the two sounds is wave A has a lower pitch.
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Which factor increases as a result of increasing altitude? Air pressure, air density, heights of sea level, space between gas molecules.
Answer:
heights of sea level and the space between gas molecules.
Explanation:
Asking that Which factor increases as a result of increasing altitude?
As the height above the sea increases, air density decreases because the molecule of the air will spread out causing air density to decrease. Also, there will be less air pressing down which will lead to a decrease in air pressure.
Therefore, the heights of sea level, and space between gas molecules are the factor that will increase as a result of increasing altitude. Not Air pressure and air density. But the heights of sea level and the space between gas molecules.
Answer:
AnswerAnswer c
Explanation:
g A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.03t3 (a) Find the velocity at time t (in ft/s). v(t) = 0.04t^3−0.09t^2 Correct: Your answer is correct. (b) What is the velocity after 1 second(s)? v(1) = -0.05 Correct: Your answer is correct. ft/s (c) When is the particle at rest? t = 0 Correct: Your answer is correct
Answer:
Explanation:
If a particle move with time and expressed according to the formula:
f(t) = 0.01t⁴ − 0.03t³
a) Velocity is the change in motion of the particle with respect to time and it is expressed as;
[tex]v(t) =\frac{d(f(t))}{dt}[/tex]
[tex]v(t) = 4(0.01)t^{4-1} - 3(0.03)t^{3-1}\\v(t) = 0.04t^3 - 0.09t^2[/tex]
Hence the velocity of the particle at time t is [tex]v(t) = 0.04t^3 - 0.09t^2[/tex]
b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:
[tex]v(t) = 0.04t^3 - 0.09t^2\\v(1) = 0.04(1)^3 - 0.09(1)^2\\v(t) = 0.04 - 0.09\\v(t) = -0.05[/tex]
Hence the velocity after 1second is -0.05
c) The particle is at rest when when the time is zero.
Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second
Which statement best describes an atom? (2 points)
оа
Protons and neutrons grouped in a specific pattern
Ob
Protons and electrons spread around randomly
ос
A group of protons and neutrons that are surrounded by electrons
Od
A ball of electrons and neutrons surrounded by protons
Answer:
A group of protons and neutrons that are surrounded by electrons I think that's the answer...
Explanation:
A man is passing barrels of water to another person from a height of 7 meters. The barrel is attached to a roof that is 10 meters above the ground. The rope holding the barrel is 8 meters long and will break if the tension exceeds 638 N. How many liters of water can the man put in the barrel without the rope breaking (assuming that the barrel is massless)? (The density of water = 1000 kg/m3 and 1 m3 = 1000 L. Density = mass/volume.)
Consider a 50-turn circular loop with a radius of 1.55 cm in a 0.35-T magnetic field. This coil is going to be used in a galvanometer that reads 45 μA for a full-scale deflection. Such devices use spiral springs which obey an angular form of Hooke's law, where the restoring torque is:
τs = -κ θ.
Here κ is the torque constant and θ is the angular displacement, in radians, of the spiral spring from equilibrium, where the magnetic field and the normal to the loop are parallel.
Required:
a. Calculate the maximum torque, in newton meters, on the loop when the full-scale current flows in it.
b. What is the torque constant of the spring, in newton meters per radian, that must be used in this device?
Complete Question
Consider a 50-turn circular loop with a radius of 1.55 cm in a 0.35-T magnetic field. This coil is going to be used in a galvanometer that reads 45 μA for a full-scale deflection. Such devices use spiral springs which obey an angular form of Hooke's law, where the restoring torque is:
τs = -κ θ.
Here κ is the torque constant and θ is the angular displacement, in radians, of the spiral spring from equilibrium, where the magnetic field and the normal to the loop are parallel.
Required:
a. Calculate the maximum torque, in newton meters, on the loop when the full-scale current flows in it.
b. What is the torque constant of the spring, in newton meters per radian, that must be used in this device? Assume the full scale deflection is 60° from the spring's equilibrium position
Answer:
a
[tex]\tau_{m} = 5.95 *10^{-7} \ N \cdot m[/tex]
b
[tex]\beta = 2.83 *10^{-7} \ N \cdot m / rad [/tex]
Explanation:
From the question we are told that
The number of turns is N = 50
The radius is r = 1.55 cm = 0.0155 m
The magnetic field is B = 0.35 T
The induced current is [tex]I = 45 \mu A = 45 *10^{-6} \ A[/tex]
Generally the area of loop is mathematically represented as
[tex]A = \pi r^2[/tex]
=> [tex]A =3.142 * 0.0155^2[/tex]
=> [tex]A =0.000755\ m^2[/tex]
Generally the maximum torque is mathematically represented as
[tex]\tau_{m} = N * B * I * A[/tex]
=> [tex]\tau_{m} = 50 * 0.35 * 45 *10^{-6} * 0.000755[/tex]
=> [tex]\tau_{m} = 5.95 *10^{-7} \ N \cdot m[/tex]
Generally the torque 60° from the spring's equilibrium position is mathematically represented as
[tex]\tau = N * B * I * A * sin (60)[/tex]
=> [tex]\tau = 50 * 0.35 * 45 *10^{-6} * 0.000755 * sin (60)[/tex]
=> [tex]\tau = 2.973 *10^{-7} \ N \cdot m [/tex]
Generally the toque constant of the spring is mathematically represented as
[tex]\beta = \frac{\tau}{60}[/tex]
=> [tex]\beta = \frac{\tau}{\frac{\pi}{3}}[/tex]
=> [tex]\beta = \frac{2.973 *10^{-7}}{\frac{\pi}{3}}[/tex]
=> [tex]\beta = 2.83 *10^{-7} \ N \cdot m / rad [/tex]
Water enters a 2 m3 tank at a rate of 6 kg/s and is withdrawn at a rate of 2 kg/s. The tank is initially half full. What type of process is this?
Answer:
This process is semicontinuous.
Explanation:
Given that,
Volume = 2 m³
Enter flow rate = 6 kg/s
Exit flow rate = 2 kg/s
The tank is initially half full.
We need to find what type of process
Using given data,
This process is not continuous because given enter and exit flow rate is not equal.
This process is semicontinuous and the water level in the tank does not reach a constant level.
Hence, This process is semicontinuous.
In which scenario is the least amount of work done on a crate?
A force of 120 N moves it 25 m.
A force of 115 N moves it 26 m.
A force of 130 N moves it 22 m.
A force of 125 N moves it 27 m.
Answer:
The answer is C. A force of 130 N moves it 22 m
The least amount of work done on a crate is when A force of 130 N moves it 22 m, the correct option is C.
What is work done?The total amount of energy transferred when a force is applied to move an object through some distance
The work done is the multiplication of applied force with the displacement.
Work Done = Force * Displacement
The work done for option A is
= 120*25= 3000 Nm
The work done for option B is
=115*26 = 2990 Nm
The work done for option C is
=130*22= 2860 Nm
The work don for option D is
= 125*27= 3375 Nm
Hence, the least amount of work is done on the crate when a force of 130 N moves it 22 m.
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An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is 7.9 times the magnitude of the tangential acceleration. What is the angle
Answer:
The angle is 3.95 rad.
Explanation:
The angle can be calculated as follows:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta [/tex]
Where:
[tex]\omega_{f}[/tex]: is the final angular speed
ω₀: is the initial angular speed = 0 (it starts from rest)
α: is the angular acceleration
θ: is the angle=?
The centripetal acceleration is:
[tex]a_{c} = \omega_{f}^{2}*r[/tex]
And the tangential acceleration is:
[tex] a_{T} = \alpha*r [/tex]
Since the magnitude of the centripetal acceleration is 7.9 times the magnitude of the tangential acceleration:
[tex]a_{c} = 7.9a_{T}[/tex]
[tex]\omega_{f}^{2}*r = 7.9*\alpha*r \rightarrow \alpha = \frac{\omega_{f}^{2}}{7.9}[/tex]
Now, the angle is:
[tex]\omega_{f}^{2} = 2(\frac{\omega_{f}^{2}}{7.9})\theta[/tex]
[tex] \theta = \frac{7.9}{2} = 3.95 rad [/tex]
Therefore, the angle is 3.95 rad.
I hope it helps you!
The angular distance traveled by the electric drill is 3.95 radians.
The given parameters;
initial angular speed, [tex]\omega_i[/tex] = 0centripetal acceleration, [tex]a_c[/tex] = 7.9aThe angular distance traveled by the electric drill is calculated as follows;
[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta[/tex]
The relationship between centripetal acceleration, tangential acceleration and angular speed is given as;
[tex]a_c = \omega ^2 r\\\\a = \alpha r\\\\a_c = 7.9a= 7.9\alpha r\\\\7.9\alpha r = \omega^2 r\\\\\alpha = \frac{\omega ^2}{7.9}[/tex]
Substitute the value of angular acceleration into the first equation;
[tex]\omega _f^2 = 0 + 2(\a (\frac{\omega _f^2}{7.9})\theta\\\\2\theta \omega_f^2 = 7.9\omega_f ^2\\\\\theta = \frac{7.9}{2} \\\\\theta = 3.95 \ rad[/tex]
Thus, the angular distance traveled by the electric drill is 3.95 radians.
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A motorboat is a lot heavier than a pebble. Why does the boat float?
Answer:
The boat has more buoyancy
Explanation:
can humans be considered carbon sinks? If so,for how long
Answer:
Humans be considered carbon sinks. Not only do humans
have a lot of carbon in them, they also use a lot of carbon.
hope this helps
Given two vectors A= 4.00i + 7.00jand B= 5.00i-2.00ja.Find the magnitude of each vector.b.Find the vector differenceC= A –B, give its magnitude and direction. c.Draw the vector A, B and C.d.Find the scalar product of A.B e.Find the vector productD= A x B, draw D and give its magnitude.
Answer:
444.99
Explanation:
Explanation:
C=A-B
=(4.00i+7.00j)-(5.00i-2.00j)
= -1.00i+9.00j
A.B=AxBx+AyBy
=(4 x 5)+(7 x -2)
=20-14
=6
If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s then the KINETIC ENERGY of the CANNON will be ("E4" means "*10^4") *
1 point
4 J
40 J
400 J
None of the above
Answer:
M V = m v conservation of momentum (Caps-cannon Small-projectile)
V = m / M * V = 2 / 2000 * 200 m/s = .2 m/s recoil velocity of cannon
KE = 1/2 M V^2 = 2000 / 2 kg * (.2 m/s)^2 = 40 kg m^2/s^2 = 40 J
A refigerator is designed to remove 3 kW from the cold space at -10C while it rejects heat to the kitchen at 25 C. The rate of the reversible work is:______
Answer:
b
Explanation:
Two cars come to a stop from the same initial speed, one braking gently and the other braking hard. Which car converts more kinetic energy to thermal energy
Answer:
Both cars convert more kinetic energy to thermal energy.
Explanation:
Given that,
Two cars come to a stop from the same initial speed, one braking gently and the other braking hard.
We know that,
The conservation of energy :
The energy of the system is always constant.
The energy can be change one form to another form.
So, we can say that when both cars come to a stop from the same speed, one braking softly and the other braking strongly then both cars convert more kinetic energy to thermal energy.
Hence, Both cars convert more kinetic energy to thermal energy.
Based on the law of consevation of energy, the car that brakes hard will convert more kinetic energy to thermal energy.
Which car will convert more kinetic energy to thermal energy?According to the law of conservation of energy, the total energy is conserved in any closed system.
Since the cars have the same mass and had the same initial speed, their kinetic energies are the same.
However, the car braking hard comes to rest in a shorter time and in a shorter distance.
Using the formula for calculating power:
Power = energy/timeThe car that braked hard will have greater power due to shorter time of action.
This power is seen as greater thermal energy, whereas some of the energy of the other car will still be used in travelling some distance before stopping.
Therefore, the car that brakes hard will convert more kinetic energy to thermal energy.
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Alejandro made 6.4 liters of punch using half apple juice and half orange juice. How many milliliters of apple juice are in the punch?
Answer:
3.2
Explanation:
I hope that this helps! Have a good day!!
HELP PLS GIVING BRAINLIEST!!
Which of the following is not true about taxes? A. Mandatory sum of money by government so that it can operate B. Due on April 15th C. largely collected to support private businesses D. Collected by the Internal Revenue Service (IRS)
What observations did the solar system this geocentric models of the solar system help to explain
Answer:
Geocentric model of the solar system helped to explain retrograde motions of planets. Explanation: Geocentric model of planets was proposed by Ptolemy. It stated that all sun, planets and stars revolve round the earth in circular orbits.
Answer:
retrograde motion
Explanation:
i reverse searched the image
If I am driving down the highway going north at 50 miles per hour, and another car is driving south at 75 miles per hour. How fast is the car coming toward me?
Its an exam >.
I WILL GIVE BRAINLIEST
A 50 kg bicyclist starts his ride down the road with an acceleration of 1m/s2 in air with a density of 1.2 kg/m3. If his velocity at a given moment is 2m/s, how much force is he exerting? Assume the area of his body is 0.5m^2.
a. The bicyclist is exerting 1.1 N of force.
b. The bicyclist is exerting 49 N of force.
c. The bicyclist is exerting 50 N of force.
d. The bicyclist is exerting 51 N of force.
Answer:
b. The bicyclist is exerting 49 N of force
Explanation:
Given;
mass of bicyclist start, m = 50 kg
acceleration, a = 1 m/s²
density of air, ρ = 1.2 kg/m³
velocity, v = 2 m/s
Area of the bicyclist body, A = 0.5 m²
The drag force on the bicyclist is given by ;
Fd = 0.5CρAv²
where;
C is drag coefficient = 0.9 for bicycle
Fd = 0.5 x 0.9 x 1.2 x 0.5 x 2²
Fd = 1.1 N
The force of the bicyclist is given by;
F = ma
F = 50 x 1
F = 50 N
The effective force exerted by the bicyclist is given by;
Fe = F - Fd = 50 N - 1.1 N
Fe = 49 N
Therefore, the force exerted by the bicyclist is 49 N
What (rather remarkable!) equation relates the speed of light to other fundamental electromagnetic constants?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The equation is [tex]c = \frac{1}{\sqrt{ \mu_o * \epsilon_o} }[/tex]
The value of c is [tex]c = 2.998 *10^{8} \ m/s [/tex]
Explanation:
From the question we are told that
Generally the equation that relates the speed of light to other fundamental electromagnetic constants is
[tex]c = \frac{1}{\sqrt{ \mu_o * \epsilon_o} }[/tex]
Here c is the speed of light
[tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]
and [tex]\epsilon_o[/tex] is the permittivity of free space with value
[tex]\epsilon_o = 8.85*10^{-12} \ C/V \cdot m[/tex]
So
[tex]c = \frac{1}{\sqrt{ 4\pi *10^{-7} * 8.85*10^{-12}} }[/tex]
=> [tex]c = 2.998 *10^{8} \ m/s [/tex]
What is an example of a scientific topics that requires knowledge of at least two branches of physics?
Explanation:
classical physics and modern physics.
Physics is the branch of science that deals with the interaction of energy and matter. Modern and Classical Physics are the two major branches of physics.
What is physics?Physics is the discipline of science that investigates the structure of matter and how the universe's fundamental constituents interact. It investigates objects ranging from the very small to the entire universe using quantum mechanics and general relativity.
One of the most crucial physics topics is kinematics. Kinematics is the study of an object 's direction of travel. Kinematics is only concerned with the motion of an object.
There is a vertical and a horizontal component to two-dimensional projectile motion, such as that of a football or other thrown object.
Throwing a rock or kicking a ball produces a projectile pattern of motion that includes both a vertical and a horizontal component.
Thus, these can be the topics that require the knowledge of physics.
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The name for this type of energy is
A.potential energy
B.motion
C.position
D.kinetic energy
Explanation:
[tex]kinetic \: energy \: is \: the \: energy \: of \: a \: body \: in \: motion. \\ that \: is \: energy \: of \: a \: body \: that \: is \: moving.\\ while \\ potential \: energy \: is \: the \: energy \: of \: a \: body \: by \: the \: virtue \: of \: its \: position \: \\ that \: is \: energy \: of \: a \: body \: that \: is \: not \: moving.[/tex]
♨Rage♨
Compare the amount of thermal energy required to MELT a solid with the amount of thermal energy released when the same liquid becomes a solid.
What is dark ⚡ energy ????
plz make it simple writting, not big big words
