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iris [78.8K]
3 years ago
12

A 300 kg piano needs to be moved to the other side of the room. The maximum static frictional force is equal to 90 N and the kin

etic frictional force is equal to 70 N. Calculate the acceleration of the piano for an applied force of 100 N.
Physics
1 answer:
riadik2000 [5.3K]3 years ago
5 0

Answer:

a = 0.1 m/s²

Explanation:

  • If the maximum static frictional force is 90 N, this means that any applied force that will overcome this force, will cause the piano to slide, so kinetic frictional force applies.
  • Under these conditions, the net force in the horizontal direction is just the difference between the applied force (which is larger that the static friction force) and the kinetic frictional force, as follows:

       F_{net}  = F_{app} -F_{frk}  = 100 N - 70 N = 30 N (1)

  • By the same token, according Newton's 2nd Law, this force is just equal to the product of the mass of the piano, times the acceleration, as follows:

       F_{net} = m* a = 300 Kg * a = 30 N (2)

  • Solving for a:

        a = \frac{F_{net}}{m} = \frac{30 N}{300kg} = 0.1 m/s2 (3)

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A cell phone weighing 80 grams is flying through the air at 15 m/s. What is its kinetic energy
Contact [7]

Answer:

The kinetic energy of the cell phone is 9J

Explanation:

The kinetic energy is the energy possessed by a body by virtue of motion.

The kinetic energy is expressed as

KE= 1/2m(v)²

Given data

Mass of cell phone m= 80g--to kg=80/1000= 0.08kg

Velocity of cell phone v= 15m/s

Substituting our given data we have

KE= 1/2*0.08(15)²

KE= (0.08*225)/2

KE=18/2

KE= 9J

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3 years ago
An astronaut has a mass of 74.0 kg. 1) how much would the astronaut weigh on mars where surface gravity is 38.0% of that on eart
NeX [460]
274.614N or 61.736lbs
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3 years ago
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Can someone explain to how to calculate this
Karo-lina-s [1.5K]

answer

option d is the correct answer

explanation

as we know frequency is equal to 1 /t

f= 457 Hz

t=1

SO, 1/457

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3 years ago
A car goes from 0 to 26.8 m/s in 6.2 s. What is the average acceleration of the car?
Nutka1998 [239]

Answer:

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Explanation:

5 0
3 years ago
A(n) 12500 lb railroad car traveling at 7.8 ft/s couples with a stationary car of 7430 lb. The acceleration of gravity is 32 ft/
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To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then

\text{Initial Momentum} = \text{Final Momentum}

m_1u_1 +m_2u_2 = m_1v_1+m_2v_2

Here,

m_{1,2}= Mass of each object

u_{1,2}= Initial velocity of each object

v_{1,2}= Final velocity of each object

When they position the final velocities of the bodies it is the same and the car is stationary then,

m_2u_2 = (m_1+m_2)v_f

Rearranging to find the final velocity

v_f = \frac{m_2u_2}{ (m_1+m_2)}

v_f = \frac{ 12500*7.8}{ 12500+7430}

v_f = 4.8921ft/s

The expression for the impulse received by the first car is

I = m_1 (v-u)

I = \frac{W}{g} (v-u)

Replacing,

I = \frac{12500}{32.2}(4.89-7.8)

I = -1129.65lb\cdot s

The negative sign show the opposite direction.

7 0
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