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Romashka [77]
3 years ago
5

ID: A

Physics
1 answer:
postnew [5]3 years ago
4 0

Answer: Velocity=8.26m/s

Explanation:

Acceleration=Finial velocity (V) - Initial velocity (u) ÷ Time

that is, a=v-u/t

a=1.2m/s², v=?, u=5.5m/s, t=2.3s

From a=v-u/t, make v the subject of the formula

v=at + u

v=(1.2* 2.3) + 5.5

v=2.76+5.5

v=8.26m/s

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PolarNik [594]
The total mechanical energy of the system at any time t is the sum of the kinetic energy of motion of the ball and the elastic potential energy stored in the spring:
E=K+U= \frac{1}{2}mv^2+ \frac{1}{2}kx^2
where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.

Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.

When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:
E=U_{max}= \frac{1}{2}k(x_{max})^2

Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):
E=K_{max}= \frac{1}{2}m(v_{max})^2

Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:
U_{max}=K_{max}
\frac{1}{2}k(x_{max})^2 =  \frac{1}{2}m(v_{max})^2
v_{max}= \sqrt{ \frac{k x_{max}^2}{m} }= \sqrt{ \frac{(12 N/m)(0.20 m)^2}{0.4 kg} }=1.1 m/s
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Answer:

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Explanation:

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A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh
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Answer:

Part a)

T_L = 155.4 N

Part b)

T_R = 379 N

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

T_L cos65 = T_R cos80

T_L = 0.41 T_R

now we will have force balance in Y direction

mg = T_L sin65 + T_Rsin80

514 = 0.906T_L + 0.985T_R

Part a)

so from above equations we have

514 = 0.906T_L + 0.985(\frac{T_L}{0.41})

514 = 3.3 T_L

T_L = 155.4 N

Part b)

Now for tension in right string we will have

T_R = \frac{T_L}{0.41}

T_R = 379 N

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How much momentum does a 10 kg object have when is moving 30 m/s
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P = m x v

P = 30 x 10
=300
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3 years ago
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