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NARA [144]
4 years ago
5

A ball of moist clay falls 15.0 m to the ground. It is in contact with the ground for 20.0 ms before stopping. (a) What is the m

agnitude of the average acceleration of the ball during the time it is in contact with the ground?
Physics
1 answer:
rosijanka [135]4 years ago
3 0

Answer:

a= -0.86 m/s²

The negative sign shows that ball down the ground or moving down

Explanation:

Vf² - Vo² = 2gS

where

Vf = velocity of clay as it hits the ground

Vo = initial velocity of clay = 0

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

S = distance travelled by clay = 15 m

Substituting appropriate values,

Vf² - 0 = 2(9.8)(15)  

Vf = 17.15 m/sec.

Formula to use is,  

V - Vf = aT

where

V = velocity of clay when it stops = 0

Vf = 17.15 m/sec (as determined above)

a = acceleration

T = 20 ms  

Put the values to find acceleration

a=(V-Vf)/T

a=(0-17.15)/20

a= -0.86 m/s²

The negative sign shows that ball down the ground

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Consider a 7 m stretched string that is clamped at both ends. What is the longest wavelength standing wave that it can support (
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A vibrating stretched string has nodes or fixed points at each end. The string will vibrate in its fundamental frequency with just one anti node in the middle - this gives half a wave.

l=\frac{\lambda }{2}

Rearranging for the wavelength

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\lambda = 14m

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You hold a bucket in one hand, and in the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.
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Answer:

3.28 m/s

Explanation:

From the question,

Applying newtons ,

F = mv²/r........................ Equation 1

Where m = mass of the rock, v = speed, r = radius of the circle, F = Centriputal force.

But,

Centriputal force is equal to the weight of the rock

F = mg................. Equation 2

Equate equation 2 into equation 1

mg = mv²/r

Therefore,

v = √gr...................... Equaion 3

Given: r = 2.2/2 = 1.1 m

Constant: g = 9.8 m/s²

Substitute into equation 3

v = √(9.8×1.1)

v = √(10.78)

v = 3.28 m/s

Hence the minimum speed is 3.28 m/s

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3 years ago
A force is a vector that has both
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3 years ago
A 30.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 6.00 cm on a frictionles
Naddika [18.5K]

Answer:

a) 72mJ

b) 0.33m/s

c) 54mJ

d) 18mJ

Explanation:

(a) The total energy of the system is given by:

E_T=\frac{1}{2}kA^2

where k is the force constant and A is the amplitude. By replacing we get:

E_T=\frac{1}{2}(40N/m)(0.06m)^2=0.072J = 72mJ

(b) we can get the speed by the conservation of energy (the kinetic energy and potential energy must equal the total energy in any place):

E_T=E_k+E_p\\\\\frac{1}{2}kA^2=\frac{1}{2}mv^2+\frac{1}{2}kx^2\\\\v=\sqrt{\frac{kA^2-kx^2}{m}}=\sqrt{\frac{(40N/m)((0.06m)^2-(0.0115m)^2)}{0.03kg}}=0.33m/s

where we have used that x=1.15cm=0.0115m; A=6.00cm=0.06m

(c) Again, by the conservation of energy:

E_k=\frac{1}{2}k(A^2-x^2)=\frac{1}{2}(40N/m)((0.06m)^2-(0.03m)^2)=0.054J=54mJ

(d) E_p=\frac{1}{2}kx^2=\frac{1}{2}(40.0N/m)(0.03m)^2=0.018J=18mJ

hope this helps!!

6 0
3 years ago
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