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Anon25 [30]
2 years ago
10

How does a push or pull affect motion?

Physics
1 answer:
STatiana [176]2 years ago
8 0

Answer:

The motion of an object acted on by a force depends partly on the strength of the push or pull. The stronger the push or pull, the faster the object will move.

Explanation:

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Explanation:

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3 years ago
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A simple pendulum consists of a ball connected to one end of a thin brass wire. The period of the pendulum "is 2.51 s". The temp
Artyom0805 [142]

Answer:

0.0034 sec

Explanation:

L = initial length

T = initial time period = 2.51 s

Time period is given as

T = 2\pi \sqrt{\frac{L}{g}}

2.51 = 2\pi \sqrt{\frac{L}{9.8}}

L = 1.56392 m

L' = new length

ΔT = Rise in temperature = 142 °C

α = coefficient of linear expansion = 19 x 10⁻⁶ °C

New length due to rise of temperature is given as

L' = L + LαΔT

L' = 1.56392 + (1.56392) (19 x 10⁻⁶) (142)

L' = 1.56814 m

T' = New time period

New time period is given as

T' = 2\pi \sqrt{\frac{L'}{g}}

T' = 2\pi \sqrt{\frac{1.56814}{9.8}}

T' = 2.5134 sec

Change in time period is given as

ΔT = T' - T

ΔT = 2.5134 - 2.51

ΔT = 0.0034 sec

5 0
3 years ago
A long string is pulled so that the tension in it increases by a factor of four. If the change in length is negligible, by what
rusak2 [61]

To solve this problem we will apply the concepts related to wave velocity as a function of the tension and linear mass density. This is

v = \sqrt{\frac{T}{\mu}}

Here

v = Wave speed

T = Tension

\mu = Linear mass density

From this proportion we can realize that the speed of the wave is directly proportional to the square of the tension

v \propto \sqrt{T}

Therefore, if there is an increase in tension of 4, the velocity will increase the square root of that proportion

v \propto \sqrt{4} = 2  

The factor that the wave speed change is 2.

3 0
3 years ago
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
rosijanka [135]

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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