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2 years ago

A baseball goes from zero to 34 m/s in 0.188 s. What is its average acceleration? Answer in units of m/s^2

Physics

1 answer:

Kaylis [27]2 years ago

6 0

Answer:

$a_{avg} = 180.85 m/s^{2}$

Given:

Initial velocity, u = 0 m/s

Final velocity, v = 34 m/s

Time interval, $\Delta t = 0.188 s$

Solution:

Acceleration is the rate at which the velocity of an object changes.

Thus

Average acceleration, $a_{avg} = \frac{\Delta v}{\Delta t}$

$a_{avg} = \frac{v - u}{t - 0} = \frac{34 - 0}{0.188} = 180.85 m/s^{2}$

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= 20m/s2 × 1sec

= 20m/s

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Answer:

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3 years ago

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Which property describes the temperature at which a liquid changes to a gas?

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3 years ago

A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i

erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

$\mu_k$ = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

u₁ = √(2 × 9.8 × 5)

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem, we have

Work done by the friction force = change in kinetic energy of the mass 2

$0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)$

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring = Kinetic energy of the mass 2

$\frac{1}{2}kx^2=\frac{1}{2}mv_3^2$

on substituting the values, we get

$\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2$

x = 2.86 m

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3 years ago