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mr Goodwill [35]

3 years ago

13

Why do all freely falling objects have the same acceleration

2 answers:

iren2701 [21]3 years ago

8 0

They are falling under the sole influence of gravity all objects<span> will </span>fall<span> with the </span>same<span> rate of </span><span>acceleration needless of there size</span>

Tanzania [10]3 years ago

7 0

Objects<span> that are said to be undergoing free </span>fall<span>, are not encountering a significant force of air resistance; they are </span>falling<span> under the sole influence of gravity. Under such conditions, </span>all objects<span> will </span>fall<span> with the </span>same<span> rate of </span>acceleration<span>, regardless of their mass.</span>

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A 2.00 kg box slides on a rough, horizontal surface, hits a spring with a speed of 1.90 m/s and compresses it a distance of 10.0

oksian1 [2.3K]

Answer:

Explanation:

Given

mass of box $m=2\ kg$

speed of box $v=1.9\ m/s$

distance moved by the box $x=10\ cm$

coefficient of kinetic friction $\mu _k=0.66$

Friction force $f_r=\mu_kN$

$f_r=0.66\times mg$

$f_r=0.66\times 2\times 9.8=12.936 \N$

Kinetic Energy of box will be utilize to overcome friction and rest is stored in spring in the form of elastic potential energy

$\frac{1}{2}mv^2=f_r\cdot x+\frac{1}{2}kx^2$

$\frac{1}{2}\times 2\times 1.9^2=12.936\times 0.1+\frac{1}{2}\times k\times (0.1)^2$

$3.61-1.2936=0.005\times k$

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A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.6 m/

jenyasd209 [6]

Answer:

$\Delta t =1.31\ s$

Explanation:

given,

coefficient of kinetic friction, μ = 0.25

Speed of sled at point A = 8.6 m/s

Speed of sled at point B = 5.4 m/s

time taken to travel from point A to B.

we know,

J = F Δ t

J is the impulse

where F is the frictional force.

t is the time.

we also know that impulse is equal to change in momentum.

$J = m(v_f - v_i)$

frictional force

F = μ N

where as N is the normal force

now,

$F\Delta t = m(v_f -v_i)$

$\mu m g \times \Delta t = m(v_f-v_i)$

$\mu g \times \Delta t = v_f-v_i$

$\Delta t =\dfrac{v_f-v_i}{\mu g}$

$\Delta t =\dfrac{8.6-5.4}{0.25\times 9.8}$

$\Delta t =1.31\ s$

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