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2 years ago

The 0.5kg soccer ball moves toward the net with a force of 4N. What is its acceleration?

Physics

1 answer:

Eduardwww [97]2 years ago

6 0

Answer:

8 m/s²

Explanation:

Given,

Force ( F ) = 4 N

Mass ( m ) = 0.5 kg

To find : -

Acceleration ( a ) = ?

Formula : -

F = ma

a = F / m

= 4 / 0.5

= 40 / 5

a = 8 m/s²

It's acceleration is 8 m/s².

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2 years ago

Without the wheels, a bicycle frame has a mass of 8.29 kg. Each of the wheels can be roughly modeled as a uniform solid disk wit

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69.66 Joule

Explanation:

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$K = \frac{1}{2}m_{f}v^{2}+ \frac{3}{2}\times m_{w}v^{2}$

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3 years ago

The acceleration of a particle traveling along a straight line isa=14s1/2m/s2, wheresis in meters. Ifv= 0,s= 1 m whent= 0, deter

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Answer:

0.78m/s

Explanation:

We are given that

Acceleration= $a=\frac{1}{4}s^{\frac{1}{2}}m/s^2$

v=0, s=1 when t=0

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We know that

$a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=\frac{dv}{ds}v$

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$\int_{0}^{v} vdv=\int_{1}^{s}0.25s^{\frac{1}{2}}ds$

$\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2})^{s}_{1}$

By using formula: $\int x^ndx=\frac{x^{n+1}}{n+1}+C$

$\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2}}-1$

Substitute s=2

$\frac{v^2}{2}=\frac{0.50}{3}((2^{1.5})-1)$

$\frac{v^2}{2}=\frac{0.50}{3}\times 1.83$

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3 years ago

A 0.30-kg object is traveling to the right (in the positive direction) with a speed of 3.0 m/s. After a 0.20 s collision, the ob

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Answer:

$I = -2.1 kg.m/s$

Explanation:

Given,

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2 years ago

The 0.5kg soccer ball moves toward the net with a force of 4N. What is its acceleration?​

The 0.5kg soccer ball moves toward the net with a force of 4N. What is its acceleration?