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3 years ago

12

You are standing on a log and a friend is trying to knock

1 answer:

WINSTONCH [101]3 years ago

3 0

Answer: catching the ball is a better choice.

Explanation:

The collision of 2 objects involves involves large impact force since the force is inversely proportional to the time in which the momentum of the object changes.

Mathematically

$F=\frac{\Delta p}{\Delta t}$

If we catch the ball we increase the time in which the momentum of the ball is decreased thus the impact force that acts on us is lower as larger time is allowed for the ball to decrease it's momentum.

If we allow the ball to hit us the momentum of the ball changes in a short period of time thus applying a large impact force on our body thus increasing the chances of toppling.

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David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2

dangina [55]

A. 441 m B: 46.0 m/s

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2 years ago

A train travels 94 kilometers in 2 hours, and then 58 kilometers in 4 hours what is it's average speed

Alisiya [41]

The answer to your question 25.333333333

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3 years ago

Find the unbalanced force required to accelerate a 6kg mass at 3ms

Mila [183]

Force (F)=Mass (m)×Acceleration (a).

∴F=ma.

∴F=(2kg)×(3ms2).

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3 years ago

You are working for a manufacturing company. Your supervisor has an idea for controlling the position of a small bead by using e

Mashutka [201]

You are working for a manufacturing company, which is mathematically given as

$m=3\sqrt{2}$
$m=\frac{15\sqrt{5}}{16}$
x=0.747a
$m/n=\frac{(x^2+a^2)3/2}{x^3}$

<h3>What is the value of m that will place the movable bead in equilibrium at x-a a ....?</h3>

a)

Generally, the equation for the force of equilibrium is mathematically given as

F=2fcos\theta

Therefore

$K(npq^2/a^2)=2\frac{kmpq^2}{a\sqrt{2}^2}0.5\\\\np=mP/ \sqrt{2}\\\\where n=3$

$m=3\sqrt{2}$

b)

By force equilibrium

$K(npq^2/(2a^2))=2*\frac{kmpq^2}{a\sqrt{5a}^2}* \frac{2a}{\sart{5a}}$

Therefore

$n/4=2/5*m*2/\sqrt{5}\\\\m= \frac{5\sqrt{5}}{16}\\\\\\\\$

$m=\frac{15\sqrt{5}}{16}$

c)

$K(npq^2/x^2)=2\frac{kmpq^2}{a\sqrt{x^2+a^2}^2}0.5*x/\sqrt{x^2+a^2}$

x^2+a^2=(14/3)^{2/3}x^2

x=a/1.338

x=0.747a

d)

By force equilibrium

$K(npq^2/x^2)=2\frac{kmpq^2n}{\sqrt{x^2+a^{3/2}}^2}\\\\n/x^2=\frac{2mx}{(x^2+a^2)^{3/2}}$

$m/n=\frac{(x^2+a^2)3/2}{x^3}$

Read more about electric fields

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3 0

2 years ago

V125BC [204]

Answer:C

Explanation:

For every action or force there is an equal and opposite reaction or force

So the force on the bat is the same as the force on the ball but in opposite directions

Now the ball will fly farther but that is a different law

6 0

3 years ago

Read 2 more answers