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2 years ago

Which of the following equations describe the 2nd law of motion?

Physics

2 answers:

Paladinen [302]2 years ago

7 0

Answer:

b. force= mass *acceleration

Explanation:

Sever21 [200]2 years ago

4 0

Answer:

b. F = m × a is the correct answer.

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4.0km at 25 west of south, then 2.0km at 15 north east

Aliun [14]

please, give the question properly.

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2 years ago

Classify each of the following chemical reactions. Upper S plus upper O Subscript 2 right arrow upper S upper O subscript 2. Upp

vodka [1.7K]

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2 years ago

When you throw a ball up in the air, it travels up and then stops instantaneously before falling back down. At the point where i

Gnoma [55]

Answer:

The ball stops instantaneously at the topmost point of the motion.

Explanation:

Assume we have thrown a ball up in the air. For that we have given a force on the ball and it acquires an initial velocity in the upward direction.

The forces that resist the motion of the ball in the upward direction are the force of gravity and air resistance. The ball will instantaneously come to rest when the velocity of the ball reduces to zero.

The two forces acting in the downward direction reduces its speed continuously and it becomes zero at the topmost point.

4 0

3 years ago

Two boxers are fighting. Boxer 1 throws his 5 kg fist at boxer 2 with a speed of 9 m/s.

Sladkaya [172]

Answer:

0.001 s

Explanation:

The force applied on an object is equal to the rate of change of momentum of the object:

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

where

F is the force applied

$\Delta p$ is the change in momentum

$\Delta t$ is the time interval

The change in momentum can be written as

$\Delta p=m(v-u)$

where

m is the mass

v is the final velocity

u is the initial velocity

So the original equation can be written as

$F=\frac{m(v-u)}{\Delta t}$

In this problem:

m = 5 kg is the mass of the fist

u = 9 m/s is the initial velocity

v = 0 is the final velocity

F = -45,000 N is the force applied (negative because its direction is opposite to the motion)

Therefore, we can re-arrange the equation to solve for the time:

$\Delta t=\frac{m(v-u)}{F}=\frac{(5)(0-9)}{-45,000}=0.001 s$

4 0

3 years ago

g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before

AnnyKZ [126]

Answer:

Time, $t=1.07\times 10^{-23}\ s$

Explanation:

Given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel $3.2\times 10^{-15}\ m$ before interacting.

Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,

$t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s$

So, the time interval is $1.07\times 10^{-23}\ s$

5 0

3 years ago