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3 years ago

Dave throws a 10 kg bowling ball straight up in the air. At the very tippy-top of its path, what is it's momentum?

Physics

1 answer:

svetoff [14.1K]3 years ago

8 0

Answer:Zero

Explanation:

Given

mass of ball $m=10\ kg$

If the ball is thrown upward then at maximum point velocity of ball is zero because ball is no longer able to move upward

Momentum(P) of a particle is given by

$P=mass\times velocity$

$P=10\times 0$

$P=0$

Therefore at the highest point momentum is zero .

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A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. Th

Taya2010 [7]

Answer:

The greatest extension of the spring is $\bf{0.055~m}$ and the maximum speed of the block is $\bf{0.695~m/s}$ .

Explanation:

Given:

The mass of the block is, $m = 0.50~kg$

The spring constant of the spring is, $k = 80~N/m$

The mechanical energy of the block is, $E = 0.12~J$

When a particle is oscillating in a simple harmonic way, its total energy is given by

$E = \dfrac{1}{2}m\omega^{2}a^{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

where $\omega$ is the angular velocity of the mass and $a$ is the amplitude of its motion.

The relation between angular velocity and spring constant is given by

$\omega = \sqrt{\dfrac{k}{m}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting equation (2) in equation (1), we have

$~~~~~~&& E = \dfrac{1}{2}ka^{2}\\&or,& a = \sqrt{\dfrac{2E}{k}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$

Substituting $0.12~J$ for $E$ and $80~N/m$ for $k$ in equation (3), we can write

$a &=& \sqrt{\dfrac{2(0.12~J)}{80~N/m}}\\~~~&=& 0.055~m$

The relation between the maximum velocity and the amplitude is given by

$v_{m} &=& \omega a\\~~~~&=& \sqrt{\dfrac{k}{m}}a~~~~~~~~~~~~~~~~~~~~~~~~~~~~(4)$

Substituting $80~N/m$ for $k$ , $0.50~kg$ for $m$ and $0.055~m$ for $a$ in equation (4), we have

$v_{m} &=& \sqrt{\dfrac{80~N/m}{0.50~kg}}(0.055~m)\\~~~&=& 0.695~m/s$

8 0

3 years ago

Two point charges, initially 3 cm apart, are moved to a distance of 1 cm apart. By what factor does the resulting electric force

Solnce55 [7]

The answer is B (1/9).

6 0

3 years ago

A teacher asks her students to jump off of the ground. Once the students complete the task, she says, "All of you just made Eart

Crazy boy [7]

Answer:

Explained below

Explanation:

A) Newton's first law of motion states that an object will remain at rest or continue in its current state of motion except it is acted upon by another force.

Now using this law, when you jump off the ground, the earth will move a tiny bit and accelerate due to the force applied by the jumping.

B) Newton's 2nd law states that the acceleration of a system is directly proportional to the net external force acting on that system, is in the same direction with it and also inversely proportional to the mass.

In this case, when one jumps, an external force is exerted on the earth and we are told it is directly proportional to the acceleration of the system which in this case it's the earth, then it means that there is some motion by the earth even though you didn't see it move.

C) Newton's third law of motion states that to every action, there is an equal and opposite reaction.

In this case the motion of the jumper will lead to an equal and opposite reaction of the earth.

8 0

3 years ago

What is the resulting velocity of the launcher if the net force on the launcher is equal to the reaction force?

xz_007 [3.2K]

Answer:

according to this question best answer is C

5 0

3 years ago

Please please help I'm stuck!!!

Korvikt [17]

Answer: iDc im in 3grade

Explanation:sorry

6 0

3 years ago

Read 2 more answers