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Vinvika [58]
3 years ago
9

What is the formula for force

Physics
1 answer:
tresset_1 [31]3 years ago
6 0

Force = Mass * Acceleration

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1. (15 points) Small cart is rolling down an inclined track and accelerating. It fires a ball straight out of the cannon as it m
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Answer:

Explanation:

This problem can be solved with the conservation of the momentum.

If the ball is fired upward, the momentum before and after the ball is fired must conserve. Hence, the speed of the ball is the same that the speed of the car just in the moment in wich the ball is fired.

Hence, the result depends of the acceleration of the car. If the change in the speed is higher than the speed of the ball, it is probably that the ball will be behind the car or it will come back to the car.

If the ball is fired forward, and if the change in the speed of the car is not enogh, the ball will be in front of the car.

HOPE THIS HELPS!!

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3 years ago
Please please help. I don’t understand this.
Kazeer [188]
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A uniform cylinder of mass 1.5 kg and radius 0.3 m rolls down a ramp inclined at an angle 0.12 radians to the horizontal. What i
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3 years ago
Which statement best explains why the overall charge on an atom is zero?
deff fn [24]
This indicates​ nucleus of an atom constantly positively charged. an atom has an nuteral overall charge because it has the same amount of electrons as protons
7 0
4 years ago
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Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and th
Evgen [1.6K]

Answer:

\omega=0.24\ rad.s^{-1}

Explanation:

Given:

mass of person, m=36\ kg

mass of merry go-round, M=300\ kg

radius of merry go-round, R=2\ m

velocity of the person running, v=4\ m.s^{-1}

<u>We consider merry go-round as a ring:</u>

Now the moment of inertial of the ring is given as,

I=M.R^2

I=300\times 2^2

I=1200\ kg.m^{-2}

<u>Moment of inertia of the person considering as a point mass:</u>

I_p=m.R^2

I_p=36\times 2^2

I_p=144\ kg.m^2

<u>Now according to the conservation of angular momentum:</u>

I.\omega=I_p.\omega_p

where:

\omega = angular velocity of the merry-go-round

\omega_p= angular velocity of the person running

1200\times \omega=144\times \frac{v}{R}

\omega=\frac{144}{1200} \times \frac{4}{2}

\omega=0.24\ rad.s^{-1}

4 0
4 years ago
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