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3 years ago

What makes up momentum?

Physics

2 answers:

amid [387]3 years ago

6 0

The momentum of a particle is defined as the product of its mass times its velocity.<span>
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natita [175]3 years ago

4 0

To calculate the momentum of an object, you need three numbers:

... the object's mass
... its speed
... the direction it's moving

The magnitude (size) of the momentum is (mass) x (speed).
The direction of the momentum vector is the direction of the motion.

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A ball is thrown straight up in the air. For which situations are both the instantaneous velocity and the acceleration zero? (Se

Gwar [14]

Answer:

Therefore, the situation in which both the instantaneous velocity and acceleration become zero, is the situation when the ball reaches the highest point of its motion.

Explanation:

When a ball is thrown upward under the free fall action of gravity, it starts to loose its Kinetic Energy as it moves upward. As the ball moves in upward direction, its kinetic energy gradually converts into its potential energy. As a result the speed of the ball starts to decrease as it moves up. Therefore, at the highest point during its motion, the velocity of ball becomes zero and it stops at the highest point for a moment, and then it starts to fall back down, under the influence of gravitational force.

Therefore, the situation in which both the instantaneous velocity and acceleration become zero, is the situation <u>when the ball reaches the highest point of its motion.</u>

6 0

3 years ago

What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a

Paha777 [63]

$E = mgh + \frac{1}2} m v^{2} + \frac{1}{2} I \omega^{2} = mgh + \frac{1}2} m r ^{2} \omega ^{2} + \frac{1}{2} I \omega^{2}$

for a solid cylinder: $I = \frac{1}{2} m r^{2}$
for a hollow cylinder: $I = mr^{2}$

I will look at the case of a hollow cylinder:

$E = mgh + I \omega ^{2} = constant \\ \\ I = \frac{mgh}{ \omega^{2} }$

That is as far as i get.

7 0

4 years ago

You spill a bucket of soapy water on a marble floor accidentally.Would it make it easier or difficult for you to walk on the flo

stiks02 [169]

Answer:

Of course harder

Explanation:

Just imagone the floor is wet and you walk on it, do u feel it hard or easy to walk? :D

7 0

3 years ago

Read 2 more answers

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

Irina-Kira [14]

Answer:

part (a) $\alpha\ =\ -2.5\ rad/s^2$

part (b) N = 79.61 rev

part (c) $\tau\ =\ 23.54\ Nm$

Explanation:

Given,

Initial speed of the wheel = $w_o\ =\ 50.0\ rad/s$
total time taken = t = 20.0 sec

part (a)

Let $\alpha$ be the angular acceleration of the wheel.

Wheel is finally at the rest. Hence the final angular speed of the wheel is 0.

$\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2$

part (b)

Let $\theta$ be the total angular displacement of the wheel from initial position till the rest.

$\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad$

We know, 1 revolution = $2\pi$ rad

Let N be the number of revolution covered by the wheel.

$\therefore N\ =\ \dfrac{\theta}{2\pi}\\\Rightarrow N\ =\ \dfrac{500}{2\times 3.14}\\\Rightarrow N\ =\ 79.61\ rev$

Hence the 79.61 revolution is covered by the wheel in the 20 sec.

part (c)

Given,

Mass of the pole = m = 4 kg
Length of the pole = L = 2.5 m
Angle of the pole with the horizontal axis = $\theta\ =\ 60^o$

Now the center of mass of the pole = $d\ =\ \dfra{L}{2}\ =\ \dfrac{2.5}{2}\ =\ 1.25\ m$

Weight component of the pole perpendicular to the center of mass = $F\ =\ mgcos\theta$

$\therefore \tau\ =\ F\times d\\\Rightarrow \tau\ =\ 4\times 9.81\times cos60^o\times 1.25\\\Rightarrow \tau\ =\ 23.54\ Nm$

3 0

4 years ago

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 94.5 cm and diameter 2.00 cm fr

Aleks [24]

The volume of a cylinder is PI*R^2H where PI is a regular, R is the radius and H the height. Consequently, the quantity of the iron rod could be (3.141)*((2.05/2)^2)*(ninety four.Three) = 311.2 cm^three. The density of Iron is 7.874 gm/cm^3, so the whole weight would be 2449 grams, or about 2.5 kilograms.

3 0

4 years ago