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Fudgin [204]
3 years ago
7

A 2 kg ball travellng to the right with a speed of 4 m/s collidees with a 5 kg ball traveling to the left with a speed of 3 m/s.

Take right to be the positive direction. What is the total momentum of the two balls before they collide? What is the total momentum of the two balls after they collide?
Physics
1 answer:
Savatey [412]3 years ago
5 0

Explanation:

Momentum before collision:

(2 kg) (4 m/s) + (5 kg) (-3 m/s) = -3 kg m/s

No external forces act on the balls, so momentum is conserved.  Therefore, momentum after collision is also -3 kg m/s.

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A bookcase has a base of 1 m long and 0.5 m wide. It has a mass of 300kg. Find the pressure it exerts on the floor in kPa.
Snezhnost [94]
1m*0.5m=0.5m^2\\
g=10\frac{m}{s^2}\\
300kg*10\frac{m}{s^2}=3000N\\
3000N--0.5m^2\\
\ \ x\ \ --\ 1\\
x=6000Pa=60kPa
4 0
2 years ago
B. Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
goblinko [34]

We have that the instantaneous velocity of the shuttlecock when it hits the ground is

V_{int}=\sqrt{U^2+19.6H}

From the question we are told

Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

shuttlecock when it hits the ground? Show your work below.

Generally the equation for acceleration  is mathematically given as

a=v \frac{dv}{dx}\\\\\Therefore\\\\\2ah=v^2-u^2

Where

acceleration is still -9.81 m/s2,

Hence,

V^2-U^2=2(-9.81)*-H

Therefore

V_{int}=\sqrt{U^2+19.6H}

For more information on this visit

brainly.com/question/12319416?referrer=searchResults

7 0
2 years ago
What is the direction of the centripetal force when applied to an object?
mihalych1998 [28]
For an object moving in a path that's a circle or a part of one,
the centripetal force acts in the direction toward the center of
the circle.  That direction is perpendicular to the way the object
is moving.
3 0
2 years ago
Read 2 more answers
Is it possible for a tennis ball to have more momentum than a cannonball? explain ​
Olenka [21]
Well you know how it
8 0
3 years ago
is dimensionally correct relation necessarily to be a correct physical relation? explain with example.​
Andreas93 [3]

Answer: hope it helps you...❤❤❤❤

Explanation: If your values have dimensions like time, length, temperature, etc, then if the dimensions are not the same then the values are not the same. So a “dimensionally wrong equation” is always false and cannot represent a correct physical relation.

No, not necessarily.

For instance, Newton’s 2nd law is  F=p˙ , or the sum of the applied forces on a body is equal to its time rate of change of its momentum. This is dimensionally correct, and a correct physical relation. It’s fine.

But take a look at this (incorrect) equation for the force of gravity:

F=−G(m+M)Mm√|r|3r  

It has all the nice properties you’d expect: It’s dimensionally correct (assuming the standard traditional value for  G ), it’s attractive, it’s symmetric in the masses, it’s inverse-square, etc. But it doesn’t correspond to a real, physical force.

It’s a counter-example to the claim that a dimensionally correct equation is necessarily a correct physical relation.

A simpler counter example is  1=2 . It is stating the equality of two dimensionless numbers. It is trivially dimensionally correct. But it is false.

4 0
3 years ago
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