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3 years ago

Please Help Quick ASAP Hurry This is Physical Science

Physics

1 answer:

krok68 [10]3 years ago

6 0

Answer:

Think it is C

Explanation:

Not sure!!!

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Given that the collision is elastic and glider 2 is initially at rest (v2,i =0), please use below Eqs. to explain why

Morgarella [4.7K]

Answer:

Explanation:

1 )

Put v2,i =0, in second equation

v2,f= (m2-m1)v2,i + 2m1v1,i/m1+m2

v2,f = 0 + 2m1v1,i/m1+m2

v2,f = 2m1v1,i/m1+m2

In this equation coefficient of v1,i is positive so v2,f and v1,i have the same sign.

2 )

Put m1 < m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

As m1-m2 is negative , v1f and v1i will have opposite sign.

3 )

Put m1 > m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

m1 - m2 is positive so v1f and v1i will have same sign.

4 )

Put m1 = m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i

= 0 because m1 = m2

So glider 1 will stop because v1,f = 0 .

5 0

3 years ago

What part of the atom takes up most of the atoms space?

Ahat [919]

Answer:

The electrons

Explanation:

3 0

3 years ago

Read 2 more answers

At what condition does a body becomes weightless at the equator?

ArbitrLikvidat [17]

Answer:

The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface

Explanation:

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3 years ago

Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th

brilliants [131]

From the case we know that:

The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

#SPJ4

8 0

1 year ago

What do you mean by focus in science

Tanzania [10]

“a point at which rays of light, heat, or other radiation meet after being refracted or reflected.” Meaning multiple light rays or heat (and other forms of radiation) are all being refracted or reflecting to a certain point

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2 years ago