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Pani-rosa [81]
2 years ago
6

Think about it: suppose a meteorite collided head-on with mars and becomes buried under mars's surface. what would be the elasti

city of this collision? explain your answer.
Physics
1 answer:
Ket [755]2 years ago
4 0

Answer:

Perfectly inelastic collision

Explanation:

There are two types of collision.

1. Elastic collision : When the momentum of the system and the kinetic energy of the system is conserved, the collision is said to be elastic. For example, the collision of two atoms or molecules are considered to be elastic collision.

2. Inelastic collision: When the momentum  the system is conserved but the kinetic energy is not conserved, the collision is said to be inelastic. For example, collision of a ball with the mud.

For a perfectly elastic collision, the two bodies stick together after collision.

Here, the meteorite collide with the Mars and buried inside it, the collision is said to be perfectly inelastic. here the kinetic energy of a body lost completely during the collision.

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A 2 kg stone moves with a speed of 1 m/s. A second 2 kg stone is moving twice as fast. Compare their kinetic energies.
alekssr [168]
D
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5 0
2 years ago
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Two particles carrying charges q1 and q2 are separated by a distance r and exert an electric force F⃗ E on each other. If q1 is
zepelin [54]

Answer:

q2 must also be doubled

r may also be halved

Explanation:

According to Coulumbs law

F= K q1 q2/r^2

If q1 is doubled, we must necessarily double q2 and r may also be halved in order to maintain F at the same value. Once the value of F is thus kept constant and E is also constant, the product FE must remain constant.

5 0
2 years ago
A proton moves along the x-axis with vx=1.0×107m/s. As it passes the origin, what are the strength and direction of the magnetic
Sunny_sXe [5.5K]

Answer:

Magnetic field will be ZERO at the given position

Explanation:

As we know that the magnetic field due to moving charge is given as

B = \frac{\mu_0 qv sin\theta}{4\pi r^2}

so here we know that for the direction of magnetic field we will use

\hat B = \hat v \times \hat r

so we have

\hat B = \hat i \times (\hat i + 0\hat j + 0\hat k)

so magnetic field must be ZERO

So whenever charge is moving along the same direction where the position vector is given then magnetic field will be zero

3 0
2 years ago
A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp
liq [111]

Answer:

W_2=\sqrt{\frac{3}{5} }W_1

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

I_1 =\frac{2}{5}MR^2

K_1 = \frac{1}{2}IW_1^2

So, replacing, we get that:

K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

At the same way, the moment of inertia and kinetic energy for second ball is:

I_2 =\frac{2}{3}MR^2

K_2 = \frac{1}{2}IW_2^2

So:

K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2

Then, K_2 is equal to K_1, so:

K_2 = K_1

\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2

\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2

Finally, solving for W_2, we get:

W_2=\sqrt{\frac{3}{5} }W_1

5 0
2 years ago
A book rests on the shelf of a bookcase. The reaction force to the force of gravity acting on the book is 1. The force of the sh
Hoochie [10]

Answer:

1. The force of the shelf holding the book up.

Explanation:

The free body diagram of the book is as follows:

1 - The weight of the book towards downwards

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Since the book is at rest, these two forces are equal to each other and according to Newton's Third Law the reaction force to the force of gravity is equal but opposite to the weight of the book. This reaction force is the one that holds the book up on the shelf.

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2 years ago
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