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aleksandrvk [35]
3 years ago
13

When a snowball turns into a puddle of

Physics
1 answer:
In-s [12.5K]3 years ago
8 0

When a snowball turns into a puddle of water, we know that (the snow ball gains energy and changes from a solid to liquid).

This is correct due to the fact that particles of the snowball are gaining speed and so it is heating up, when the solid's temperature reaches the melting point, it will become a liquid.

Therefore, D is the correct answer.

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The first step in responding to any sports injury is to __________.
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Kamir and Alexis are studying the properties of water. They conducted a variety of experiments to determine its physical and che
solniwko [45]

The answer is D) neutral water reacts with carbon dioxide to form an acid solution

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Two spacecraft are both 10 million kilometers from a star. The total power output of the star is 4 x 1025 W. Spacecraft 1 has a
Ksenya-84 [330]

The concept of power is given by the relationship between intensity and area, that is to say that power is defined as

P = A*I

Our values are given under the condition of,

r_1 = 18m

r_2 = m

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\text{Power} \propto r^2

For both panels we would have to

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A block of mass m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia I, which is free
Hoochie [10]

Answer:

Explanation:

I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.

The relation between the linear acceleration and the angular acceleration is

a = R α   .... (1)

According to the diagram,

T x R = I x α

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T = I x a / R²      .... (2)

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Substitute the value of T from equation (2) in equation (3)

mg - \frac{Ia}{R^{2}}=ma

a=\frac{mg}{m+\frac{I}{R^{2}}}

T is the acceleration in the system

Substitute the value of a in equation (2)

T = \frac{I}{R^{2}}\times \frac{mg}{m+\frac{I}{R^{2}}}

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4 0
3 years ago
A toroidal coil of N turns has a central radius b and a square cross section of side a. Find its self-inductance.
Xelga [282]

Answer:

L = \frac{\mu_0 N^2 (a^2)}{2\pi b}

Explanation:

As we know that magnetic field due to torroid is given as

B = \frac{\mu_0 N i}{2\pi b}

this is approximately constant magnetic field along the axis of the torroid

now the flux linked with one coil of the torroid is given as

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So we know that

L = \frac{\phi}{i}

L = \frac{\mu_0 N^2 (a^2)}{2\pi b}

3 0
3 years ago
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