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2 years ago

A race car accelerates at a constant rate from 20 m/s to 60 m/s in a time of 2 seconds. What is the car’s acceleration?

Physics

1 answer:

Vlada [557]2 years ago

5 0

$\huge\fbox{Answer ☘}$

Given ,

initial velocity , u = 20 m/s

final velocity , v = 60 m/s

time taken = 2 seconds

Now ,

$\bold{acceleration = \frac{v - u}{t} } \\ \\ = > \bold{ \frac{60 - 20}{2} } \\ \\ = > \bold{\frac{40}{2} } \\ \\ \bold{ => 20 \: ms {}^{ - 2}}$

hope helpful~

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HELPPPP

Arisa [49]

Answer:

A. nuclear fusion reactions

C. it's still hot from the big bang

Explanation:

The inside of the earth is hot due to some reasons. This heat provides the internal energy the drives processes within the earth interior. Here are some of the ways in which the heat has accumulated:

Nuclear reactions within the earth interior by fusion and other radioactive processes releases a large amount of heat.
Some heat accreted during the early formation of the earth and has not been lost till this day.
Heating due to friction

These are some of the sources of the earth's internal heat.

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3 years ago

A ping pong ball with a dent in it can be put into a pan of boiling water. After a short amount of time, the dent will pop out.

kifflom [539]

heat from water goes into air in ball

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4 years ago

Read 2 more answers

You hear a sound with a frequency of 256 Hz. The amplitude of the sound increases and decreases periodically: it takes 2 seconds

german

To solve this problem it is necessary to take into account the concepts related to frequency and period, and how they are related to each other.

The relationship that defines both agreements is given by the equation,

$f_{beat}=\frac{1}{T}$

Then the frequency for the previous period given (2sec) is

$f_{beat}=\frac{1}{2}$

$f_{beat} = 0.5Hz$

The beat frequency of two frequencies is equal to the difference between the two frequencies, then

$f_{beat} = |f_1-f_2|\\f_{beat} = |256Hz-2Hz|\\f_{beat} = 254Hz$

Hence option A is incorrect.

We can do this process for 254Hz as $f_1$ and 258 Hz for $f_2$ , then

$f_{beat} =|254Hz-258Hz|$

$f_{beat} = 4Hz$

Hence option B is incorrect.

We can also do this process for 255Hz as $f_1$ and 257 Hz for $f_2$ , then

$f_{beat} =|255Hz-257Hz|$

$f_{beat} = 2Hz$

Hence option C is incorrect.

We can also do this process for 255.5Hz as f_1 and 256.5 Hz for f_2, then

$f_{beat} =|255.5Hz-256.5Hz|\\f_{beat} = 1Hz$

Hence option D is incorrect.

We can also do this process for 255.75Hz as $f_1$ and 256.25 Hz for $f_2$ , then

$f_{beat} =|255.75Hz-256.25Hz|\\f_{beat} = 0.5Hz$

Hence option E is incorrect.

Therefore the sum of the frequencies in the sound wave would be 256.25Hz and 255.75Hz

3 0

3 years ago

In a certain time period a coil of wire is rotated from one orientation to another with respect to a uniform 0.38-T magnetic fie

Evgesh-ka [11]

Answer:

8.4 V

Explanation:

induced emf, e1 = 5.8 V

Magnetic field, B1 = 0.38 T

magnetic field, B2 = 0.55 T

induced emf, e2 = ?

As we know that the induced emf is directly proportional to the magnetic field strength.

When the other parameters remains constant then

$\frac{e_{1}}{e_{2}}=\frac{B_{1}}{B_{2}}$

$\frac{5.8}{e_{2}}=\frac{0.38}{0.55}$

e2 = 8.4 V

Thus, the induced emf is 8.4 V.

4 0

3 years ago

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift i

WINSTONCH [101]

Answer:

Total work done = = 29811.60 J

Explanation:

Since the person is being moved upward, the person’s potential and kinetic energy are increasing.

To determine the increase in potential energy, the following equation is used;

∆ PE = m * g * ∆ h

m = 78.0 Kg, g = 9.8 m/s, ∆ h = 13.0 m

∆ PE = 78.0 * 9.8 * 13.0 = 9937.20 J

Increase in kinetic energy is given by the following equation;

∆ KE = ½ * m * (vf² – vi²)

vf = 3.4, vi = 0

∆ KE = ½ * 78.0 * 3.4² = 450.84 J

Total work = 9937.20 + 450.84 = 10388.04 J

(b) He is then lifted at the constant speed of 3.40 m/s

Velocity is constant, therefore, there is no increase in kinetic energy. The only work done is the increase of potential energy

∆ PE = 78.0 * 9.8 * 13.0

∆ PE = 9937.20 J

Since his velocity decreased from 3.40 m/s to 0 m/s, his kinetic energy decreased.

∆ KE = ½ * 78.0 * (0² - 3.4²)

∆ KE = - 450.84 J

∆ PE = 78.0 * 9.8 * 13.0 = 9937.20 J

Total work = 9937.20 - 450.84 = 9486.36 J

Sum of works = 10388.04 + 9937.20 + 9486.36 = 29811.60 J

8 0

3 years ago