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ValentinkaMS [17]
3 years ago
11

A worker drives a 0.562 kg spike into a rail tie with a 2.26 kg sledgehammer. The hammer hits the spike with a speed of 64.4 m/s

. If one third of the hammer’s kinetic energy is converted to the internal energy of the hammer and spike, how much does the total internal energy increase? Answer in units of J
Physics
1 answer:
zubka84 [21]3 years ago
4 0

Answer:

Explanation:

Given that,

Mass of sledge hammer;

Mh =2.26 kg

Hammer speed;

Vh = 64.4 m/s

The expression fot the kinetic energy of the hammer is,

K.E(hammer) = ½Mh•Vh²

K.E(hammer) = ½ × 2.26 × 64.4²

K.E ( hammer) = 4686.52 J

If one forth of the kinetic energy is converted into internal energy, then

ΔU = ¼ × K.E(hammer)

∆U = ¼ × 4686.52

∆U = 1171.63 J

Thus, the increase in total internal energy will be 1171.63 J.

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A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 13.62 cm and mass 1.199 kg that is attached
Feliz [49]

Answer:

Explanation:

Expression for time period of a pendulum is as follows

T = 2\pi\sqrt{\frac{l}{g} }

l is length of pendulum from centre of bob and g is acceleration due to gravity

Given

Time period T = 1.583

g = 9.846

Substituting the values

1.583 = 2\pi\sqrt{\frac{l}{9.846} }

l = \frac{(1.583)^2\times9.846}{4\times(\frac{22}{7})^2 }

l = .6244 m

= 62.44 cm

Length of rod  = length of pendulum - radius of bob

= 62.44 - 13.62

= 48.82 cm

= .488 m

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6 0
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Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars
zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

T^2 \propto R^3

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

T^2 \propto R^3

T^2 = kR^3

Substituting the values, we get the value of constant k for mars.

687^2 = k\times (2.279 \times 10^{11})^3

k = 3.92 \times 10^{-29}

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

365^3 = 3.92\times 10^{-29} R^3

R^3 = 3.39 \times 10^{33}

R= 1.50 \times 10^{11}\;\rm m

Hence we can conclude that the distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0
2 years ago
N what direction does the medium move relative to the direction of the wave? Explain.
Mashcka [7]
Two types of mechanical waves: longitudinal<span> waves and </span>transverse<span> waves; the medium movement differs between the two.

</span>In a longitudinal wave the medium particle movement is parallel to the direction of wave propagation; example is sound wave in air.

I<span>n a transverse wave the medium particle movement is perpendicular to the direction of wave propagation; example is mechanical wave on a string.
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3 0
3 years ago
Read 2 more answers
The height of the upper falls at Yellowstone Falls is 33 m. When the water reaches the bottom of the falls, its speed is 26 m/s.
Klio2033 [76]

Answer:

Speed of water at the top of fall = 5.40 m/s

Explanation:

We have equation of motion

v^2=u^2+2as

Here final velocity, v = 26 m/s

a = acceleration due to gravity

a=9.8m/s^2 \\

displacement, s = 33 m

Substituting

26^2=u^2+2\times 9.8 \times 33\\\\u^2=29.2\\\\u=5.40m/s \\

Speed of water at the top of fall = 5.40 m/s

3 0
3 years ago
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