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postnew [5]
2 years ago
9

Two stones, one of mass m and the other of mass 2m, are thrown directly upward with the same velocity at the same time from grou

nd level and feel no air resistance. Which statement about these stones is true?
a. The heavier stone will go twice as high as the lighter one because it initially had twice as much kinetic energy.
b. Both stones will reach the same height because they initially had the same amount of kinetic energy.
c. At their highest point, both stones will have the same gravitational potential energy because they reach the same height.
d. At Its highest point, the heavier stone will have twice as much gravitational potential energy as the lighter one because it is twice as heavy.
e. The lighter stone will reach its maximum height sooner than the heavier one.
Physics
1 answer:
brilliants [131]2 years ago
8 0

Answer:

c.  At their highest point, both stones will have the same gravitational potential energy because they reach the same height.

Explanation:

The above statement is true going by the fact that, the gravitational potential energy of both stones is a function of its mass (weight) as well as the height at which it reach. <em>Despite the fact that, both stones has different weights, it would reach different heights. At their various maximum height, the will have same potential energy.</em>

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2 years ago
A diver jumps off a diving platform that is 20 meters long. Describe the transfer of energy that occurs during the fall.
kobusy [5.1K]

Answer: gravitational potential energy is converted into kinetic energy

Explanation:

When the diver stands on the platform, at 20 m above the surface of the water, he has some gravitational potential energy, which is given by

E=mgh

where m is the man's mass, g is the gravitational acceleration and h is the height above the water. As he jumps, the gravitational potential energy starts decreasing, because its height h above the water decreases, and he acquires kinetic energy, which is given by

K=\frac{1}{2}mv^2

where v is the speed of the diver, which is increasing. When he touches the water, all the initial gravitational potential energy has been converted into kinetic energy.

8 0
3 years ago
7. A 1.0 kg metal head of a geology hammer strikes a solid rock with a velocity of 5.0 m/s. Assuming all the energy is retained
marta [7]

The increase in temperature of the metal hammer is 0.028 ⁰C.

The given parameters:

  • <em>mass of the metal hammer, m = 1.0 kg</em>
  • <em>speed of the hammer, v = 5.0 m/s</em>
  • <em>specific heat capacity of iron, 450 J/kg⁰C</em>

The increase in temperature of the metal hammer is calculated as follows;

Q = K.E\\\\mc \Delta T = \frac{1}{2}  mv^2\\\\\Delta T = \frac{v^2}{2 c}

where;

<em>c is the </em><em>specific heat capacity</em><em> of the metal hammer</em>

<em />

Assuming the metal hammer is iron, c = 450 J/kg⁰C

\Delta T = \frac{5^2}{2 \times 450} \\\\\Delta T = 0.028 \ ^0C

Thus, the increase in temperature of the metal hammer is 0.028 ⁰C.

Learn more about heat capacity here: brainly.com/question/16559442

8 0
2 years ago
Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the
anygoal [31]

Answer:

The value is the temperature of the air inside the tire T_{2} = 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet P_{1} = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet P_{2} = 3.2 + 1 = 4.2 atm

Temperature at inlet T_{1} = 300 K

(a) Volume of the system is constant so  pressure is directly proportional to the temperature.

\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }

\frac{T_{2} }{300}  = \frac{4.2}{3.7}

T_{2} = 340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

m_{1}  T_{1} =  m_{2}  T_{2}

\frac{m_{2} }{m_{1}  } = \frac{T_{1} }{T_{2} }

\frac{m_{2} }{m_{1}  } = \frac{300}{354.54}

\frac{m_{2} }{m_{1}  } =0.00294

value of  the original mass of air in the tire should be released is  \frac{m_{2} - m_{1}}{m_{1}}

\frac{0.00294-1}{1}

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

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2 years ago
The acronym laser stands for light amplification by ____ emission of radiation
AleksAgata [21]
<span>light amplification by stimulated emission of radiation </span>
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3 years ago
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