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Gre4nikov [31]
3 years ago
7

As a rocket rises, its kinetic energy changes. At the time the rocket reaches its highest point, most of the kinetic energy of r

ocket has been
A permanently destroyed.
B stored in bonds between its atoms.
C changed to thermal energy due to friction.
D transformed into gravitational potential energy.
Physics
2 answers:
Ulleksa [173]3 years ago
7 0
<span>D transformed into gravitational potential energy.</span>
postnew [5]3 years ago
6 0

Answer:

D transformed into gravitational potential energy.

Explanation:

Kinetic energy is the energy associated with movement and potential energy is the energy associated with position in a system. Energy, in general, is the ability to do a job.

Both kinetic and potential energy represent the two fundamental types of existing energy. Any other energy is a different version of kinetic or potential energy or a combination of both. For example, mechanical energy is the combination of kinetic and potential energy.

Kinetic energy is the type of energy that is associated with movement. Anything that is moving has kinetic energy. It can be calculated using the formula:

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

Where:

m=mass\hspace{3}of\hspace{3}the\hspace{3}object\\v=velocity\hspace{3}of\hspace{3}the\hspace{3}object

Potential energy is the type of energy that is associated with the relative position within a system, that is, the position of an object with respect to another.  It can be calculated using the formula:

E_p=mgh

Where:

g=gravitational\hspace{3}acceleration\hspace{3}constant\\m=mass\hspace{3}of\hspace{3}the\hspace{3}object\\h=It\hspace{3}is\hspace{3}the\hspace{3}distance\hspace{3}(height\hspace{3}in \hspace{3}meters) between\hspace{3}the\hspace{3}Earth\hspace{3}and\hspace{3}the\hspace{3}object.

So, at the highest point, the velocity of the rocket would be zero, and the distance respect to the launch point would be maximum. Hence according to the previous equation its kinetic energy would zero and its potential energy would be the highest. in another words the kinetic energy of rocket has been transformed into gravitational potential energy.

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If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you
Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0
3 years ago
Earth, the Sun, and billions of stars are contained within
Nitella [24]
I think it is the Milky Way.
3 0
3 years ago
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Convert to the fractional equivalent and reduce 21.12
nignag [31]

The decimal point is placed after two digits starting from the end. For each decimal place, we can write the number divided by 100.

21.12 can be written as \frac{2112}{100}.

Divide the numerator and denominator by 2:

\frac{2112}{100}= \frac{156}{50}

The numerator and denominator can be divided by 2 again:

\frac{78}{25}

There is no other common factor between numerator and denominator other than 1. Hence, it is the reduced form.




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⦁ A car going 50 m/s is brought to rest in a distance of 20.0 m as it strikes a pile of dirt. How large an average force is exer
gtnhenbr [62]

Answer:

the average force exerted by seatbelts on the passenger is 5625 N.

Explanation:

Given;

initial velocity of the car, u = 50 m/s

distance traveled by the car, s = 20 m

final velocity of the after coming to rest, v = 0

mass of the passenger, m = 90 kg

Determine the acceleration of the car as it hit the pile of dirt;

v² = u² + 2as

0 = 50² + (2 x 20)a

0 = 2500 + 40a

40a = -2500

a = -2500/40

a = -62.5 m/s²

The deceleration of the car is 62.5 m/s²

The force exerted on the passenger by the backward action of the car is calculated as follows;

F = ma

F = 90 x 62.5

F = 5625 N

Therefore, the average force exerted by seatbelts on the passenger is 5625 N.

8 0
3 years ago
José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo
satela [25.4K]

Answer:

179 kg

Explanation:

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