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Monica [59]
3 years ago
9

Which of the following laws best define this statement? the total amount of energy in a closed system stays the same.

Physics
1 answer:
TEA [102]3 years ago
7 0
That would be t<span>he law of conservation of energy.</span>
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If two equal charges are separated by a certain distance, the force of repulsion is F. Given F1, if each charge in F1 is doubled
Alisiya [41]

Answer:F_2=16\times F_1

Explanation:

Given

Force of repulsion between two charge particle is given by force F

Electrostatic force is given by

F=\frac{kq_1q_2}{r^2}

where q_1 and q_2 is the charges  of particle

r=distance between charge particle

=\frac{kq^2}{r^2}when charges are doubled and distance is reduced to half

i.e. q become 2 q and r becomes 0.5 r

F_2=\frac{k(2q)^2}{(0.5r)^2}

F_2=\frac{kq^2}{r^2}\times 4\times 4

F_2=16\times F_1

         

6 0
3 years ago
When two pool balls collide, what happens to the momentum of each one?
Schach [20]

Answer:

They slow down.

Explanation:

The collided so the slow down till they stop.

7 0
3 years ago
Balance the equation​
mote1985 [20]
Is this a chemistry question ?
7 0
2 years ago
A 21 g ball is shot from a spring gun whose spring has a force constant of 579 N/m. The spring can be compressed 1 cm. How high
vodka [1.7K]

Answer:

The answer  to the question is

The ball will go 0.14 meters high if the gun is aimed vertically

Explanation:

The energy in the spring → Energy, E = \frac{1}{2}kx^2

Where E = energy in the spring

k = Spring constant

x = Spring compression or stretch

Therefore E = \frac{1}{2} 579*0.01^2 = 0.02895 J

The spring energy is transferred  to the ball as kinetic energy based on the first law of thermodynamics which states that energy is neither created nor destroyed

Kinetic energy = KE = \frac{1}{2}mv^2

From which v = \sqrt{\frac{2*KE}{m} } = \sqrt{\frac{2*0.02895}{0.021} } = 1.66 m/s

from v² =u² - 2·a·S

Where v = final velocity = 0 m/s

u = initial velocity = 1.66 m/s

a = g = Acceleration due to gravity

S = height

Therefore 0 = 1.66² - 2×9.81×S

or S = 1.66² ÷ (2×9.81) = 0.14 m

7 0
3 years ago
A 50 kg person steps off a diving platform that is 10 meters above the water below (Olympic height). With what speed do they hit
Nookie1986 [14]

Answer:

14 m/s

Explanation:

The following data were obtained from the question:

Mass = 50 kg

Initial velocity (u) = 0 m/s

Height (h) = 10 m

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) =?

The velocity (v) with which the person hit the water can be obtained as shown below:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 10)

v² = 0 + 196

v² = 196

Take the square root of both side

v = √196

v = 14 m/s

Therefore, he will hit the water with a speed of 14 m/s

5 0
2 years ago
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