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3 years ago

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According to the picture, how would the gravitational potential energy of the car change if its weight was doubled? A) The gravi

tational potential energy would be doubled. B) The gravitational potential energy would stay the same C) The gravitational potential energy would be cut in half. D) The gravitational potential energy would be increased by 750.

2 answers:

Vera_Pavlovna [14]3 years ago

8 0

Answer:

the answer is (A)

Explanation:

Lorico [155]3 years ago

3 0

Answer is A

Samuriguy125 on YT

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Suppose the tip of the minute hand of a clock is two inches from the center of the clock. Determine the distance traveled by the

sashaice [31]

1) The minute travels the circumference of a 4 inches circle is 60 minutes:

Circumference = 2πr = 2π(2in) = 4π in
Time = 60 minutes

2) Constant angular velocity => (4π / 60) = (x / 20) => x = 4π / 3 inches

x = 4.19 inches

Answer: 4.19 inches

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3 years ago

un jugador de beisbol batea un foul recto en el aire.la pelota deja el bate con una rapidez de 120 km/h. en ausencia de resisten

lina2011 [118]

Answer:

120 km/h

Explanation:

This is a case of a parabolic motion. For this kind of motion the horizontal component of the velocity does not change in the complete trajectory. That is:

$v_x=constant$

Then, the velocity of the ball at the moment in which the catcher caught the ball is:

$v=120\frac{km}{h}$

the velocity of the ball is 120 km/h.

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3 years ago

3.Three resistors of 25.0Ω, 30.0Ω, and 40.0Ω are in a series circuit with a 6.0-volt battery. What is the current in the circuit

AURORKA [14]

Answer:

Current = 0.063 Amperes

Explanation:

Let the three resistors be R1, R2, and R3 respectively.

Given the following data;

R1 = 25.0Ω,

R2 = 30.0Ω

R3 = 40.0Ω

Voltage = 6 Volts

First of all, we would determine the equivalent or total resistance;

Total resistance (in series) = R1 + R2 + R3

Total resistance = 25.0Ω + 30.0Ω + 40.0Ω

Total resistance = 95 Ω

Next, we find the current flowing through the circuit;

Voltage = current * resistance

Substituting into the formula, we have;

6 = current * 95

Current = 6/95

Current = 0.063 Amperes

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3 years ago

Which direction will the box move?

Yuki888 [10]

Net force

F1-F2-F3
5-10-20
5-30
-25N

As it's negative the box will move left

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2 years ago

Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at

andrew-mc [135]

Answer:

1. the electric potential energy of the electron when it is at the midpoint is - 2.9 x $10^{-17}$ J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x $10^{-17}$ J

Explanation:

given information:

$q_{1}$ = 3 nC = 3 x $10^{-9}$ C

$q_{2}$ = 2 nC = 2 x $10^{-9}$ C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is at the midpoint

potential energy of the charge, F

F = k $\frac{q_{e}q}{r}$

where

k = constant (8.99 x $10^{9} Nm^{2} /C^{2}$ )

electron charge, $q_{e}$ = - 1.6 x $10^{-19}$ C

since it is measured at the midpoint,

r = $\frac{0.5}{2}$

= 0.25 m

thus,

F = $F_{1}+ F_{2}$

= k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $\frac{kq_{e} }{r}$ ( $q_{1} +q_{2}$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ +2 x $10^{-9}$ )/0.25

= - 2.9 x $10^{-17}$ J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

$r_{1}$ = 10 cm = 0.1 m

$r_{2}$ = 0.5 - 0.1 = 0.4 m

F = k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $kq_{e}$ ( $\frac{q_{1} }{r_{1} }$ + $\frac{q_{2} }{r_{2} }$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ /0.1+2 x $10^{-9}$ /0.4)

= - 5.04 x $10^{-17}$ J

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3 years ago