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mrs_skeptik
1 month ago
10

Calculate the gravitational potential energy of the interacting pair of the Earth and a 11 kg block sitting on the surface of th

e Earth. You would need to supply the absolute value of this result to move the block to a location very far from the Earth (actually, you would need to use even more energy than this due to the gravitational potential energy associated with the Sun-block interacting pair).
Physics
1 answer:
Softa [3K]1 month ago
5 0

Answer:

The gravitational potential energy (GPE) is calculated as 107.8J

Explanation:

The formula for gravitational potential energy (GPE) is GPE = mgh

Where the mass(m) is 11kg

The acceleration due to gravity(g) is 9.8m²/s

The height is considered to be 1m

The force(F) can be represented as mg

Calculating force(F) gives us 11×9.8 = 107.8N

Thus, the gravitational potential energy (GPE) is 107.8×1

which equals 107.8J

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f a stadium pays $11,000 for labor and $7,000 for parking, what would the stadium's parking revenue have to be if the stadium is
Yuliya22 [3333]

Answer:

Explanation:

a

7 0
3 months ago
Read 2 more answers
Wire A has the same length and twice the radius of wire B. Both wires are made of the same material and carry the same current.
serg [3582]

Answer:

V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}

V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}

Consequently, we find that:

V_A = \frac{1}{4} V_B

Thus, the most suitable answer would be:

a. vA = vB/4

Explanation:

In this situation, we can establish the following conditions:

L_A = L_B =L both wires share the same length

both wires carry an identical currentI_A = I_B =I

Both wires are constructed of the same material, indicating that the electron density (n) remains constant across both wires

n_A = n_B =n

We also know that r_A = 2 r_B where r signifies the radius.

Given that wires are cylindrical in shape, we can determine the area for each case:

A_A= \pi r^2_A = \pi (2r_B)^2 = 4 \pi r^2_B= 4 A_B

A_B = \pi r^2_B

Thus, we conclude that

A_A = 4 A_B

Now we are aware that the drift velocity of an electron in a wire can be described by:

v_d = \frac{I}{neA}

Where I denotes the current, n is the electron density, e represents the electron charge, and A signifies the area.

By substituting, we arrive at:

V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}

V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}

So we observe that:

V_A = \frac{1}{4} V_B

Thus, the most fitting answer is:

a. vA = vB/4

6 0
1 month ago
Compare the time period of two simple pendulums of length 4m and 16m at a place.
Ostrovityanka [3204]

Answer:

The period of the pendulum measuring 16 m is double that of the 4 m pendulum.

Explanation:

Recall that the period (T) of a pendulum with length (L) is defined by:

T=2\,\pi\,\sqrt{ \frac{L}{g} }

where "g" denotes the local gravitational acceleration.

Since both pendulums are positioned at the same location, the value of "g" will be consistent for both, and when we compare the periods, we find:

T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2

Thus, the duration of the 16 m pendulum is two times that of the 4 m one.

5 0
2 months ago
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