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3 years ago

9

A boulder with a weight of 780 N is resting at the edge of a cliff that rises 123 m above the ground. What is the gravitational

potential energy of the boulder?

1 answer:

-Dominant- [34]3 years ago

3 0

Potential energy U = mgh

Given h = 123 m,
mg = F = 780 N

Then
U = (123)(780)
= 95940
= 9.59 x 10^4

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dsp73

°C = (5/9) · (°F-32)

The "wet" thermometer is the upper one ... you can see the wet cloth wrapped around the bulb at the end. It's reading 70° F.

°C = (5/9) · (38) = 21.1° C

The "dry" thermometer is the lower one. It's reading 80° F.

°C = (5/9) · (48) = 26.7° C

So it looks like choice-A is your answer.

6 0

3 years ago

A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud

IceJOKER [234]

Answer: 1.51 km

Explanation:

<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or, $\vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}$

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

<u>So,</u>

$\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514 \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}$

Therefore, the separation between the two charges (r) = 1.51 km

3 0

2 years ago

(a) Calculate the acceleration due to gravity on the surface of the Sun.

Ira Lisetskai [31]

<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

a) Acceleration due to gravity

$g=\frac{GM}{r^2}$

Here we need to find acceleration due to gravity of Sun,

G = 6.67259 x 10⁻¹¹ N m²/kg²

Mass of sun, M = 1.989 × 10³⁰ kg

Radius of sun, r = 6.957 x 10⁸ m

Substituting,

$g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2$

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²

b) Acceleration due to gravity in earth = 9.81 m/s²

Ratio of gravity = 274.21/9.81 = 27.95

Weight = mg

Factor of increase in weight = 27.95

8 0

3 years ago

At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force t

Salsk061 [2.6K]

Answer:

The work done on the athlete is approximately 2.09 J

Explanation:

From the definition of the work done by a variable force:

$\displaystyle{\int_{x_i}^{x_j}F(x)dx}$

and substituting with the function of our problem:

$\displaystyle{\int_{0}^{0.19}(140x-190x^2)dx\approx2.09\mathrm{J}}$

5 0

4 years ago

The location where an earthquake begins

Ivahew [28]

Answer:

b

Explanation:

the location where an earthquake begin is called

3 0

3 years ago