All categories

3 years ago

Can we see the back side of the moon from earth?

Physics

2 answers:

AVprozaik [17]3 years ago

6 0

No, since the earth is always rotating around the same time as earth we can’t see the other side of the moon

Olegator [25]3 years ago

5 0

No. The moon always keeps the same side facing us. Its rotation and revolution periods are equal.

You might be interested in

Ocean currents are caused by water's density differences. The density differences in the ocean water are due to different salt c

kap26 [50]

The density differences in the ocean water are due to different salt concentration and differences in temperature. B) temperature

6 0

3 years ago

Read 2 more answers

two bowling balls each have a mass of 8kg. if they are 2 m apart, what is the gravitational force between them?

Scilla [17]

Answer:

1 x 10^-9 N

Explanation:

F = Gm²/d² = 6.674e-11(8²)/2² = 1.06784e-9

4 0

3 years ago

Camera flashes charge a capacitor to high voltage by switching the current through an inductor on and off rapidly. In what time

Ber [7]

Answer:

The value is $\Delta t = 4.0 *10^{-7} \ s$

Explanation:

From the question we are told that

The current is $\Delta I = 0.100 \ A$

The inductor is $L = 2.0mH = 2.0*10^{-3} \ H$

The voltage induced is $\epsilon = 500 V$

Generally the emf induced is mathematically represented as

$\epsilon = L * \frac{\Delta I }{\Delta t }$

Here $\Delta t$ is the time taken

=> $\Delta t = \frac{L * \Delta I }{\epsilon }$

=> $\Delta t = \frac{2*10^{-3} * 0.100 }{500 }$

=> $\Delta t = 4.0 *10^{-7} \ s$

8 0

3 years ago

All else equal, the payback period for a project will decrease whenever the_______.

telo118 [61]

Answer:

(B) cash inflows are moved earlier in time.

Explanation:

The payback period stated time-frame during which the initial amount of investment should be recovered. It is expressed in the year form

The formula to compute the payback period is shown below:

Payback period = Initial investment ÷ Net cash flow

where,

The net cash flow = annual net operating income + depreciation expenses

The payback period of the project decreases when the accumulated starting year cash flows increases that results the movement of the cash inflows earlier in time

3 0

3 years ago

Read 2 more answers

A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force

Tamiku [17]

Answer:

(a) 91 kg (2 s.f.) (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

$s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}$

Subsequently,

$F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})$

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

$v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}$

According to Newton's First Law which states objects will remain at rest

or in uniform motion (moving at constant velocity) unless acted upon by

an external force. Hence, the block of ice by the end of the first 5

seconds, experiences no acceleration (a = 0) but travels with a constant

velocity of 4.4 $m \ s^{-1}$ .

$s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}$

Therefore, the ice block traveled 22 m in the next 5 seconds after the

worker stops pushing it.

4 0

3 years ago