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3 years ago

I really don’t know what it is to me they r all the same

Physics

2 answers:

Inessa05 [86]3 years ago

4 0

The moon lacks an atmosphere compared to the Earth. Hope this helps!

aleksandr82 [10.1K]3 years ago

3 0

The Moon IS solid, it HAS gravity, and it gets loads of sunlight,
just like the Earth.

But the Earth has air all around it, miles deep.
The Moon has NO air. (Choice-C)

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A second order measuring system has a known natural frequency of 2000 Hz and damping ratio of 0.8. Would it be satisfactory to u

Anna007 [38]

Answer:

The expected dynamic error is 0.019

The phase shift is -23.10°C

Explanation:

The explanation is shown on the first uploaded image

7 0

3 years ago

Which of the following is an example of work being done on an object?

tensa zangetsu [6.8K]

Answer:

d. All of these

Explanation:

work is said to be done when a force is applied to an object through a certain distance. the SI unit of workdone is joules or newton per meter

mathematically

workdone = force x distance.

from the answers, work is being done because there is force applied in a certain distance.

from wagon is used to carry vegetables from a garden.
pulley is used to get water from a well.
hammer is used to remove a nail from a wall.

4 0

3 years ago

Read 2 more answers

How long does it take to change a flat plain into mountains? help

Len [333]

Im not for sure but i think it takes a couple hundred years (or according to the climate)

5 0

3 years ago

Read 2 more answers

Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum

Alina [70]

Answer:

Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
Uncertainty in momentum (∆P) = ?
Planck's constant (h) = 6.26 × 10⁻³⁴ Js

$\longrightarrow \: \: \sf\Delta x .\Delta p = \dfrac{h}{4\pi}$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} {4 \times \frac{22}{7} }$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} { \frac{88}{7} }$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34} \times 7} { 8 }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 8 \times 24 \times {10}^{ - 15} }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 192 \times {10}^{ - 15} }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} \times {10}^{15} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ -19} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 2} \times {10}^{ -19} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 21} } { 192}$

$\longrightarrow \: \: \sf\Delta p = 22.822\times {10}^{ - 21}$

$\longrightarrow \: \: \sf\Delta p = 2.2822 \times {10}^{1} \times {10}^{ - 21}$

$\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 \times {10}^{ - 20} \: kg/ms}}}$

4 0

2 years ago

A -1.12 μC charge is placed at the center of a conducting spherical shell, and a total charge of +8.65 μC is placed on the shell

Lisa [10]

Answer: 7.53 μC

Explanation: In order to explain this problem we have to use the gaussian law so we have:

∫E.dS=Qinside/εo we consider a gaussian surface inside the conducting spherical shell so E=0

Q inside= 0 = q+ Qinner surface=0

Q inner surface= 1.12μC so in the outer surface the charge is (8.65-1.12)μC=7.53μC

7 0

3 years ago