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sukhopar [10]
3 years ago
5

Which of the following provides evidence that Earth's Moon is rotating and revolving at the same rate?

Physics
1 answer:
kkurt [141]3 years ago
6 0

B. We can see only one side of the Moon from Earth.


( we only see one side of the moon because the moon rotates around the Earth)

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A baseball is dropped off the side of a cliff. It free-falls to the ground in 8 seconds. During the third second, the ball is tr
kozerog [31]
The acceleration of gravity is 9.8 m/s² .  That means that a falling object
is always falling 9.8 m/s faster than it was falling 1 second earlier.

If an object is not slowed by air resistance, and has far enough to go
so that it's still falling after three whole seconds, then at the end of
three seconds it's falling at

           (9.8 m/s²) x (3 sec)  =  29.4 m/s
3 0
2 years ago
Read 2 more answers
What will happen when the sun blows up?
RSB [31]
We will all chaotically burn to death!



Hope this helps!!

8 0
3 years ago
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A vertical spring (ignore its mass), whose spring constant is 1070 N/m, is attached to a table and is compressed 0.100 m.
harina [27]

I can not solve the problem if I do not have the mass.

3 0
3 years ago
You stand on a merry-go-round which is spinning at f = 0:25 revolutions per second. You are R = 200 cm from the center. (a) Find
ivanzaharov [21]

Answer:

(a) ω = 1.57 rad/s

(b) ac = 4.92 m/s²

(c) μs = 0.5

Explanation:

(a)

The angular speed of the merry go-round can be found as follows:

ω = 2πf

where,

ω = angular speed = ?

f = frequency = 0.25 rev/s

Therefore,

ω = (2π)(0.25 rev/s)

<u>ω = 1.57 rad/s </u>

(b)

The centripetal acceleration can be found as:

ac = v²/R

but,

v = Rω

Therefore,

ac = (Rω)²/R

ac = Rω²

therefore,

ac = (2 m)(1.57 rad/s)²

<u>ac = 4.92 m/s² </u>

(c)

In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:

Centripetal Force = Frictional Force

m*ac = μs*R = μs*W

m*ac = μs*mg

ac = μs*g

μs = ac/g

μs = (4.92 m/s²)/(9.8 m/s²)

<u>μs = 0.5</u>

7 0
2 years ago
Using 0.500 g of nichrome, you are asked to fabricate a wire with uniform cross-section. The resistance of the wire is 0.673 Ω.
mojhsa [17]

Explanation:

Given that,

Mass of Nichrome, m = 0.5 g

The resistance of the wire, R = 0.673 ohms

Resistivity of the nichrome wire, \rho=10^{-6}\ \Omega -m

Density, d=8.31\times 10^3\ kg/m^3

(A) The length of the wire is given by using the definition of resistance as :

Volume,

V=A\times l\\\\A=\dfrac{V}{l}\\\\Since, V=\dfrac{m}{d}\\\\V=\dfrac{m}{d}\\\\V=\dfrac{0.5\times 10^{-3}}{8.31\times 10^3}\\\\V=6.01\times 10^{-8}\ m^3

Area,

A=\dfrac{V}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{l}....(1)

R=\rho \dfrac{l}{A}\\\\l=\dfrac{RA}{\rho}\\\\l=\dfrac{0.673\times 6.01\times 10^{-8}}{l\times 10^{-6}}\\\\l=0.201\ m

(b)  Equation (1) becomes :

A=\dfrac{6.01\times 10^{-8}}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{0.201}\\\\\pi r^2=3\times 10^{-7}\\\\r=\sqrt{\dfrac{3\times 10^{-7}}{\pi}} \\\\r=3.09\times 10^{-4}\ m

Hence, this is the required solution.                                                                  

5 0
3 years ago
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