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4 years ago

PLS HELP! What is the gravitational field strength on the moon?

Physics

1 answer:

astra-53 [7]4 years ago

5 0

The gravitational field strength on the moon is $\frac{1}{6}$ of the gravitational field strength on the earth

On the earth:

the gravitational field strength (g) is determined as:

$g=G\frac{m}{r^2}$

where:

G: the gravitational constant
m: the mass of the earth
r: the radius of the earth

$g=9.8 N/kg = 9.8 m/s^2$

Then on the moon:

$g=[tex]\frac{1}{6}$ *9.8=1.6 N/kg[/tex]

The answer will be : A.1.6 N/kg

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A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040

Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force ( $F_{B}$ ), as follows:

$F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB$ (1)

Where:

m: is the proton's mass = 1.67*10⁻²⁷ kg

v: is the proton's velocity

r: is the radius of the proton's orbit

q: is the proton charge = 1.6*10⁻¹⁹ C

B: is the magnetic field = 0.040 T

Solving equation (1) for r, we have:

$r = \frac{mv}{qB}$ (2)

By conservation of energy, we can find the velocity of the proton:

$K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V$ (3)

Where:

K: is kinetic energy

U: is electrostatic potential energy

ΔV: is the potential difference = 1.0 kV

Solving equation (3) for v, we have:

$v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s$

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

$r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m$

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!

5 0

3 years ago

Imagine a simple pendulum swinging in an elevator. If the cable holding the elevator up was to snap, allowing the elevator to go

Lady_Fox [76]

One of the components that affect the period is gravity (the other is length). This gravity is basically the value of the effective acceleration that acts on the body due to gravity. When the elevator is over free fall, the effective gravity becomes zero. Mathematically this can be visualized as,

$T = 2\pi \sqrt{\frac{l}{\dot{g_{eff}}}}$

Since this value is zero, the period would tend to be infinite,

$T \rightarrow \infty$

Therefore the frequency that is inversely proportional to the period would be defined as

$f = \frac{1}{T}$

$f = \frac{1}{\infty}$

$f \approx 0$

In this way there is no frequency on the body which will not generate any oscillation on the body

5 0

3 years ago

Label the diagram Please

Nitella [24]

The earth tilt about 23 degrees which leads to the formation of seasons on earth.

<h3>At what degree the earth tilt?</h3>

This is tilted at 23 degrees from the vertical. The northern hemisphere is in the winter season.

This is the path as the earth revolves around the sun. It takes 1 year to complete this path.

This is the path as the earth rotates around its.

So we can conclude that the earth tilt about 23 degrees which leads to the formation of seasons on earth.

Learn more about season here: brainly.com/question/25870256

#SPJ1

8 0

2 years ago

What is the use of scientific knowledge and research for an action

koban [17]

Experience research study. experiment explain scientific process useful for research for an action . Albert enistein . He was found all the scientific knowledge. more scientist same method

4 0

3 years ago

When developing a question for a scientific inquiry, the question will ideally

Lisa [10]

When developing a question for a scientific inquiry, the question will ideally :

- be measurable
- Will have various variables that could be used for further testing
- not just simply ask why

hope this helps

7 0

3 years ago

Read 2 more answers