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sergiy2304 [10]
3 years ago
12

A rocket has landed on Planet X, which has half the radius of Earth. An astronaut onboard the rocket weighs twice as much on Pla

net X as on Earth. If the escape velocity for the rocket taking off from Earth is v0 , then its escape velocity on Planet X is:
Physics
1 answer:
Burka [1]3 years ago
7 0

Explanation:

Let acceleration due to Gravity for a planet is given by:

g_X=GM/R^2

Here,g_X = 2g

Escape velocity is given by:

v =\sqrt{ \frac{2GM} {R}} = \sqrt{2aR}

Here, R=R_earth/2

and g_X = 2g

Therefore,v=\sqrt(2(2g)(R/2))=v_0

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A proton travels at a velocity of 3.0x10^6 m/s in a direction perpendicular to a uniform magnetic field.
Virty [35]

Answer:

0.239 T

Explanation:

Applying,

F = Bvqsin∅................ Equation 1

Where F = magnetic force, B = magnetic Field, q = charge of a proton, v = velocity of proton, ∅ = angle between the velocity and the magnetic field.

make B the subject of the equation

B = F/(vqsin∅)................. Equation 2

From the question,

Given: F = 1.15×10⁻¹³ N, v = 3.0×10⁶ m/s, ∅ = 90°(perpendicular)

Constant: q = 1.602 x 10⁻¹⁹ C

Substitute into equation 2

B =  1.15×10⁻¹³ /(3.0×10⁶×1.602 x 10⁻¹⁹×sin90°)

B = 1.15×10⁻¹³/(4.806×10⁻¹³)

B = 0.239 T.

Hence the magnetic field = 0.239 T

7 0
3 years ago
I will give 14 points & make you the brainiest
Virty [35]

Answer:

B. silicate rocks and metals

8 0
3 years ago
A woman is 1.6 m tall and has a mass of 50 kg. She moves past an observer with the direction of the motion parallel to her heigh
eduard

To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.

By definition we know that the relative velocity of an object with reference to the Light, is defined by

V_0 = \frac{V}{\sqrt{1-\frac{V^2}{c^2}}}

Where,

V = Speed from relative point

c = Speed of light

On the other hand we have that the linear momentum is defined as

P = mv

Replacing the relative velocity equation here we have to

P = \frac{mV}{\sqrt{1-\frac{V^2}{c^2}}}

P^2 = \frac{m^2V^2}{1-\frac{V^2}{c^2}}

P^2 = \frac{P^2V^2}{c^2}+m^2V^2

P^2 = V^2 (\frac{P^2}{c^2}+m^2)

V^2 = \frac{P^2}{\frac{P^2}{c^2}+m^2}

V^2 = \frac{(2.1*10^10)^2}{\frac{(2.1*10^10)^2}{(3.8*10^8)^2}+50^2}

V = 2.81784*10^8m/s

Therefore the height with respect the observer is

l = l_0*\sqrt{1-\frac{V^2}{c^2}}

l = 1.6*\sqrt{1-\frac{(2.81*10^8)^2}{(3*10^8)^2}}

l = 0.56m

Therefore the height which the observerd measure for her is 0.56m

8 0
3 years ago
A 4-kg ball has a momentum of 12 kg∙m/s. what is the ball's speed?
Karolina [17]
M= 4kg
v= ?
momentum = Mv
so
12kgms-¹ = 4kg×v
v= 12÷4 = 3ms-¹
7 0
3 years ago
Read 2 more answers
The core of the Sun is Group of answer choices
kap26 [50]

Answer:

2. much hotter and much denser than its surface

Explanation:

We know that the temperature around the center of the Sun is about 1.57×10⁷ K and its density is about 162 g/cm³.

Now, the temperature and the density decrease as one moves outward from the center of the Sun, the temperature at the surface of the sun is about 5×10³ K and the density as an average in the surface is about 1.4 g/cm³.

Therefore the answer is:

2. much hotter and much denser than its surface.  

I hope it helps you!

Have a nice day!

3 0
4 years ago
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