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Deffense [45]
1 year ago
8

If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________

Physics
1 answer:
siniylev [52]1 year ago
8 0

If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be V_e = \sqrt{\dfrac{GM}{r}}.

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

V_e=\sqrt{\dfrac{2GM}{r}}

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

V_e=\sqrt{\dfrac{2GM}{r\times 2}}

V_e = \sqrt{\dfrac{GM}{r}}.

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be V_e = \sqrt{\dfrac{GM}{r}}.

To know more about escape velocity follow

brainly.com/question/14042253

#SPJ4

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Increase in sea water pollution
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3 years ago
A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the
insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}

Req =3.03Ω , R1 =12.18Ω

\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}

\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}

\frac{1}{R2}=0.2479

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

3 0
3 years ago
What is anything that has mass and volume?
Reika [66]
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3 0
3 years ago
As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.33 m l
deff fn [24]

Answer: 62 μT

Explanation:

Given

Length of rod, l = 1.33 m

Velocity of rod, v = 3.19 m/s

Induced emf, e = 0.263*10^-3 V

Using Faraday's law, the induced emf of a rod can be gotten by the formula

e = blv where,

e = induced emf of the rod

b = magnetic field of the rod

l = length of the rod

v = velocity of the rod. On substituting, we have

0.263*10^-3 = b * 1.33 * 3.19

0.263*10^-3 = b * 4.2427

b = 0.263*10^-3 / 4.2427

b = 0.0000620 T

b = 62 μT

Thus, the strength of the magnetic field is 62 μT

8 0
3 years ago
Read 2 more answers
A high jumper jumps over a bar that is 2 m above the mat. With what velocity does the jumper strike the mat in the landing area?
docker41 [41]

Answer:

The velocity with which the jumper strike the mat in the landing area is 6.26 m/s.

Explanation:

It is given that,

A high jumper jumps over a bar that is 2 m above the mat, h = 2 m

We need to find the velocity with which the jumper strike the mat in the landing area. It is a case of conservation of energy. let v is the velocity. it is given by :

v=\sqrt{2gh}

g is acceleration due to gravity

v=\sqrt{2\times 9.81\ m/s^2\times 2\ m}

v = 6.26 m/s

So, the velocity with which the jumper strike the mat in the landing area is 6.26 m/s. Hence, this is the required solution.

8 0
3 years ago
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