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Paul [167]
3 years ago
13

An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.

Physics
1 answer:
Georgia [21]3 years ago
7 0

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

The \ weight, W  =G\dfrac{M \times m}{R^{2}} = 800 \ N

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

W_1   =G\dfrac{\dfrac{M}{2}  \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2}  \times m \times 4}{ R ^{2}} = 2 \times G \times  \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

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Alana is skateboarding at 19 km/h and throws a tennis ball at 11 km/h to her friend Oliver who is behind her leaning against a w
NNADVOKAT [17]

The speed of the tennis ball is 8 km/h when Alana throws the ball behind at the speed of 11 km/h.

<h3>What is the frame of reference?</h3>

It is a point from where the motion is observed. The speed changes with the frame of reference.

For the person standstill on the sidewalk, If  Alana throw the ball in the front direction, The speed will be,

19+11 = 30 km/h

It Alana just leave the ball, the speed,

19 +0 = 19 km/h

If Alana throws the ball behind her at the speed of 19 km/h. The speed of the ball for a sidewalk observer,

19 +(-19 ) =  0 km/h

Thus, when Aana throw the ball behind the velocity will be negative,

So,

19 + (-11 ) = 8 km/h

Therefore, the speed of the tennis ball is 8 km/h when Alana throws the ball behind at the speed of 11 km/h.

Learn more about Frame of reference:

brainly.com/question/9820962

8 0
2 years ago
________ law is called the law of partial pressure
Studentka2010 [4]
Dalton's <span>law of partial pressures
</span>
6 0
3 years ago
Currents circulate in a piece of metal that is pulled through a magnetic field. What are these currents called?
MariettaO [177]
<h3><u>Answer;</u></h3>

Eddy currents

<h3><u>Explanation</u>;</h3>
  • <em><u>Eddy currents are currents which circulate in conductors like swirling eddies in a stream. These currents are induced by changing magnetic fields and flow in closed loops, perpendicular to the plane of the magnetic field. </u></em>
  • They are created as a result of change in the intensity or direction of a magnetic field on a conductor, this occurs either when a conductor is moving through a magnetic field, or when the magnetic field surrounding a stationary conductor is varying
4 0
3 years ago
Two point charges of +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the intensity of electric field E mid
algol13

Answer:

The intensity of the net electric field will:

E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C

Explanation:

Here we need first find the electric field due to the first charge at the midway point.

The electric field equation is given by:

|E_{1}|=k\frac{q_{1}}{d^{2}}

Where:

  • k is Coulomb's constant
  • q(1) is 20.00 μC or 20*10⁻⁶ C
  • d is the distance from q1 to the midpoint (d=10.0 cm)

So, we will have:

|E_{1}|=(9*10^{9})\frac{20*10^{-6}}{0.1^{2}}

|E_{1}|=1.8*10^{7}\: N/C

The direction of E1 is to the right of the midpoint.

Now, the second electric field is:

|E_{2}|=k\frac{q_{2}}{d^{2}}

|E_{2}|=(9*10^{9})\frac{8*10^{-6}}{0.1^{2}}

|E_{2}|=7.2*10^{6}\: N/C

The direction of E2 is to the right of the midpoint because the second charge is negative.

Finally, the intensity of the net electric field will:

E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C

I hope it helps you!

5 0
3 years ago
You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought g
TiliK225 [7]

Answer:

Only 9% weaker

Explanation:

Because this is where most stuff that people do in space takes place. So, um, here we're at a radius of the earth plus 300 kilometers. You may already be seeing why this isn't going to have much effect if this were except the 6.68 times, 10 to the sixth meters. And so the value of Gout here. You know, Newton's gravitational constant times, the mass of the Earth divided by R squared for the location we're looking at. And so this works out to be 8.924 meters per second squared, which is certainly less than it is at the surface of the earth. However, this is only 9% less than acceleration for gravity at the surface. So the decrease in the gravity gravitational acceleration of nine percent not really going toe produces a sensation of weightlessness.

4 0
3 years ago
Read 2 more answers
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