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MatroZZZ [7]
3 years ago
9

Jupiter has a mass about 300x that of Earth, and its radius is about 11x that of Earth. What would be the approximate weight of

a 1 kg rock on the surface of Jupiter?
Physics
1 answer:
erica [24]3 years ago
4 0

Answer:

W = 24.01 N

Explanation:

given,

mass of Jupiter = 300 x

radius = 11 x

weight of 1 Kg on Jupiter = ?

using equation of acceleration of gravity

 g =\dfrac{GM}{r^2}

mass of Jupiter = 300 M

radius of the Jupiter = 11 R

now, acceleration due to gravity

 g' =\dfrac{G(300M)}{(11R)^2}

 g' =\dfrac{300}{121}\dfrac{GM}{R^2}

 g' =2.45 g

acceleration due to gravity of Jupiter

 g' = 24.01 m/s²

Weight of the rock  in Jupiter

W = m g'

W = 1 x 24.01

W = 24.01 N

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A negatively charged rod is placed near two neutral metal spheres, as shown below. Which statements describe the charging method
grin007 [14]
<h3>Answer;</h3>
  • <em>The spheres develop opposite charges. </em>
  • <em>Electrons move from Sphere A to Sphere B. </em>
  • <em>The spheres are charged through induction.</em>
<h3><u>Explanation;</u></h3>
  • <u><em>When a negatively charged rod is placed near two neutral metal spheres, the spheres will develop opposite charges, because the neutral metal spheres have both negative and positive charges. </em></u>From the basic law of electrostatics unlike charges attracts and like charges repel.
  • Thus, <em><u>the sphere will develop opposite charges, electrons will move from Sphere A to sphere B,</u></em> hence we say that the spheres will be charged by induction such that sphere A will acquire a positive charge while sphere B will acquire negative charge.
3 0
3 years ago
Read 2 more answers
If a rock were dropped from a building it would hit the ground with a certain kinetic energy and velocity. If it fell from a roo
STALIN [3.7K]

Answer:

no

Explanation:

because if you test it they hit the ground at the same time.

5 0
3 years ago
A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If
Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s

The induced emf in the shorter coil is calculated as;

E = NA\frac{\delta B}{\delta t}

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

E = NA\frac{\delta B}{\delta t}

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0
3 years ago
A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff
Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

4 0
2 years ago
I will mark you brainlist please help!
Studentka2010 [4]

Answer:

C.) The Distance DH = 1.5 lambda

Explanation:

This statement C.) is false, because it does not count as the 1.5 wavelength, it is less than 1 wavelength.

7 0
2 years ago
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