Ask question

Ask question

All categories

2 years ago

7

Calculate the magnitude of the acceleration of Io due to the gravitational force exerted by Jupiter. [Show all work, including t

he equation and substitution with units.]
ag = Fg
m

1 answer:

galina1969 [7]2 years ago

4 0

The magnitude of the acceleration due to the gravitational force exerted by Jupiter is 25.91 m/s².

<h3>What is acceleration due to gravity?</h3>

This is the acceleration of a body under a free fall due to influence of gravity.

The magnitude of the acceleration due to the gravitational force exerted by Jupiter is calculated as follows;

$F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}$

where;

M is mass of Jupiter
R is the radius of Jupiter
G is universal constant

$g = \frac{6.67 \times 10^{-11} \times 1.8986 \times 10^{27}}{(69,911 \times 10^3)^2} \\\\g = 25.91 \ m/s^2$

Thus, the magnitude of the acceleration due to the gravitational force exerted by Jupiter is 25.91 m/s².

Learn more about acceleration due to gravity here: brainly.com/question/88039

You might be interested in

If Phoenix, Arizona, experiences a cool, wet day in June, does that mean the regions climate is changing?

jenyasd209 [6]

It doesn't mean that the climate is changing, it is probably just the morning dew.

5 0

2 years ago

a ultrasonic wave at 8x10^4 Hz is emitted into a vien where the speed of sound in blood is 1570 m/s. the wave reflects off the r

Aneli [31]

Answer: 0.392 m/s

Explanation:

The Doppler shift equation is:

$f'=\frac{V+V_{o}}{V-V_{s}} f$

Where:

$f=8(10)^{4} Hz$ is the actual frequency of the sound wave

$f'=8.002(10)^{4} Hz$ is the "observed" frequency

$V=1570 m/s$ is the speed of sound

$V_{o}=0 m/s$ is the velocity of the observer, which is stationary

$V_{s}$ is the velocity of the source, which are the red blood cells

Isolating $V_{s}$ :

$V_{s}=\frac{V(f'-f)}{f'}$

$V_{s}=\frac{1570 m/s(8.002(10)^{4} Hz-8(10)^{4} Hz)}{8.002(10)^{4} Hz}$

Finally:

$V_{s}=0.392 m/s$

3 0

2 years ago

A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant

vovangra [49]

Answer:

emf induced in the loop, at the instant when 9.0s have passed = 1.576 * 10 ⁻² V.

Direction is counter clockwise.

Explanation:

See attached pictures.

6 0

3 years ago

PLEASE HELP ME PLS PLS Imagine an object held at some height above the ground. It is released and falls toward the ground. Ignor

Stels [109]

Option e is true. The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

<h3>What is the law of conservation of energy?</h3>

According to the Law of conservation of energy. Although energy cannot be generated or destroyed, it may be transferred from one form to another.

The following statements are true;

a)All of its energy must be potential energy before it falls.

b)At the conclusion of its fall, all of its energy must be transformed to kinetic energy.

c)During its fall, the sum of its kinetic and potential energy must match the initial quantity of potential energy.

d)Total energy = Kinetic Energy + Potential Energy.

Hence, option e is correct.

To learn more about the law of conservation of energy, refer ;

brainly.com/question/2137260

#SPJ2

8 0

1 year ago

Read 2 more answers

At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?

sergey [27]

1) Initial upward acceleration: $6.0 m/s^2$

2) Mass of burned fuel: $0.10\cdot 10^4 kg$

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude $F_g = mg$ , in the downward direction, where

$m=1.9\cdot 10^4 kg$ is the rocket's mass

$g=9.8 m/s^2$ is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

$T=3.0\cdot 10^5 N$

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

$T-mg=ma$ (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

$a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2$

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

$a'=6.9 m/s^2$

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

$T-m'g=m'a'$

where

$m'$ is the new mass of the rocket

Re-arranging the equation and solving for m', we find

$m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg$

And since the initial mass of the rocket was

$m=1.9 \cdot 10^4 kg$

This means that the mass of fuel burned is

$\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg$

3 0

3 years ago