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Daniel [21]
3 years ago
15

on the surface of jupiter the acceleration due to gravity is about 3 times that on earth. how much would a 100kg rock weigh on j

upiter?
Physics
2 answers:
snow_lady [41]3 years ago
6 0

Answer:

2940 N

Explanation:

Given at Jupiter acceleration due to gravity(g_j) is equal to 3 times of acceleration due to gravity on the Earth(g_e).

That is,

 g_j=3g_e

We know, g_e=9.8m/s^2

So,  g_j=3\times9.8=29.4m/s^2

Now the gravitational force on Jupiter (weight on Jupiter) is given as

W = mg_j

Here m = 100 kg is the mass and it remain same on both earth and Jupiter.

Substitute the given values, we get

W=100 kg\times 29.4m/s^2

W = 2940 N.

Thus, the weight of 100 kg rock on the Jupiter is 2940 N.

madam [21]3 years ago
4 0

Answer:

Weight on Jupiter will be equal to 2940 N

Explanation:

We have given given acceleration due to gravity on Jupiter is 3 times of acceleration due to gravity on earth

Acceleration due to gravity on earth g=9.8m/sec^2

So acceleration due to gravity on Jupiter = g'=3\times 9.8=29.4m/sec^2

Mass is given m = 100 kg

We have to find the weight

Weight is equal to W = mg, here m is mass and a is acceleration

So weight W=100\times 29.4=2940N

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A​ heavy-duty shock absorber is compressed 4 cm from its equilibrium position by a mass of 700nbspkg. How much work is required
tia_tia [17]

Answer:

Explanation:

A mass of 700 kg will exert a force of

700 x 9.8

= 6860 N.

Amount of compression x = 4 cm

= 4 x 10⁻² m

Force constant K = force of compression / compression

= 6860 / 4 x 10⁻²

= 1715 x 10² Nm⁻¹.

Let us take compression of r at any moment

Restoring force by spring

= k r

Force required to compress = kr

Let it is compressed  by small length dr during which force will remain constant.

Work done

dW =  Force x displacement

= -kr -dr

= kr dr

Work done to compress by length d

for it r ranges from 0 to -d

Integrating on both sides

W  = \int\limits^{-4}_0 {kr} \, dr

= [ kr²/2]₀^-4

= 1/2 kX16X10⁻⁴

= .5 x 1715 x 10² x 16 x 10⁻⁴

= 137.20 J

3 0
3 years ago
Which is true about Pluto
andrew-mc [135]

Answer:

D.)it orbits near the Kepler belt

Explanation:

The Kuiper belt is an area similar to the asteroide belt extending from the orbit of Neptune to about 50 AU from the Sun. It mainly consists of icy asteroids and dwarf planets, which are rocky objects big enough to be defined as planet but that do not have enough gravity to clear their orbit from other obejcts.

Pluto was discovered in 1930 - initially it was classified as a planet, although it is much smaller than the other 8 planets of the Solar System. However, it has been recently de-classified to dwarf planet because its gravity is not enough to clear its orbit from other objects (asteroids). Pluto is located inside the Kuiper belt, so option D is correct. Other dwarf planets in the Kuiper belt are for instance Haumea and Makemake.

8 0
3 years ago
Read 2 more answers
A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the c
icang [17]

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

W = F x = \mu N x = \mu\ m\ g\ x

W = 150 \times 6

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

\mu = \frac {F}{m\timesg}

= \frac{150}{40\times 9.8}

= .383

We simply applied the above formulas so that each one part could calculate

3 0
3 years ago
A simple pendulum is 1.15 meters long and has a period of 6.29 seconds. The pendulum is on an unknown planet. What is the accele
Nastasia [14]

Answer:

The value of the acceleration of gravity of the Unknown Planet = 1.14 \frac{m}{s^{2} }

Explanation:

length of the pendulum (L)= 1.15 m

Time period (T)= 6.29 seconds

We know that time period  of a simple pendulum is given by

⇒ T = 2\pi × \sqrt\frac{L}{g}

put the values in the above formula we get

⇒ T = 2\pi × \sqrt \frac{1.15}{g}

⇒ 6.29 = 2\pi × \sqrt \frac{1.15}{g}

By solving the above equation we get

⇒ g = 1.14 \frac{m}{s^{2} }

This is the value of the acceleration of gravity of the Unknown Planet.

7 0
4 years ago
During which stage of prophase i does crossing over take place?
Rom4ik [11]
Do you mean mitosis? If so metaphase. 
3 0
3 years ago
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