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3 years ago

What do we call a group of objects that orbit a star?

Physics

1 answer:

Reika [66]3 years ago

8 0

A group of objects that orbit a star is called a planet.

A small sized object that enlightens in the night sky in the infinite universe.

The group of objects is a large sized object which orbits a star called Sun.

This group of object is called planets.

Thus, a group of objects that orbit a star is called a planet.

Learn more about star.

brainly.com/question/18930464

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You might be interested in

A 2.0 kg mass weighs 10 Newtons on planet X. what is the acceleration due to gravity on planet X? Show the work.

KengaRu [80]

Answer: The gravitational acceleration on planet X is 5 N/kg

On Earth (with the gravitational accelartion g_E) the mass of 2kg will correspond to

$F_E = m\cdot g_E = 2.0 \mbox{kg}\cdot 9.8 \frac{N}{kg} = 19.6 N$

On planet X we are told the same measure is only 10N. Since there is a proportional relationship between g and F, we can calculate g_X:

$\frac{F_E}{g_E}=\frac{F_X}{g_X} \implies\\g_x = \frac{g_E\cdot F_X}{F_E}=\frac{9.8 N/kg \cdot 10 N }{19.6N}=5 \frac{N}{kg}$

8 0

3 years ago

An arrow of mass 20 g is shot horizontally into a bale of hay, striking the hay with a velocity of 60 m/s. It penetrates a depth

goblinko [34]

From the question, The kinetic energy of the fired arrow is equal to the work done by the bale of hale in stopping the arrow.

We make use of the following formula

mv²/2 = F'd................... Equation 1

Where

m = mass of the arrow
v = velocity of the arrow
F' = average stopping force acting on the arrow
d = distance of penetration

Make F' the subject of the equation

F' = mv²/2d.................. Equation 2

From the question,

Given:

m = 20 g = 0.02 kg
v = 60 m/s
d = 40 cm = 0.4 m

Substitute these values into equation 2

F' = 0.02(60²)/(0.4×2)

F' = 72/0.8

F' = 90 N

Hence, The average stopping force acting on the arrow is 90 N

Learn more about average stooping force here: brainly.com/question/13370981

5 0

2 years ago

A car collides into a concrete wall going 25.0 m/s . It stops in 0.141 seconds and has a change in momentum of 39,400. What is t

vodka [1.7K]

Answer:

Mass of the car is 1576 kg.

Explanation:

Let the mass of the car be $m$ kg.

Given:

Initial velocity of the car is, $u=25\ m/s$

As the car stops, final velocity of the car is, $v=0\ m/s$

Change in momentum is, $\Delta p=39400$

Now, we know that, momentum is given as the product of mass and velocity.

So, change in momentum is given as:

$\Delta p=m(u-v)\\39400=m(25-0)\\39400=25m\\m=\frac{39400}{25}\\m=1576\ kg$

Therefore, the mass of the car is 1576 kg.

4 0

3 years ago

Carlos pushes a 3 kg box with a force of 9 newtons. The force of friction on the box is 3 newtons in the opposite direction. Wha

Ivenika [448]

The 'net' force acting on the box is (9 - 3) = 6 newtons
in the direction that Carlos is pushing.

Force = (mass) x (acceleration)

6 = (3) x (acceleration)

Divide each side by 3 :

<em>2 m/s² = acceleration</em>

4 0

4 years ago

Read 2 more answers

A thin walled spherical shell is rolling on a surface. What

ExtremeBDS [4]

Answer:

$=\frac{1/3}{5/6} = 0.4$

Explanation:

Moment of inertia of given shell $= \frac{2}{3} MR^2$

where

M represent sphere mass

R -sphere radius

we know linear speed is given as $v = r\omega$

translational $K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2$

rotational $K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{2}{3} MR^2 \omega^2$

total kinetic energy will be

$K.E = \frac{1}{2} m(r\omega)^2 + \frac{1}{2} \frac{2}{3} MR^2 \omega^2$

$K.E =\frac{5}{6} MR^2 \omega^2$

fraction of rotaional to total K.E

$=\frac{1/3}{5/6} = 0.4$

8 0

3 years ago