All categories

3 years ago

In the picture below, each object is positioned the same distance from the star. The mass of each object is given.

Physics

1 answer:

Lisa [10]3 years ago

4 0

The answer is 100 kg because the gravitational attraction is greater in objects that have more mass ;)

You might be interested in

neptune is an average distance of 4.5×10^12m from the sun. Estimate the length of the Neptunian year.

Vikentia [17]

As per Kepler's third law we know that

$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

now here we know that

$T_1$ = year of Neptune

$T_2$ = year of Earth

$R_1$ = distance of Neptune from Sun

$R_2$ = Distance of Earth from Sun

so now we will have

$\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}$

$T_1^2 = 27000$

$T_1 = 164.3 years$

so length of year of Neptune is 164.3 years

6 0

3 years ago

An atom has an atomic number 17 and atomic mass number 35. How many neutrons are present in its nucleus?

Anna [14]

The number of neutrons that are present in its nucleus is 18. The correct option among all the options that are given in the question is the second option or option "B". Mass number minus the atomic number gives the number of neutrons. I hope that this is the answer that has come to your help.

7 0

3 years ago

Read 2 more answers

A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origi

Liula [17]

Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

a) calculating work done by the force over the journey of the train

F = mx + b ------ ( 1 )

m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8 ----- ( 2 )

hence to determine the work done by the force

W = $\int\limits^7_0 { ( -0.373) x + 2.8 )} \, dx$ Note: the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

b) calculate the speed of the train at the end of its journey

we will apply the work energy theorem

W = 1/2 m*v^2 - 1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )

V^2 = 12.11

hence V = 3.48 m/s

6 0

3 years ago

Animal studies indicate that

notka56 [123]

Answer: mammals may be relatively better at solving problems than birds.

3 0

3 years ago

A 10kg sphere hits a stationary 8kg sphere. After the collision the 8kg sphere moves off in the positive direction at 4m/s. If t

Bumek [7]

M = mass of the first sphere = 10 kg

m = mass of the second sphere = 8 kg

V = initial velocity of the first sphere before collision = 10 m/s

v = initial velocity of the second sphere before collision = 0 m/s

V' = final velocity of the first sphere after collision = ?

v' = final velocity of the second sphere after collision = 4 m/s

using conservation of momentum

M V + m v = M V' + m v'

(10) (10) + (8) (0) = (10) V' + (8) (4)

100 = (10) V' + 32

(10) V' = 68

V' = 6.8 m/s

7 0

3 years ago