Ask question

Ask question

All categories

3 years ago

15

Show by algebraic reasoning that your gravitational acceleration toward an object of mass M a distance d away is a = GM/d2 and t

herefore doesnÍt depend on your mass.

1 answer:

trasher [3.6K]3 years ago

8 0

Gravitational force is
GmM / r^2

Force is also m x a

When equating both of them, m cancels out, which leaves us with GM/r^2

Hope this helps

You might be interested in

What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T ma

Sindrei [870]

Answer:

E = 1.50 × $10^{8}$ V/m

Explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

$\mu _b = \frac{B}{2\mu _o}$ ...............1

and

energy density due to electric filed is

$\mu _e = \frac{\epsilon _o E^2}{2}$ ...............2

and here $\mu _b = \mu _ e$

so that

E = $\frac{B}{\sqrt{\mu _o \times \epsilon _o}}$ ...................3

put here value and we get

$E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}$

E = 3 × $10^{8}$ × 0.50

E = 1.50 × $10^{8}$ V/m

6 0

2 years ago

A car is accelerating at a rate of 35 m/s2 and has a resulting velocity increase from 17 m/s to 26 m/s. How long did it take to

harina [27]

Answer:The answer is (60 mph - 0 mph) / 8s = (26.8224 m/s - 0 m/s) / 8s = 3.3528 m/s 2 (meters per second squared) average acceleration. That would be 27,000 miles per hour squared.

Explanation:

6 0

2 years ago

The change in momentum experienced by a object is equivalent to the...

Papessa [141]

Answer: C (impulse acting on the object)

The momentum is defined as it is the impulse acting on the force . Change in momentum is known as Impulse. Impulse is used to increase or decrease the momentum of object.

From Newtons II law

F = m. a

= m. v/t <em>since a = rate of change of velocity.</em>

<em> </em>F . t = m . v

F . t is known as impulse momentum

8 0

2 years ago

To get employees to work longer hours, employers often offer __________ in the form of extra pay.

Dovator [93]

C- Incentives it’s like a reward ~!

6 0

2 years ago

Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their

geniusboy [140]

Answer:

a. 11 m/s at 76° with respect to the original direction of the lighter car.

Explanation:

In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

$m_{1} v_{x1}+m_{2}v_{x2}=(m_{1}+m_{2})v_{fx}\\m_{1} v_{y1}+m_{2}v_{y2} =(m_{1}+m_{2})v_{fy}$

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

$1*13 + 4*0 = (1+4) v_{fx}\\v_{fx}=\frac{13}{5} =2.6\\1*0 + 4*13 = (1+4) v_{fy}\\v_{fy}=\frac{13*4}{5} =10.4\\$

The magnitude of the final velocity of the wreck can be found as:

$v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72$

The final velocity has an intensity of roughly 11 m/s

As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

$\theta = cos^{-1}(\frac{v_{fx} }{v_{f}} )\\\theta = cos^{-1}(\frac{2.6}{10.7} )\\\theta = 76^{o}$

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.

5 0

3 years ago