What force is needed to accelerate a 4.2 kg mass with an acceleration of 9.3 m/s/s?

The moon is 3x10^5 km away from Nepal and the mass of the moon is 7x10^22 kg. Calculate the force with which the Moon pulls ever

hammer [34]

Answer:

Approximately $5.19 \times 10^{-5}\; \rm N$ .

Explanation:

Let $G$ denote the gravitational constant. ( $G \approx 6.67 \times 10^{-11} \; \rm N \cdot kg^{-2} \cdot m^{2}$ .)

Let $M$ and $m$ denote the mass of two objects separated by $r$ .

By Newton's Law of Universal Gravitation, the gravitational attraction between these two objects would measure:

$\displaystyle F = \frac{G \cdot M \cdot m}{r^{2}}$ .

In this question: $M = 7 \times 10^{22}\; \rm kg$ is the mass of the moon, while $m = 1\; \rm kg$ is the mass of the water. The two are $r = 3\times 10^{5}\; \rm km$ apart from one another.

Important: convert the unit of $r$ to standard units (meters, not kilometers) to reflect the unit of the gravitational constant $G$ .

$\displaystyle r = 3 \times 10^{5}\; \rm km \times \frac{10^{3}\; \rm m}{1\; \rm km} = 3 \times 10^{8}\; \rm m$ .

$\begin{aligned} F &= \frac{G \cdot M \cdot m}{r^{2}} \\ &= \frac{6.67 \times 10^{-11}\; \rm N \cdot kg^{-2}\cdot m^{2} \times 7 \times 10^{22}\; \rm kg \times 1\; \rm kg}{(3 \times 10^{8}\; \rm m)^{2}} \\ &\approx 5.19 \times 10^{-5} \; \rm N\end{aligned}$ .

5 0

3 years ago

A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take t

marin [14]

Answer:

$T=66262.4s$

Explanation:

From the question we are told that:

Altitude $A=2.90 *10^7$

Mass $m=5.97 * 10^{24} kg$

Radius $r=6.38 *10^6 m.$

Generally the equation for Satellite Speed is mathematically given by

$V=(\frac{GM}{d} )^{0.5}$

$V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}$

$V=3354.83m/s$

Therefore

Period T is Given as

$T=\frac{2 \pi *a}{V}$

$T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}$

$T=66262.4s$

4 0

3 years ago

What is this ???????????

babymother [125]

A, something to remember is that if the numbers are even the answer too will be even. Every time hope this helps :>

8 0

3 years ago

Read 2 more answers

What is the order of magnitude of the gravitational force between two 1.0 kilogram charges that are positioned 1.0 meter apart?

stiks02 [169]

The answer is -11

Enjoy!

6 0

3 years ago

A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca

igomit [66]

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

$f=mg\ sin\theta$

$f=1100\ kg\times 9.8\ m/s^2\ sin(4)$

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

3 0

3 years ago