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3 years ago

All matter in the Universe consists of many substances called elements. true or faluse

Physics

2 answers:

Mrrafil [7]3 years ago

8 0

The answer to this statement is true!

mr Goodwill [35]3 years ago

6 0

Ma'am your answer is true i do believe

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A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.

Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately $\displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}$ .

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately $\displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s$ .

c) The orbital radius required would be approximately $\rm 2.04 \times 10^7\; m$ .

d) The escape velocity from the surface of that planet would be approximately $\rm 5.01\times 10^3\; m \cdot s^{-1}$ .

Explanation:

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ ,

where

$G$ is the constant of universal gravitation.
$M$ is the mass of the first mass. (In this case, let $M$ be the mass of the planet.)
$m$ is the mass of the second mass. (In this case, let $m$ be the mass of the satellite.)
$r$ is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

$\displaystyle \frac{m \cdot v^{2}}{r}$ ,

where

$m$ is the mass of the object (in this case, that's the mass of the satellite.)
$v$ is the orbital speed of the satellite.
$r$ is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for $v$ :

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

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

$\displaystyle v = \sqrt{\frac{G \cdot M}{r}}$ .

Take $G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2}$ , Simplify the expression $v$ :

$\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}$ .

Since the orbit is a circle of radius $R$ , the distance traveled in one period would be equal to the circumference of that circle, $2 \pi R$ .

Divide distance with speed to find the time required.

$\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx 9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}$ .

Convert $24.6\; \rm \text{hours}$ to seconds:

$24.6 \times 3600 = 88560\; \rm s$

Solve the equation for $R$ :

$9.62 \times 10^{-7}\, R^{3/2}= 88560$ .

$R \approx 2.04 \times 10^7\; \rm m$ .

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

$\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r}$ (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

$\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}$ .

Solve for $v$ . The value of $m$ shouldn't matter, for it would be eliminated from both sides of the equation.

$\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0$ .

$\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}$ .

5 0

3 years ago

How much energy (in Joules) is released when 12.0 g of water cools from 20.0 °C to 11.0 °C? This is a grade 10 question from the

KATRIN_1 [288]

Answer: - 452.088joule

Explanation:

Given the following :

Mass of water = 12g

Change in temperature(Dt) = (11 - 20)°C = - 9°C

Specific heats capacity of water(c) = 4.186j/g°C

Q = mcDt

Where Q = quantity of heat

Q = 12g × 4.186j/g°C × - 9°C

Q = - 452.088joule

7 0

3 years ago

A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho

creativ13 [48]

Answer:

The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.

Explanation:

Given:

Mass of the piano $(m)$ = 190 kg

Inclined angle $(\theta)$ = 18 degree

Considering gravity, $g$ = 9.8 $ms^-^2$

And

Using, $sin(18) =0.30$ and $cos(18)=0.95$

<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

When the man pushes it parallel to the incline.

Balancing the forces as $\sum F=0$ .

⇒ $F+mgsin(\theta) =0$

⇒ $F=-mgsin(\theta)$

⇒ Here it is negative as the force is acting downward.

⇒ Plugging the values of mass $(m)$ and angle $(\theta)$ .

⇒ $F=190\times 9.8\times sin(18)$

⇒ $F=575.38$ N

When the force is parallel to the floor.

⇒ $Fcos(\theta)=mgsin(\theta)$

⇒ $F=\frac{mgsin(\theta)}{cos(\theta)}$

⇒ Plugging the values.

⇒ $F=\frac{190\times 9.8\times sin(18)}{cos(18)}$

⇒ $F=605$ N

So,

The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

6 0

3 years ago

A 50 kg archer standing on frictionless ice shoots a 100g arrow at a speed of 100 m/s what uis the recoild pseed of the archer?

nexus9112 [7]

Answer:

v= 0.2 m/s

Explanation:

Given that

m₁ = 50 kg

m₂ = 100 g = 0.1 kg

u =10 0 m/s

If there is no any external force on the system then the total linear momentum of the system will be conserve.

Initial linear momentum = Final momentum

m₁u₁ + m₂ u₂ =m₂ v₂ +m₁v₁

50 x 0 + 0.1 x 100 = 50 v + 0

0+ 10 = 50 v

$v=\dfrac{10}{50}\ m/s$

v= 0.2 m/s

Therefore the recoil speed will be 0.2 m/s.

7 0

3 years ago

Discuss the objectives of business 8 marks I need answers help me

Dima020 [189]

Answer:

* to make more income

*to became popular on stock markers

*to be known as a good business

4 0

2 years ago