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3 years ago

A 2400-kg satellite is in a circular orbitaround a planet. The

Physics

1 answer:

Fofino [41]3 years ago

4 0

Answer

given,

mass of satellite = 2400 Kg

speed of the satellite = 6.67 x 10³ m/s

acceleration of satellite = ?

gravitational force of the satellite will be equal to the centripetal force

$F = \dfrac{mv^2}{r}$

$F = \dfrac{2400\times (6.67\times 10^3)^2}{r}$

Assuming the radius of circular orbit = 8.92 x 10⁶ m

now,

$F = \dfrac{2400\times (6.67\times 10^3)^2}{8.92\times 10^6}$

F = 11970.11 N

acceleration,

$a = \dfrac{F}{m}$

$a = \dfrac{11970.11}{2400}$

a = 4.98 m/s²

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Using a rope that will snap if the tension in it exceeds 387 N, you need to lower a bundle of old roofing material weighing 449

Anni [7]

Answer:

(a) $a\approx1.4 m.s^{-2}$

(b) $v\approx 4.133 m.s^{-1}$

Explanation:

Given:

Limiting tension of snapping of the rope, T= 387 N

Weight of the object to be lifted, $w=449 N$

∴mass, $\Rightarrow m= 45.8163 kg$

height of letting down the weight, h = 6.1 m

(a)

Now,

The force to be compensated for being on the verge of snapping:

(T-w) = 62 N

Therefore, we need to produce and acceleration equivalent to the above force.

∵ $F=m.a$

$62=45.8163\times a$

$a= \frac{62}{45.8163}$

$a\approx 1.4 m.s^{-2}$

(b)

From the equation of motion ,we have:

$v^{2} =u^{2} +2a.s$ ....................(2)

where:

u= initial velocity= 0 (here, starting from rest)

v= final velocity = ?

$a= 1.4 m.s^{-2}$

s= displacement =h =6.1 m

Now, putting the values in eq. (2)

$v^2= 0^2 + 2\times 1.4\times 6.1$

$v\approx 4.133 m.s^{-1}$ is the velocity with which the body will hit the ground in the given conditions.

7 0

4 years ago

HELP PLEASE ASAP

kherson [118]

I am pretty sure the correct answer is C.

3 0

3 years ago

Read 2 more answers

On the moon, a feather is dropped from a height of 1.40 m. The acceleration of gravity on the moon is -1.67 m/s2. Determine the

Molodets [167]

Answer:

1.29 s

Explanation:

Height = 1.40m
Acclⁿ due to gravity = 1.67 m/s²
Time of descent = ?

As we know that ,

$\sf\longrightarrow Time_{descent}= \sqrt{\dfrac{2H}{g_{(moon)}}}\\\\\sf\longrightarrow t_d = \sqrt{\dfrac{ 2\times 1.4}{1.67}} \\\\\sf\longrightarrow t_d =\sqrt{1.67} s\\\\\sf\longrightarrow \boxed{\red{\sf Time_{descent}= 1.29 s }}$

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3 years ago

What happens to the amount of energy present when it is moved from one form to another

Jet001 [13]

The amount of energy is ALWAYS conserved based on the law of conservation of energy. Therefore, the amount of energy will remain the same.

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3 years ago

Please help me on this position-time graph assignment!!!

saw5 [17]

$\text{Hello there! :)}$

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$\text{Given: From 0 - 8 sec, velocity of 10 m/s.}\\\\\text{From 8 - 18 sec, velocity of -5 m/s. (In opposite direction because of the negative sign)}\\\\\\\text{From 18 - 22 sec, velocity of 0 m/s. (Object at rest)}\\\\\text{From 22 - 24 sec, velocity of 5 m/s.}\\\\\text{Graph the position vs. time graph using the velocities as the slope of the line }$

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4 years ago