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3 years ago

PLEASE HELP!

Physics

1 answer:

SashulF [63]3 years ago

3 0

Answer:

They gon' hate me regardless, that's why I do what I do (what I do)

See me in person i'm flawless (I'm flawless girl), i might just snatch up your dude (hahahaha)

Explanation:

i just want to be....appreciated

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A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent

tensa zangetsu [6.8K]

Answer:

(a): The car's relative position to the base of the cliff is x= 32.52m.

(b): The lenght of the car in the ir is tfall= 1.78 sec.

Explanation:

Vo= 0

V= ?

d= 50m

h= 30m

a= 4 m/s²

t= √(2*d/a)

t= 5 sec

V= a*t

V= 20 m/s

Vx= V * cos(24º)

Vx= 18.27 m/s

Vy= V* sin(24º)

Vy= 8.13 m/s

h= Vy*t + g*t²/2

clearing t:

tfall= 1.78 sec (b)

x= Vx * tfall

x= 32.52 m (a)

4 0

3 years ago

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago

Which of the following is an example of technology influencing science?

FinnZ [79.3K]

Answer:A and C

Explanation:

8 0

3 years ago

14.A 90 kg quarterback gets tackled by being hit by a 120 kg lineman backwards

quester [9]

The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

F = Ma.

The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

So, the acceleration can be calculated using the formula,

500N = 90kg x a

a = 5.55 m/s².

Now, the weight if the lineman is 120kg, the force applied by him is 500N.

So, from the formula, his acceleration A will be,

500N = 120Kg x A

A = 4.16 m/s².

both of them will have acceleration in the same direction,

To know more about Force, visit,

brainly.com/question/25239010

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8 months ago

Electrical power is transmitted from power plants to consumers–sometimes over very long distances– through conducting power line

sergeinik [125]

There is more wire to travel through,farther distance, and a higher possibility of other disruptions. Please Mark Brainliest!!!

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3 years ago