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Sergeeva-Olga [200]

3 years ago

11

Suppose an object is in orbit around the earth, the earth's gravitational pull some how becomes stronger than the objects forwar

d motion, what will happen to the object?

1 answer:

Dahasolnce [82]3 years ago

5 0

The object will fall towards Earth.

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An outdoor Wi-Fi unit for a picnic area has a 110 mW output and a range of about 38 m. What output power (in mW) would reduce it

valkas [14]

Answer:

The output power is 24.68 mW.

Explanation:

Given that,

Power = 110 mW

Range = 38 m

Reduced range = 18 m

We need to calculate the power

Using formula of intensity

$I=\dfrac{P}{A}$

$I=\dfrac{P}{\pi r^2}$

As intensity is constant

$P\propto r^2$

So, $\dfrac{P_{1}}{P_{2}}=\dfrac{r_{1}^2}{r_{2}^{2}}$

$P_{2}=\dfrac{r_{2}^2}{r_{1}^{2}}\timesP_{1}$

Put the value into the formula

$P_{2}=\dfrac{18^2}{38^2}\times110\times10^{-3}$

$P_{2}=24.68\times10^{-3}\ W$

$P_{2}=24.68\ mW$

Hence, The output power is 24.68 mW.

8 0

4 years ago

A police car moving at 36.0 m/s is chasing a speeding motorist traveling at 30.0 m/s. The police car has a siren that emits soun

maxonik [38]

Answer:

The frequency heard by the motorist is 4313.2 Hz.

Explanation:

let f1 be the frequency emited by the police car and f2 be the frequency heard by the motorist, let v1 be the speed of the police car and v2 be the speed of the motorist and v = 343 m/s be the speed of sound.

because the police car is moving towards the motorist at a higher speed, then the motorist will hear a increasing frequency and according to Dopper effect, that frequency is given by:

f1 = [(v + v2/(v - v1))]×(f2)

= [( 343 + 30)/(343 - 36)]×(3550)

= 4313.2 Hz

Therefore, the frequency heard by the motorist is 4313.2 Hz.

5 0

3 years ago

Which statement is true about a planet’s orbital motion?

lana66690 [7]

Answer:

Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

Explanation:

The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.

When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.

The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

The velocity of the orbit is given by the relation,

$V = \sqrt{\frac{GM}{R + h} }$

Where

V - velocity of the orbit at a height h from the surface

R - Radius of the second object

G - Gravitational constant

h - height from the surface

The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

$1/2 mV^{2} = GMm/R$

Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

3 0

3 years ago

An object with a mass of 10 kg is rolled down a frictionless ramp from a height of 3 meters. If a factory worker at the bottom o

ruslelena [56]

Answer:

The amount of work the factory worker must to stop the rolling ramp is 294 joules

Explanation:

The object rolling down the frictionless ramp has the following parameters;

The mass of the object = 10 kg

The height from which the object is rolled = 3 meters

The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp

Where;

The potential energy P.E. = m × g × h

m = The mass of the ramp = 10 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the object rolls down = 3 m

Therefore, we have;

P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules

The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules

8 0

3 years ago

A snail and an inchworm are in a race. Their race track heads north for a distance of 2 m. If the inchworm comes to the end of t

olga2289 [7]

Well the basic equation for velocity is v=d/t where d is distance and t is time. So v=2m/50s and the answer is v=0.04meter/second.

7 0

3 years ago