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postnew [5]
3 years ago
15

Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of app

roximately 4.0 cm/yr. Assuming this to be a constant rate, how many years will pass before the radius of the Moon's orbit increases by 3.84x 10^7 m (10%)?
Physics
1 answer:
melomori [17]3 years ago
7 0

Answer:

967500000 years

Explanation:

The Speed at which the radius of the orbit of the Moon is increasing is 4 cm/yr

Converting to m

1 m = 100 cm

1\ cm=\frac{1}{100}\ m

4\ cm\y=\frac{4}{100}=0.04\ m/yr

The distance by which the radius increases is 3.84×10⁷ m

Time = Distance / Speed

\text{Time}=\frac{3.87\times 10^7}{0.04}\\\Rightarrow \text{Time}=967500000\ yr

967500000 years will pass before the radius of the orbit increases by 10%.

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A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find
sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge q_{1}=9\times10^{-6}\ C at origin

Second charge q_{2}=6\times10^{-6}\ C

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

E=\dfrac{kq}{r^2}

0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}

\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}

\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1

\dfrac{1}{d}=1.82

d=\dfrac{1}{1.82}

d=0.55\ m

Hence, The coordinates of the point is (0,0.55).

3 0
3 years ago
A 65-kg skier grips a moving rope that is powered by an engine and is pulled at a constant speed to the top of a 230 hill. The s
Sveta_85 [38]

Answer:

The required power by the engine is 33.0 hp

Explanation:

Solution

Newton's second law says that, the net force Fnet on an object of mass m will accelerates the object

Where

Fnet = ma

a = acceleration

θ = angle of incline,

m = mass of the 30 skiers,

f = frictional force

N = normal force

mg sinθ, mg cos θ are components of weight skier

F = the force applied by engine

Now,

The skier mass  is 65 kg

We calculate the mass of the 30 skier

m = 30 (65kg) = 1950 kg

Calculate the net force acting on the skiers along the x-axis

Fnet, x=ma

Now,

F-mg sin θ - f = 0

F= mg sin θ + f -----(1)

The kinetic frictional force is denoted by

f = μk N ------(2)

μk = The coefficient of the kinetic friction

We now, calculate the net force acting on the skiers along y axis

Fnet, y = ma

N- mg cos θ = 0

so,

N = mg cos θ

This value is  substituted in equation (2)

f = μk mg cos θ

we substitute the value for equation (1)

F = mg sin θ + μk mg cos θ

mg =  sinθ + μk cos θ)-----(3)

The next step is to calculate the work done by the engine in pulling the skiers, the incline top by applying the equation 3

W = Fx

= mg ( sinθ + μk cos θ)x

x = the displacement

we now substitute 1950 kg for m, 23° for θ, 0.10 for μk and 320m for x

so,

W = mg ( sinθ + μk cos θ)x

= (1950 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (320 m)

= 2.99 * 10 ^6 J

Then,

The time from minute to s is converted

t =(2.0min) ( 60sec/1.0min) = 120 sec

Now we calculate the power needed by the engine to pull the skiers at the incline top

Thus,

P = W/t

we substitute  2.955 * 10 ^6 J for W and  120 s for t

we have,

P = 2.955 * 10 ^ 6 J/ 120 s

= ( 2.4625 * 10 ^ 4 W) (1.0 hp/746 W)

= 33.0 hp

In conclusion the required power by the engine is 33.0 hp

3 0
3 years ago
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IRISSAK [1]
Plant trees around the perimeter of his fields
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3 years ago
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mass/volume

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3 years ago
Please explain the method of making electricity from non renewable sources like coal?
Leno4ka [110]
The coal is  urned to heat up water. this produces steam. the steam turns a turbine that turns a generator which provided energy yhat can be transferred into electrisity
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3 years ago
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