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xeze [42]
3 years ago
7

Ocean waves are hitting a beach at a rate of 3.5 Hz. What is the period of the waves?

Physics
1 answer:
Radda [10]3 years ago
5 0

Answer:

the correct answer is 0.286 s

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Earthquake information would most likely be shown on which type of map?
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Information about Earthquakes would normally be shown on a map called "geologic hazards"

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Why gravitational force at the center og earth is zero
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A car of mass 500 kg is moving at a speed of 1.2 m/s. A man pushes the car, increasing the speed to 2 m/s. How much work did the
Vikki [24]

Answer:

Option B - 360 J

Explanation:

One of the formulas used to find out how much work was done is :

E = 0.5×m×v² (where m is the mass and v is the velocity)

Now we substitute values from the question and simplify :

E = 0.5×500×1.2²

E = 250×1.2²

E = 250×1.44

E = 360 J

So our answer is Option B

Hope this helped and have a good day

5 0
2 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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