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kozerog [31]
3 years ago
9

What is the difference between epicenter and focus?

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
6 0

Answer:

The hypocenter is the point within the earth where an earthquake rupture starts. The epicenter is the point directly above it at the surface of the Earth. Also commonly termed the focus.

Explanation:

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Why is it important to know the direction of the force applied to a moving object and the direction in which the object is movin
erastova [34]

Answer

(C).

When there is an angle between the two directions, the cosine of the angle must be considered.


Step by step Solution

The work done by a force is defined as the product of the force and the distance traveled in the direction of motion.

The first answer "Only the component of the force perpendicular to the motion is used to calculate the work" is wrong because, the force perpendicular to motion does no work.

The second choice "If the force acts in the same direction as the motion, then no work is done" is wrong because the work in the direction of the force is W=F\times d.

Fourth answer "A force at a right angle to the motion requires the use of the sine of the angle" is wrong because the sin(90)=0 meaning that there is no work done in the direction perpendicular to the motion.

The third answer" When there is an angle between the two directions, the cosine of the angle must be considered." is correct because the work is calculated using the force in the direction of the motion. The magnitude of this force is F\times d\times \cos(\theta).




4 0
3 years ago
Read 2 more answers
A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
Which substance(s) will sink if placed in a fluid that has a density of 1.19 g/ml? (List all)
olga2289 [7]

Answer:

Corn Syrup

Explanation:

7 0
3 years ago
Not a question, but I need help on the last two questions I posted, preferably no links, please...
Andrei [34K]

Answer:

ok...

Explanation:

7 0
3 years ago
Read 2 more answers
A large container contains a large amount of water. A hole is drilled on the wall of the container, at a vertical distance h = 0
barxatty [35]

Answer:

Velocity = 3.25[m/s]

Explanation:

This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.

In point 1, (surface of the water) we have the atmospheric pressure and at point 2 the water is coming out also at atmospheric pressure, therefore this members in the Bernoulli equation could be cancelled.

The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"

h2 is located at point 2 and it will be zero.

(P_{1} +\frac{v_{1}^{2} }{2g} +h_{1} )=(P_{2} +\frac{v_{2}^{2} }{2g} +h_{2} )\\P_{1} =P_{2} \\v_{1}=0\\h_{2} =0\\v_{2}=\sqrt{0.54*9.81*2}\\v_{2}=3.25[m/s]

4 0
3 years ago
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