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KatRina [158]
2 years ago
7

f an earthquake’s magnitude is measured based on the richter scale, how is this abbreviated? a. mr b. ms c. ml d. mb

Physics
1 answer:
zhannawk [14.2K]2 years ago
5 0
The answer is A. Hope this helps!
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The answer is true. The table does show an object moving with changing speed.
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Given the nuclear equation:
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This is a nuclear fission reaction, in which a larger nucleus is bombarded with a neutron to make it break down into two smaller nuclei and release energy.
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What is the momentum of a 0.1-kg mass moving with a speed of 5 m/s
Andrej [43]
This question is wrong because in momontum we will write acceleration instead of speed.     suppose acceleration is 5m/s2 then 
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4 0
3 years ago
A ball is thrown straight up. What are the velocity and acceleration of the ball at the highest point in its path?
zubka84 [21]

Answer:

b. v = 0, a = 9.8 m/s² down.

Explanation:

Hi there!

The acceleration of gravity is always directed to the ground (down) and, near the surface of the earth, has a constant value of 9.8 m/s². Since the answer "b" is the only option with an acceleration of 9.8 m/s² directed downwards, that would solve the exercise. But why is the velocity zero at the highest point?

Let´s take a look at the height function:

h(t) = h0 + v0 · t + 1/2 g · t²

Where

h0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity

Notice that the function is a negative parabola if we consider downward as negative (in that case "g" would be negative). Then, the function has a maximum (the highest point) at the vertex of the parabola. At the maximum point, the slope of the tangent line to the function is zero, because the tangent line is horizontal at a maximum point. The slope of the tangent line to the function is the rate of change of height with respect to time, i.e, the velocity. Then, the velocity is zero at the maximum height.

Another way to see it (without calculus):

When the ball is going up, the velocity vector points up and the velocity is positive. After reaching the maximum height, the velocity vector points down and is negative (the ball starts to fall). At the maximum height, the velocity vector changed its direction from positive to negative, then at that point, the velocity vector has to be zero.

8 0
3 years ago
34. A train, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. A passenger standing on the platf
ira [324]

Answer:

4.08 s

Explanation:

Let the passenger took "t" time to catch the train

so in this case the total distance moved by the train + 5 m = total distance moved by the passenger

so we will have

distance moved by train is given as

d_1 = \frac{1}{2}(0.6) t^2

also the distance moved by passenger

d_2 = \frac{1}{2}(1.2) t^2

so we will have

d_1 + 5 = d_2

0.3 t^2 + 5 = 0.6 t^2

0.3 t^2 = 5

t = 4.08 s

3 0
3 years ago
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