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nikitadnepr [17]
3 years ago
15

I need help on 6, 7, 8, and 9

Physics
1 answer:
Ivahew [28]3 years ago
5 0
6 is b. part B on 6 is a. 7 is a. partB ON 7 b
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what is the the wavelength if a wave cycles up and down three times per second and the distance between each wave is 1.7m?
DENIUS [597]

Answer:5.1 meters

Explanation:

wavelength=3 x1.7

Wavelength=5.1 meters

8 0
3 years ago
How can two tectonic plates move relative to each other
yarga [219]
They are doing the transform type of tectonic plate movement. This is a process where the plates move sideways. But they move at a rate of a few inches a year. Did that help you?
7 0
3 years ago
I WILL GET BRAINLIEST PLS HELP A student attempts to push a box across the floor. The student decides to create a free-body diag
MissTica

Answer: The gravitational

Explanation: The student is pushing the box so u have to have gravitational force so it could move

5 0
2 years ago
A skydiver jumps out of a hovering helicopter. A few seconds later, another diver jumps out, so they both fall along the same ve
Sergio039 [100]

Answer:

distance difference would a) increase

speed difference would f) stay the same

Explanation:

Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.

Their equations of motion for distance and velocities are

s_2 = gt^2/2

s_1 = g(t + \Delta t)^2/2

v_2 = gt

v_1 = g(t + \Delta t)

Their difference in distance are therefore:

\Delta s = s_1 - s_2 = g(t + \Delta t)^2/2 - gt^2/2

\Delta s = g/2((t + \Delta t)^2 - t^2)

\Delta s = g/2(t + \Delta t - t)(t + \Delta t + t) (AsA^2 - B^2 = (A-B)(A+B)

\Delta s = g\Delta t/2(2t + \Delta t)

So as time progress t increases, Δs would also increases, their distance becomes wider with time.

Similarly for their velocity difference

\Delta v = v_1 - v_2 = g(t + \Delta t) - gt

\Delta v = gt + g\Delta t - gt = g\Delta t

Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.

This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.

8 0
3 years ago
Read 2 more answers
Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.
mr_godi [17]

Answer:

The magintude of the acceleration for both objects is 3.11m/s^2

Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

T_A-m_A*g=m_A*a

and for mB (descending):

T_B-m_B*g=-m_B*a

T_A=T_B

because the two boxes has the same acceleration because they are attached together:

m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2

So the magintude of the acceleration for both objects is 3.11m/s^2

5 0
3 years ago
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