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zzz [600]
3 years ago
6

What happens to a wave when it moves from one medium to another?

Physics
1 answer:
Nina [5.8K]3 years ago
3 0
One side of the wave changes speed before the other side, causing the wave to move
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Meme on newtons law of motion (should me made ur self not from any searh engine)
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He will be a pilot and he will fly the plane over bridges fewwww
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1 year ago
If a builder of mass 75kg climbs a vertical ladder of 25m how much energy has she gained ?
erica [24]
So,

GPE (graviational potential energy) = mass x g x height

GPE is depends on where zero height is defined.  In this situation, we define h = 0 as the initial height.

GPE = 75 \ kg*9.8 \ \frac{m}{s^2}*25 \ m

GPE = 18,375 \frac{kg*m^2}{s^2}

GPE = 18,375 \ joules(J) \ or \ 18.375 \ kilojoules(kJ)

The builder has gained 18.375 kJ of PE.
4 0
3 years ago
Which of the following are vector quantities?
pshichka [43]

Answer:

Vector Quantity: A physical quantity is said to be a vector quantity when it has both magnitude and direction. The scalar quantities are distance, mass, time, volume, density, speed, temperature, and energy, The vector quantities are weight, velocity, acceleration, and force.

Explanation:

Mark me brainleist PLZZZZ

8 0
3 years ago
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Two forces act on the screw eye. If F = 600 N, determine the magnitude of the resultant force and the angle θ if the resultant f
asambeis [7]

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7 0
2 years ago
A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch
ANTONII [103]

Answer:

A.    A = 0.913 m

B.    amax = 132.24m/s^2

C.    Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

v_{max}=A\omega          (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

\omega=\sqrt{\frac{k}{m}}           (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}

Next, you solve the equation (1) for A and replace the values of vmax and w:

A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0
3 years ago
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