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dybincka [34]
1 year ago
13

What is the top of a wave called?

Physics
1 answer:
ohaa [14]1 year ago
7 0

Answer:

The highest part of the wave is called the crest.

Explanation:) hope this helps

The highest part of the wave is called the crest. The lowest part is called the trough. The wave height is the overall vertical change in height between the crest and the trough and distance between two successive crests (or troughs) is the length of the wave or wavelength.

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A constant 20 N force is applied to a 7 kg box to push it along the ground. How
Elis [28]

Answer:

40 joules

Explanation:

Work Done=Force*Distance

6 0
3 years ago
The entropy of an isolated system must be conserved, so it never changes.a. Trueb. Fasle
Snowcat [4.5K]

Answer:

B: False

Explanation:

The second law of thermodynamics states that: the entropy of an isolated system will never decrease because isolated systems always tend to evolve towards thermodynamic equilibrium which is a state with maximum entropy.

Thus, it means that the entropy change will always be positive.

Therefore, the given statement in the question is false.

6 0
3 years ago
A watt is a measure of power (the rate of energy change) equal to 1 j>s. (a) calculate the number of joules in a kilowatt- ho
GalinKa [24]
Before solving this, you must know the definition of these units of measurement. Watt is a measurement of power which is the amount of energy per unit time in seconds. Energy, on the other hand, is expressed through the SI unit Joules. Thus, power is the amount of energy in Joules per second.

From here, you can use the dimensional analysis technique. Also, you should know that 1 kilowatt is equal to 1000 watts, and 24 hours is equal to 86,400 seconds. Then,

100 = Energy/86,400
Energy = 8,640,000 Joules

4 0
3 years ago
A spinning wheel has a rotational inertia of 2 kg•m². It has an angular velocity of 6.0 rad/s. An average counterclockwise torqu
ira [324]

Answer:

-20.0 kg m^2/s

Explanation:

The angular momentum of an object in rotation is given by

L=I \omega

where

I is the moment of inertia

\omega is the angular speed

In this problem, initially we have

I=2 kg m^2 is the moment of inertia of the wheel

\omega_i = 6.0 rad/s is the initial angular speed

So the initial angular momentum is

L_i = I\omega_i = (2)(6.0)=12 kg m^2/s

Later, a counterclockwise torque of

\tau=-5.0 Nm is applied

So the angular acceleration of the wheel is:

\alpha = \frac{\tau}{I}=\frac{-5.0}{2}=-2.5 rad/s^2 in the direction opposite to the initial rotation.

As a result, the final angular velocity of the wheel will be:

\omega_f = \omega_i + \alpha t

where

t = 4.0 is the time interval

Solving,

\omega_f = +6.0 +(-2.5)(4.0)=-4 rad/s

which means that now the wheel is rotating in the counterclockwise direction.

Therefore, the new angular momentum of the wheel is:

L_f = I\omega_f =(2)(-4.0)=-8.0 kg m^2/s

So, the change in angular momentum is:

\Delta L=L_f - L_i = (-8.0-(12))=-20.0 kg m/s^2

7 0
3 years ago
A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at
pav-90 [236]

Answer:

Explanation:

Given

mass of drop m=2 kg

height of fall h=1 m

ball leaves the foot with a speed of 18 m/s at an angle of 55^{\circ}

Velocity of ball just before the collision with the floor

u^=2gh

u=\sqrt{2gh}

u=\sqrt{2\times 9.8\times 1}=4.42 m/s

Impulse delivered in Y direction

J_y=m(v\sin (55)-(-u))

J_y=2(18\sin (55)+4.42)

J_y=38.32 kg-m/s

Impulse in x direction

J_x=m\times v\cos (55)

J_x=2\times 4.42\cos (55)=20.646

J_{net}=\sqrt{J_x^2+J_y^2}

J_{net}=\sqrt{(38.32)^2+(20.64)^2}

J_{net}=43.52 kg-m/s

at an angle of \tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}

\phi =tan^{-1}(1.856)

\phi =61.7^{\circ}  

7 0
3 years ago
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