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Katyanochek1 [597]
3 years ago
10

What is the momentum of a 110kg fullback running at 860 m/s? *

Physics
1 answer:
Natalka [10]3 years ago
8 0
M/s is 860 your anser is 47miles
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When sound travels through air, the particles _____. a. vibrate along the direction the wave travels b. vibrate but not in any f
Mandarinka [93]
Vibrate along the direction the wave travels
8 0
3 years ago
Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line
Vilka [71]

Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

vDE = 7.5 m/s

vPE = 20.2 m/s

Required:  vDP

Assume that

vDE to be in direction of - j

vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

vDP = vDE - vPE     (I)

Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s =  21.7454 m/s

θ = Arctan (- 7.5/- 20.2) = 20.3693°

θ is in 3rd quadrant so add 180°

θ = 20.3693° + 180° = 200.3693°

4 0
3 years ago
In a light wave, which properties tell you the color of light?
Alenkasestr [34]

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<em><u>Wavel</u></em><em><u>ength</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>distan</u></em><em><u>ce</u></em><em><u> between</u></em><em><u> </u></em><em><u>one</u></em><em><u> </u></em><em><u>crest</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>one</u></em><em><u> </u></em><em><u>through</u></em><em><u> </u></em><em><u>,</u></em><em><u> </u></em><em><u>also</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>dist</u></em><em><u>ance</u></em><em><u> </u></em><em><u>after</u></em><em><u> </u></em><em><u>which</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>wave</u></em><em><u> </u></em><em><u>repe</u></em><em><u>at</u></em><em><u> </u></em><em><u>its</u></em><em><u>elf</u></em><em><u> </u></em><em><u>!</u></em>

<em><u>It's</u></em><em><u> </u></em><em><u>SI</u></em><em><u> </u></em><em><u>unit</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>meter</u></em><em><u> </u></em><em><u>!</u></em><em><u> </u></em>

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<em><u>Diff</u></em><em><u>erent</u></em><em><u> </u></em><em><u>Wavelength</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>light</u></em><em><u> </u></em><em><u>have</u></em><em><u> </u></em><em><u>diff</u></em><em><u>erent</u></em><em><u> </u></em><em><u>col</u></em><em><u>or</u></em><em><u> </u></em><em><u>!</u></em><em><u>!</u></em>

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5 0
2 years ago
ILI
Nikolay [14]

The original kinetic energy will be 0 J and the final kinetic energy will be 7500 J and the amount of work utilized will be similar to the final kinetic energy i.e., 7500 J.

<u>Explanation:</u>

As it is known that the kinetic energy is defined as the energy exhibited by the moving objects. So the kinetic energy is equal to the product of mass and square of the velocity attained by the car. Thus,

                  \text {Kinetic energy}=\frac{1}{2} m v^{2}

So the initial kinetic energy will be the energy exerted by the car at the initial state when the initial velocity is zero. Thus the initial kinetic energy will be zero.  

The final kinetic energy is

\text {Kinetic energy}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 600 \times 5 \times 5 = 7500 J

As the work done is the energy required to start the car from zero velocity to 5 m/s velocity.  

                       Work done = Final Kinetic energy - Initial Kinetic energy

Thus the work utilized for moving the car is  

                         Work done = 7500 J - 0 J = 7500 J

Thus, the initial kinetic energy of the car is zero, the final kinetic energy is 7500 J and the work utilized by the car is also 7500 J.

7 0
3 years ago
An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m
Aneli [31]

Answer:

\displaystyle y_m=3.65m

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of v_o and \theta\\ as the initial speed and angle, then we have

\displaystyle v_x=v_o\ cos\theta

\displaystyle v_y=v_o\ sin\theta-gt

\displaystyle x=v_o\ cos\theta\ t

\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}

If we want to know the maximum height reached by the object, we find the value of t when v_y becomes zero, because the object stops going up and starts going down

\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt

Solving for t

\displaystyle t=\frac{v_o\ sin\theta }{g}

Then we replace that value into y, to find the maximum height

\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2

Operating and simplifying

\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}

We have

\displaystyle v_o=20\ m/s,\ \theta=25^o

The maximum height is

\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}

\displaystyle y_m=3.65m

7 0
2 years ago
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