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andrew11 [14]
3 years ago
9

A 3.0-kilogram cart possesses 96 joules of kinetic energy. Calculate the speed of the car.

Physics
1 answer:
Dvinal [7]3 years ago
5 0

Answer:

8.0 m/s

Explanation:

KE = ½ mv²

96 J = ½ (3.0 kg) v²

v = 8.0 m/s

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When I wave a charged golf tube at the front of the classroom with a frequency of two oscillations per second, I produce an elec
borishaifa [10]

To solve the exercise it is necessary to take into account the concepts of wavelength as a function of speed.

From the definition we know that the wavelength is described under the equation,

\lambda = \frac{c}{f}

Where,

c = Speed of light (vacuum)

f = frequency

Our values are,

f = 2Hz

c = 3*10^8km/s

Replacing we have,

\lambda = \frac{c}{f}

\lambda = \frac{3*10^8km/s}{2Hz}

\lambda = 1.5*10^8m

<em>Therefore the wavelength of this wave is 1.5*10^{8}m</em>

8 0
3 years ago
What is true about X-rays and microwaves?
erma4kov [3.2K]
I believe the correct answer from the choices listed above is option C. X-rays have greater frequency than microwaves. In a electromagnetic spectrum, the order in increasing frequency is as follows:

radio waves,microwaves, terahertz radiation, infrared radiation, visible light, ultraviolet radiation,X-rays<span> and gamma </span>rays<span>.</span>
7 0
3 years ago
A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.
kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight=\sqrt{\frac{2H}{g}}

substituting value

  • =\sqrt{\frac{2*25}{9.8}}
  • =2.26seconds

<h3><u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>

b. How far from the building does the ball strike the ground?

<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?

we have

Horizontal range=u*\sqrt{\frac{2H}{g}}

  • =8.25*2.26
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<h3><u>The ball strikes 18.63m far from building</u>. </h3>
7 0
2 years ago
Where in the motion is the magnitude of the force from the spring on the object zero? Where in the motion is the magnitude of th
kkurt [141]

<em></em>

Answer:

1. The magnitude of the force from the spring on the object is zero on <em>Equilibrium.</em>

2. The magnitude of the force from the spring on the object is a maximum on <em>The top and bottom.</em>

3. The magnitude of the net force on the object is zero on <em>The Bottom.</em>

4. The magnitude of the force on the object is a maximum on <em>the Top.</em>

Explanation:

<em>1. Because the change in position delta X is zero.</em>

<em>2. Because of delta X.</em>

<em>3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.</em>

<em>4. Because, the force of the spring from compressiom and the force of gravity both act on the mass.</em>

8 0
3 years ago
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dexar [7]
U=120 \text{ V}\\&#10;I=8\text{ A}\\&#10;P=U\cdot I\\\\&#10;P=120\cdot8=960 \text{ W}&#10;
7 0
3 years ago
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