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UNO [17]
3 years ago
7

A 7 kg sled is initially at rest on a horizontal road. the sled is pulled a distance of 2.9 m by a force of 42 n applied to the

sled at an angle of 30o to the horizontal. find the change in the kinetic energy of the sled.
Physics
1 answer:
Artemon [7]3 years ago
6 0
I have no idea on this question.
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I NEED THIS ANSWER TODAY PLEASE HELP ME
Anastaziya [24]

The answer is "B" - If there are no windows then there will be no light coming in, and therefore you don't have to worry about what time of day you do the experiment at.

7 0
3 years ago
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An electromagnetic wave has a frequency of 4.0 x 10^18 Hz. What is the wavelength of the wave?
LenaWriter [7]

Answer:

7.5 × 10^-11 m

Explanation:

Hope this helps !

3 0
3 years ago
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Match the terms to the correct descriptions.
LenaWriter [7]

1. Energy Conversion

2. Light Energy

3. Mechanical Energy

4. Kinetic Energy

5. Potential Energy

6. Sound Energy

7. Electrical Energy

8. Chemical Energy

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8 0
3 years ago
An MRI scanner is based on a solenoid magnet that produces a large magnetic field. The magnetic field doesn't stop at the soleno
GaryK [48]

Answer:

The maximum change in  flux is \Delta \o = 0.1404 \ Wb

The average  induced emf     \epsilon =0.11232 V

Explanation:

   From the question we are told that

             The speed of the technician is v = 0.80 m/s

              The distance from the scanner is d = 1.0m

              The  initial magnetic field is  B_i = 0T

               The final magnetic field is B_f = 6.0T

                 The diameter of the loop is  D = 19cm = \frac{19}{100} = 0.19 m

The area of the loop is mathematically represented as

        A  =  \pi [\frac{D}{2} ]^2

             = 3.142 \frac{0.19}{2}

             = 0.02834 m^2

At maximum the change in magnetic field is mathematically represented as

            \Delta \o = (B_f - B_i)A

  =>      \Delta  \o = (6 -0)(0.02834)

                  \Delta \o = 0.1404 \ Wb

The  average induced emf is mathematically represented as

           \epsilon =  \Delta \o v

              = 0.1404 * 0.80

             \epsilon =0.11232 V

7 0
3 years ago
If a spring has a spring constant of 5 N/m and it is stretched 20 cm, what is the force of the Spring
Nastasia [14]

Answer:

1 N

Explanation:

First the equation is momentum = Force / distance

20 cm = 0.2 m

5 N/m = F / 0.2 m

F = 1 N

7 0
3 years ago
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