The power developed is 500 W ( to the nearest Watt)
Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.
Therefore,
Substitute the given values of force , displacement and time
Thus the Power can be rounded off to the nearest value of 500 W
Answer:
A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t]y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t], where the origin is at the left end of the string, the x-axis is along the string, and the y-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation y(x, t) for this string if it were vibrating in its eighth harmonic?
The emf is induced in the wire will be 1.56 ×10 ⁻³ V. The induced emf is the product of the magnetic field,velocity and length of the wire.
<h3>What is induced emf?</h3>
Emf is the production of a potential difference in a coil as a result of changes in the magnetic flux passing through it.
When the flux coupling with a conductor or coil changes, electromotive Force, or EMF, is said to be induced.
The given data in the problem is;
B is the magnitude of the magnetic field,= 5.0 ×10⁻⁵ T
V(velocity)=125 M/SEC
L(length)=25 cm=0.25 m
The maximum emf is found as;
E=VBLsin90°
E=125 × 5.0 × 10⁻⁵ ×0.25
E=1.56 ×10 ⁻³ V
Hence, the emf is induced in the wire will be 1.56 ×10 ⁻³ V
To learn more about the induced emf, refer to the link;
brainly.com/question/16764848
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<u>Given </u><u>:</u><u>-</u>
- An elevator is moving vertically up with an acceleration a.
<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>
- The force exerted on the floor by a passenger of mass m .
<u>Solution</u><u> </u><u>:</u><u>-</u>
As the man is in a accelerated frame that is <u>non </u><u>inertial</u><u> frame</u><u> </u>, we would have to think of a pseudo force .
- The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
For the FBD refer to the attachment . From that ,
<u>Hence</u><u> </u><u>option</u><u> </u><u>d </u><u>is </u><u>correct</u><u> </u><u>choice </u><u>.</u>
<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em>