Answer:
Gamma rays
Explanation:
Gamma rays is at the end of the electromagnetic spectrum, and has the highest energy. It propagates through space at 3x10^8 m/s and has the smallest wavelength and the highest frequency. It is given off by atoms of element as they undergo nuclear disintegration.
The given question is incomplete. The complete question is as follows.
The block has a weight of 75 lb and rests on the floor for which = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.
Determine the output of the motor at the instant .
Explanation:
We will consider that equilibrium condition in vertical direction is as follows.
N - W = 0
N = W
or, N = 75 lb
Again, equilibrium condition in the vertical direction is as follows.
= 0
=
= 30 lb
Now, the equilibrium equation in the horizontal direction is as follows.
or, T =
=
=
= 17.32 lb
Now, we will calculate the output power of the motor as follows.
P = Tv
=
=
= 0.189 hp
or, = 0.2 hp
Thus, we can conclude that output of the given motor is 0.2 hp.
Answer:
kinetic energy at that point is 3500 MJ
Explanation:
Given data
potential energy = 5000 MJ
kinetic energy = 4500 MJ
satellite’s potential energy is 6000 MJ
to find out
kinetic energy
solution
we know that total energy is sum of kinetic energy and potential energy
so kinetic energy will be
kinetic energy = total energy - potential energy ................1
here
total energy = 5000 + 4500 = 9500 MJ
so now put all value in equation 1
kinetic energy = total energy - potential energy
kinetic energy = 9500 - 6000 = 3500
so kinetic energy at that point is 3500 MJ
Answer:
The focal length of the lens is 2.5 cm
Explanation:
Use the two equations for thin lenses combined: the one for magnification (m), and the one that relates distances of object , of image , and focal length;
Since we know the value of the magnification (m), we can write the image distance in terms of the object distance, and then use it to replace the image distance in the second equation:
then, solving for the focal distance knowing that the object distance is 1.5 cm: