Answer:
The magnification of an astronomical telescope is -30.83.
Explanation:
The expression for the magnification of an astronomical telescope is as follows;

Here, M is the magnification of an astronomical telescope,
is the focal length of the eyepiece lens and
is the focal length of the objective lens.
It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.
Put
and
in the above expression.

M=-30.83
Therefore, the magnification of an astronomical telescope is -30.83.
Answer:
The resistance that will provide this potential drop is 388.89 ohms.
Explanation:
Given;
Voltage source, E = 12 V
Voltage rating of the lamp, V = 5 V
Current through the lamp, I = 18 mA
Extra voltage or potential drop, IR = E- V
IR = 12 V - 5 V = 7 V
The resistance that will provide this potential drop (7 V) is calculated as follows:
IR = V

Therefore, the resistance that will provide this potential drop is 388.89 ohms.
For fundamental frequency of a string to occur, the length of the string has to be half the wavelength. That is,
1/2y = L, where L = length of the string, y = wavelength.
Therefore,
y = 2L = 2*0.75 =1.5 m
Additionally,
y = v/f Where v = wave speed, and f = ferquncy
Then,
v = y*f = 1.5*220 = 330 m/s
Answer: Normal Force of the ground pushing the athlete up.
Explanation: I did this on Khan Academy and got the answer right.