100 g lead target. Lead has a very high density which means even a very small volume and weight quite a lot. So because the lead target weighs only 100 g it's going to be small in size compared to other targets which are all made of lesser dense materials than lead.
Answer:
The electric force is 
Explanation:
From the question we are told that
The Bohr radius at ground state is 
The values of the distance between the proton and an electron 
The electric force is mathematically represented as

Where n and p are charges on a single electron and on a single proton which is mathematically represented as

and k is the coulomb's constant with a value

substituting values
![F = \frac{9*10^{9} * [(1.60*10^{-19} ]^2)}{(2.63 * 0.529 * 10^{-10})^2}](https://tex.z-dn.net/?f=F%20%3D%20%20%5Cfrac%7B9%2A10%5E%7B9%7D%20%2A%20%20%5B%281.60%2A10%5E%7B-19%7D%20%5D%5E2%29%7D%7B%282.63%20%2A%200.529%20%2A%2010%5E%7B-10%7D%29%5E2%7D)

Answer:
The percentage of the weight supported by the front wheel is A= 19.82 %
Explanation:
B] Let the mass of plane be m, force on backwheels be Nb and nosewheel be Nn
Torque equation about nose wheel,
mg*(2.58-0.8) - Nb *(3.02-0.8) = 0
Nb = mg*(2.58-0.8)/(3.02-0.8) = 0.8018 mg
Nn = mg - Nb = (1-0.8018) mg = 0.1982mg
Weight percentage supported by front wheel = 19.82% answer
Answer:
The magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
Explanation:
Given;
radius of the wire, r = 0.45 m
current on the loop, I = 2.4 A
angle of inclination, θ = 36⁰
torque on the coil, τ = 1.5 N.m
The torque on the coil is given by;
τ = NIBAsinθ
where;
B is the magnetic field
Area of the loop is given by;
A = πr² = π(0.45)² = 0.636 m
τ = NIBAsinθ
1.5 = (1 x 2.4 x 0.636 x sin36)B
1.5 = 0.8972B
B = 1.5 / 0.8972
B = 1.67 T
Therefore, the magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
Answer:
The frequency is the same
Explanation:
When a wave is created by a source which is vibrating at a certain frequency, the frequency of the wave itself is equal to the frequency of the source.
This occurs with every kind of wave. For instance, if we consider the radio waves produced by an antenna, the frequency of the radio waves is equal to the frequency of the antenna.
In this case, the waves are created by the vibrating bug. The bug is vibrating with a certain frequency
: as a consequence, the frequency
of the waves produced by the bug will be equal to the frequency of vibration of the bug:
.