Answer:
option a.
Explanation:
We can think of an atom as a nucleus (where the protons and neutrons are) and some electrons orbiting it.
We also know that the mass of an electron is a lot smaller than the mass of a proton or the mass of an electron.
So, if all the protons and electrons of an atom are in the nucleus, we know that most of the mass of an atom is in the nucleus of that atom.
Then we define the mass number, which is the total number of protons and neutrons in an atom. Such that the mass of a proton (or a neutron) is almost equal to 1u
Then if we define A as the total number of protons and neutrons, and each one of these weights about 1u
(where u = atomic mass unit)
Then the weight of the nucleus is about A times 1u, or:
A*1u = A atomic mass units.
Then the correct option is:
The mass of the nucleus is approximately EQUAL to the mass number multiplied by __1__ Atomic Mass unit.
option a.
Answer:
a= 92. 13 m/s²
Explanation:
Given that
Amplitude ,A= 0.165 m
The maximum speed ,V(max) = 3.9 m/s
We know that maximum velocity in the SHM given as
V(max) = ω A
ω=Angular speed
A=Amplitude
ω=23.63 rad/s
The maximum acceleration given as
a = ω² A
a= (23.63)² x 0.165 m/s²
a= 92. 13 m/s²
Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².
Answer:
19.1 deg
Explanation:
v = speed of the proton = 8 x 10⁶ m/s
B = magnitude of the magnetic field = 1.72 T
q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C
F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N
θ = Angle between proton's velocity and magnetic field
magnitude of magnetic force on the proton is given as
F = q v B Sinθ
7.20 x 10⁻¹³ = (1.6 x 10⁻¹⁹) (8 x 10⁶) (1.72) Sinθ
Sinθ = 0.327
θ = 19.1 deg
Answer:
-2.26×10^-4 radians
Explanation:
The solution involves a right angle triangle
Length is z while the horizontal is the height x
X^2+ 100^2=z^2
Taking the derivatives
2x(dx/dt)=Z^2(dz/dt)
Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11
dz/dt= 1100sqrt3/200 = 9.53
Sin a= 100/a
Taking derivatives in terms of t
Cos a(da/dt)=100/z^2 dz/dt
a= 30°
Cos (30°)da/dt= (-100/40000×9.5)
a= -2.26×10^-4radians
Answer:
<em>B) 1.0 × 10^5 V</em>
Explanation:
<u>Electric Potential Due To Point Charges
</u>
The electric potential produced from a point charge Q at a distance r from the charge is
The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.
We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is
where a is the length of the side.
The distance from any corner to the center is half the diagonal, thus
The total potential is
Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of 
. Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.
The total potential is
