Answer:
a = (v2 - v1) / t
From A to B (8 - 4) m/s / 1 s = 4 m / s^2
From A to D ( 7 - 4) m/s / 5 s = .6 m / s^2
Note these equations hold for "uniform" values
They say nothing about the acceleration at intermediate points - the equation just says that his average speed increased from 4 m/s to 7 m/s during a 5 sec period
Answer:
The y-component of the electric force on this charge is 
Explanation:
<u>Given:</u>
- Electric field in the region,
- Charge placed into the region,
where, 
are the unit vectors along the positive x and y axes respectively.
The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,
Thus, the y-component of the electric force on this charge is
Answer:
1. 57.99V
2. 37.5Hz
3. 0.7072A
4. 82 ohms
5. 5.18x10^-5F
Explanation:
In answer to this question, we have the Standard equation of AC emf to be
V = V0 x sin ωt
We have
V0 = 82V,
ω = 75π
1.
RMS Voltage =
V0/√2 = 82 /√2
= 82/1.414
= 57.99V
2.
ω = 2π* f
75π = 2πf
Frequency,f = 75π/2π
= 235.5/6.28
= 37.5 Hz
3.
RMS current
= Imax/√2
= 1.00/1.414
= 0.7072A
4.
Reactance
= Vrms/ Irms
= 57.99/0.7072
= 81.999
= 82.0 Ω
5.
Reactance = 1/ ω x C
Reactance = 82
ω = 75π
We put these values into the equation above and make c the subject of the formula
C = 1/82.0 x 75π
C = 1/ 82.0 x 75 x 3.14
C = 1/19311
Capacitance = 5.18x10^-5F
Answer:
Option (1), option (4) and option (5)
Explanation:
The main observations of Ernest Rutherford's experiment are given below:
1. most of the positively charged particles pass straight, it means there is an empty space in the atom.
2. Very few positively charged particles retraces their path.
So,
The positively charged particles were deflected because like charges repel, that means they are deflected by protons.
Almost all the positively charge concentrate in a very small part which is called nucleus.
