Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The tension is 
b
The time taken is 
c
The position for maximum velocity is
S = 0
d
The maximum velocity is
Explanation:
The free body for this question is shown on the second uploaded image
From the question we are told that
The mass of the bob is 
The angle is 
The length of the string is 
The tension on the string is mathematically represented as

substituting values


The motion of the bob is mathematically represented as

=> 
Where
is the angular speed
and
is the phase change
At initial position S = 0
So 

Generally
can be mathematically represented as

Where T is the period of oscillation which i mathematically represented as

So



substituting values


Looking at the equation

We see that maximum velocity of the bob will be at S = 0
i. e 
The maximum velocity is mathematically represented as

Where A is the amplitude which is mathematically represented as

So

Recall 
substituting values
Answer:
16/22
Explanation:
you add red and blue together
Answer:
16.8ohms
Explanation:
According to ohm's law which states that the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.
Mathematically, V = IRt where;
V is the voltage across the circuit
I is the current
R is the effective resistance
For a series connected circuit, same current but different voltage flows through the resistors.
If the initial current in a circuit is 19.3A,
V = 19.3R... (1)
When additional resistance of 7.4-Ω is added and current drops to 13.4A, our voltage in the circuit becomes;
V = 13.4(7.4+R)... (2)
Note that the initial resistance is added to the additional resistance because they are connected in series.
Equating the two value of the voltages i.e equation 1 and 2 to get the resistance in the original circuit we will have;
19.3R = 13.4(7.4+R)
19.3R = 99.16+13.4R
19.3R-13.4R = 99.16
5.9R = 99.16
R= 99.16/5.9
R = 16.8ohms
The resistance in the original circuit will be 16.8ohms
Answer:
0.2 T
Explanation:
Magnetic field is inversely proportional to the distance from wire since the distance is halved therefore magnetic field will be doubled.
You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.
ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
<em>ΔH = 4,373.04 J</em>