Answer:
Acceleration due to gravity will be 
Explanation:
We have given length of pendulum l = 55 cm = 0.55 m
It is given that pendulum completed 100 swings in 145 sec
So time taken by pendulum for 1 swing 
We have to find the acceleration due to gravity at that point
We know that time period of pendulum;um is given by

So 

Squaring both side


So acceleration due to gravity will be 
Answer:
<h2> $1.50</h2>
Explanation:
Given data
power P= 2 kW
time t= 15 min to hours = 15/60= 1/4 h
cost of power consumption per kWh= 10 cent = $0.1
We are expected to compute the cost of operating the heater for 30 days
but let us computer the energy consumption for one day
Energy of heater for one day= 2* 1/4 = 0.5 kWh
the cost of operating the heater for 30 days= 0.5*0.1*30= $1.50
<u><em>Hence it will cost $1.50 for 30 days operation</em></u>
Correct question:
A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?
Answer:
the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
Explanation:
Given;
length of solenoid, L= 0.35 m
diameter of the solenoid, d = 0.04 m
current through the solenoid, I = 5.0 A
magnetic field in the center of the solenoid, 2.8 x 10⁻² T
The number of turns per meter for the solenoid is calculated as follows;

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
Answer:
is 3 and 2
Explanation: the firth one is 3 and the 2
Complete question:
A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?
Answer:
The current in the circuit 7 ms later is 0.2499 A
Explanation:
Given;
Ideal inductor, L = 45-mH
Resistor, R = 60-Ω
Ideal voltage supply, V = 15-V
Initial current at t = 0 seconds:
I₀ = V/R
I₀ = 15/60 = 0.25 A
Time constant, is given as:
T = L/R
T = (45 x 10⁻³) / (60)
T = 7.5 x 10⁻⁴ s
Change in current with respect to time, is given as;

Current in the circuit after 7 ms later:
t = 7 ms = 7 x 10⁻³ s

Therefore, the current in the circuit 7 ms later is 0.2499 A