Answer: 22.6 hours
Explanation:
The power is the measure of the rate of energy.
In this problem, the 12.0 V battery is rated at 51.0 Ah, which means it delivers 51.0 A of current in a time of t = 1 h = 3600 s. The power delivered by the battery can be written as
where
I is the current
V = 12.0 V is the voltage of the battery
So the energy delivered by the battery can be written as
Where
So the energy delivered is
At the same time, the headlight consumes 27.0 W of power, so 27 Joules of energy per second; Therefore, it will remain on for a time of:
I’m am a mother no drama !!!!
Boom
F=ma
a=F/m
=825N/75kg
=825kg*m/75kg*s^2
=11m/s^2 in the direction of the force (ans)
Answer:
0.21%
Explanation:
We are given;
Mass; m = 100 kg
Diameter; d = 2.2 mm = 2.2 × 10^(-3) m
Young's modulus; E = 12.5 x 10^(10) N/m².
Formula for area is;
A = πd²/4
A = (π/4) x (2.2 x 10^(-3))²
A = 3.8 x 10^(-6) m²
Force; F = mg
g is acceleration due to gravity and has a constant value of 9.8 m/s²
F = 100 × 9.8
F = 980 N
Formula for young's modulus is;
E = Stress/strain
Formula for stress = F/A
Formula for strain = ΔL/L
Thus;
E = (F/A)/(ΔL/L)
Making ΔL/L the subject, we have;
ΔL/L = (F/A)/E
Plugging in the relevant values;
ΔL/L = 980/(3.8 x 10^(-6) × 12.5 × 10^(10))
ΔL/L = 0.0021
Then percentage increase in length of a wire = 0.0021 × 100% = 0.21%
Answer:
Intensity of the next bright fringe will remain same.
Explanation:
The question is based on Young's double slit experiment, since its about bright fringe, the interference here is constructive.
Young's condition for constructive interference is given by:
dsin
where,
d = slits distance from eachother or width of the slits
= wavelength
n = interferance order
Also, we know that in Young's experiment, the fringe intensity is given by:
where,
= phase difference
Therefore, in absence of phase difference i.e., 
, the intensity of the next bright fringe will not change and it will remain same.
