Answer:
Explanation:
mass of 1 L water = 1 kg .
200⁰F = (200 - 32) x 5 / 9 = 93.33⁰C .
260.928 K = 260.928 - 273 = - 12.072⁰C .
water is at higher temperature .
Let the equilibrium temperature be t .
Heat lost by water = mass x specific heat x fall of temperature
= 1 x 4.2 x 10³ x ( 93.33 - t )
Heat gained by copper
= .25 x .385 x 10³ x ( t + 12.072 )
Heat lost = heat gained
1 x 4.2 x 10³ x ( 93.33 - t ) = .25 x .385 x 10³ x ( t + 12.072 )
93.33 - t = .0229 ( t + 12.072)
93.33 - t = .0229 t + .276
93.054 = 1.0229 t
t = 90.97⁰C .
Answer:
As the capacitor is discharging, the current is increasing
Explanation:
Lets take
C= Capacitance
L=Inductance
V=Voltage
I= Current
The total energy E given as
We know that total energy E is conserved so when electric energy 1/2 CV² decreases then magnetic energy 1/2 IL² will increases.
It means that when charge on the capacitor decreases then the current will increase.
As the capacitor is discharging, the current is increasing
Answer:
what
Explanation:
Racket sports include tennis, badminton, squash or any other sport where you use rackets to hit a ball or shuttlecock to play. They can be played competitively or just for fun and are a great form of physical activity.
Answer:
v = 10 m/s
Explanation:
given,
Mass of Mercedes engine = 2000 Kg
Power delivered = 100 kW
angle made with horizontal = 30°
acceleration due to gravity = 10 m/s²
largest speed car can sustain = ?
we know,
Power = Force x velocity
P = F x v
P = mg sinθ x v
P = mg sin 30° x v
P = 0.5 mg x v
v = 10 m/s
hence, the maximum velocity is equal to v = 10 m/s
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
