<span>1 w = 1N/s the change in total energy =mgh = 600*4 = 2400N. 2400N/200N/s = 12seconds
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<span>Short-range forecasts are more accurate than longer range ones. Short term forecasts may use mathematical techniques such as moving averages and, exponential smoothing. Longer term forecasts not only use different methodologies, such as qualitative vs. quantitative, they also tend to consider different issues.</span>
As we know that range of the projectile motion is given by
here we know that range will be same for two different angles
so here we can say the two angle must be complementary angles
so the two angles must be
so it is given that one of the projection angle is 75 degree
so other angle for same range must be 90 - 75 = 15 degree
so other projection angle must be 15 degree
Answer:
a) I = 13.04 A
b) R = 8.82 ohms
c) 1291.87 kilocalories are generated an hour.
Explanation:
let P be the power of the heater, V be the voltage of the heater, I be the current of the heater, R be the resistance.
a) we know that:
P = I×V
I = P/V
= (1500)/(115)
= 13.04 A
Therefore, the current of the heater is 13.04 A
b) we now have voltage and current, according to Ohm's law:
R = V/I
= (115)/(13.04)
= 8.82 ohms
Therefore, the resistance of the heating coil is 8.82 ohms.
c) the number of kilocalories generated in one hour by the heater is just the energy the heater produces in one hour which is given by:
E = P×t
= (1500)(1×60×60)
= 5400000 J
since 1 calorie = 4.81 J
1 kilocalorie = 0.001 calories
E = 5400000/4.18 ≈ 1291866.029 calories ≈1291.87 kilocalories
Therefore, 1291.87 kilocalories are produced/generated in one hour.
Answer:
53.63 μA
Explanation:
radius of solenoid, r = 6 cm
Area of solenoid = 3.14 x 6 x 6 = 113.04 cm^2 = 0.0113 m^2
n = 17 turns / cm = 1700 /m
di / dt = 5 A/s
The magnetic field due to the solenoid is given by
B = μ0 n i
dB / dt = μ0 n di / dt
The rate of change in magnetic flux linked with the solenoid =
Area of coil x dB/dt
= 3.14 x 8 x 8 x 10^-4 x μ0 n di / dt
= 3.14 x 64 x 10^-4 x 4 x 3.14 x 10^-7 x 1700 x 5 = 2.145 x 10^-4
The induced emf is given by the rate of change in magnetic flux linked with the coil.
e = 2.145 x 10^-4 V
i = e / R = 2.145 x 10^-4 / 4 = 5.36 x 10^-5 A = 53.63 μA
