Answer:
Explanation:
First, let's find the voltage through the resistor using ohm's law:
AC power as function of time can be calculated as:
(1)
Where:
Because of the problem doesn't give us additional information, let's assume:
Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):
Answer:
a) 12.8212 N
b) 12.642 N
Explanation:
Mass of bucket = m = 0.54 kg
Rate of filling with sand = 56.0 g/ sec = 0.056 kg/s
Speed of sand = 3.2 m/s
g= 9.8 m/sec2
<u>Condition (a);</u>
Mass of sand = Ms = 0.75 kg
So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg
== > total weight = 1.29 × 9.8 = 12.642 N
Now impact of sand = rate of filling × velocity = 0.056 × 3.2 = 0.1792 kg. m /sec2=0.1792 N
Scale reading is sum of impact of sand and weight force ;
i-e
scale reading = 12.642 N+0.1792 N = 12.8212 N
<u>Codition (b);</u>
bucket mass + sand mass = 0.54 +0.75=1.29 kg
==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)
Answer:he formula for average speed is (total distance/total time)
the y-component does not matter in this problem. so do 6.26(cos45)=4.43m/s to find the x-component velocity which is constant throughout the duration of the flight. the total distance is 2L because he travels distance L twice.
the total time is ((time in water)+(time out of water)) since you dont have time you must eliminate it. to do this you need (distance)/(time)=velocity
solve for time and you get T=D/V
time in water is L/3.52 and time out of water is L/4.43
add them together and you get (4.43L+3.52L)/(15.59) = 7.95L/15.59
that value is your total time
divide you total distance (2L) by total time (7.95L/15.59) and the Ls cancel out and you get
(31.18)/(7.95) = 3.92 m/s = Average Speed
Explanation:
I claim that it's d moving electrons.
Answer:
8050 J
Explanation:
Given:
r = 4.6 m
I = 200 kg m²
F = 26.0 N
t = 15.0 s
First, find the angular acceleration.
∑τ = Iα
Fr = Iα
α = Fr / I
α = (26.0 N) (4.6 m) / (200 kg m²)
α = 0.598 rad/s²
Now you can find the final angular velocity, then use that to find the rotational energy:
ω = αt
ω = (0.598 rad/s²) (15.0 s)
ω = 8.97 rad/s
W = ½ I ω²
W = ½ (200 kg m²) (8.97 rad/s)²
W = 8050 J
Or you can find the angular displacement and find the work done that way:
θ = θ₀ + ω₀ t + ½ αt²
θ = ½ (0.598 rad/s²) (15.0 s)²
θ = 67.3 rad
W = τθ
W = Frθ
W = (26.0 N) (4.6 m) (67.3 rad)
W = 8050 J
