Answer:
40.25 V
Explanation:
Parameters given:
Initial magnetic field, B1 = 0.65 T
Final magnetic field, B2 = 0 T
Number of turns, N = 73
Radius of loop, r = 18 cm = 0.18 m
Time taken, Δt = 0.12 s
The magnitude of the EMF induced in a loop of wire due to a changing magnetic field is given as:
V = | [-N * A * (B2 - B1)] / Δt |
Where A = area of loop
Area of loop = πr²
Therefore, the magnitude of the EMF induced will be:
V = | [-N * πr² * (B2 - B1)] / Δt |
V = | [-73 * π * 0.18² * (0 - 0.65)] / 0.12 |
V = |40.25| V
V = 40.25 V
This is the magnitude of the EMF induced in the loop of wire.
<h3><u>Answer;</u></h3>
<em>Work = 125 joules </em>
<h3><u>Explanation and solution</u>;</h3>
- Work is the product of force and the distance covered. Therefore, Work = force × distance.
- Work is measured in joules.
- Work is also a change in energy, such that work is done when energy changes, so when kinetic energy, or potential energy changes the there is work being done.
Thus; kinetic energy = work done
Kinetic energy = 1/2mv²
= 1/2 × 10× 5²
= 5 × 25
= 125 joules
Hence, work done is 125 joules.
Answer:
a) dh/dt = -44.56*10⁻⁴ cm/s
b) dr/dt = -17.82*10⁻⁴ cm/s
Explanation:
Given:
Q = dV/dt = -35 cm³/s
R = 1.00 m
H = 2.50 m
if h = 125 cm
a) dh/dt = ?
b) dr/dt = ?
We know that
V = π*r²*h/3
and
tan ∅ = H/R = 2.5m / 1m = 2.5 ⇒ h/r = 2.5
⇒ h = (5/2)*r
⇒ r = (2/5)*h
If we apply
Q = dV/dt = -35 = d(π*r²*h/3)*dt
⇒ d(r²*h)/dt = 3*35/π = 105/π ⇒ d(r²*h)/dt = -105/π
a) if r = (2/5)*h
⇒ d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π
⇒ (4/25)(3*h²)(dh/dt) = -105/π
⇒ dh/dt = -875/(4π*h²)
b) if h = (5/2)*r
Q = dV/dt = -35 = d(π*r²*h/3)*dt
⇒ d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π
⇒ (5/2)*(3*r²)(dr/dt) = -105/π
⇒ dr/dt = -14/(π*r²)
Now, using h = 125 cm
dh/dt = -875/(4π*h²) = -875/(4π*(125)²)
⇒ dh/dt = -44.56*10⁻⁴ cm/s
then
h = 125 cm ⇒ r = (2/5)*h = (2/5)*(125 cm)
⇒ r = 50 cm
⇒ dr/dt = -14/(π*r²) = - 14/(π*(50)²)
⇒ dr/dt = -17.82*10⁻⁴ cm/s
Answer:
The question that Ellen most likely investigating is "How does distance affect the gravitational force on objects?"
Explanation:
