The force needed to keep the space shuttle moving at constant speed is 0.
The given parameters;
- <em>weight of the space shuttle, F = 750,000 N</em>
- <em>constant speed of the space shuttle, v = 28,000 km/h</em>
The mass of the space shuttle is calculated as follows;
The force needed to keep the space shuttle moving at constant speed is calculated as follows;
where;
a is the acceleration of the space shuttle
At a constant speed, acceleration is zero.
F = 76,530.61 x 0
F = 0
Thus, the force needed to keep the space shuttle moving at constant speed is 0.
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The magnitude of the induced emf is given by:
ℰ = |Δφ/Δt|
ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time
The magnetic field is perpendicular to the loop, so the magnetic flux φ is given by:
φ = BA
B = magnetic field strength, A = loop area
The area of the loop A is given by:
A = πr²
r = loop radius
Make a substitution:
φ = B2πr²
Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:
Δφ = ΔB2πr²
ΔB = change in magnetic field strength
Make another substitution:
ℰ = |ΔB2πr²/Δt|
Given values:
ΔB = 0.20T - 0.40T = -0.20T, r = 0.50m, Δt = 2.5s
Plug in and solve for ℰ:
ℰ = |(-0.20)(2π)(0.50)²/2.5|
ℰ = 0.13V
P=4800kgm/s
As
p=mΔv
where p is momentum, m is mass and v is velocity
Given values is
m =1200kg
Δv= 17m/s-13m/s=4m/s
Now
p=mΔv
p=(1200kg)*(4m/s)
p=4800kgm/s
The correct answer is B. Fuse.
Fuses were commonly used in houses up to recent ages, when circuit breakers became the more popular options, as they don't have to be changed every time they melt, since they don't melt.
