Answer:
A) ( - 200t + 40 ) volts
B) b) anticlockwise , c) anticlockwise , d) clockwise , e) clockwise
Explanation:
Given data:
magnetic flux (Φm) = 5.0t^2 − 2.0t
number of turns = 20
<u>a) determine induced emf </u>
E = - N 
= - N ( 10t - 2 ) = - 20 ( 10t - 2 )
= - 200t + 40 volts
<u>b) Determine direction of induced current </u>
i) at t = 0
E = - 0 + 40 ( anticlockwise direction )
ii) at t = 0.10
E = -20 + 40 = 20 ( anticlockwise direction )
iii) at t = 1
E = - 200 + 40 = - 160 ( clockwise direction)
iv) at t = 2
E = -400 + 40 = - 360 ( clockwise direction )
First, foremost, and most critically, you must look at the graph, and critically
examine its behavior from just before until just after the 5-seconds point.
Without that ability ... since the graph is nowhere to be found ... I am hardly
in a position to assist you in the process.
Answer:
v ≈ 7900 m/s
Explanation:
centripetal force will equal gravity force
mv²/R = mg
v²/R = g
v² = Rg
v = √(Rg)
v = √(6.4e6(9.8))
v = 7.91959...e+3
v ≈ 7900 m/s
of course, at those velocities and that deep into the atmosphere, the satellite would quickly burn up, slow down, and cause tremendous damage to buildings etc. with the sonic boom shock wave. It would also have to avoid a lot of mountains as 4000 m is not that high.
Answer:
c) may also be conserved
Explanation:
Momentum is conserved in both elastic and inelastic type of collisions.
But the differences is that:
In an ELASTIC type of collisions, KINETIC ENERGY IS ALSO CONSERVED.
whereas, In an INELASTIC type of collision, KINETIC ENERGY IS NOT CONSERVED.
So unless until type of collision is specified, we can not say anything about the conservation of kinetic energy after collision.
Hence, may also be conserved is the appropriate option here.
Answer:
Explanation:
The force of gravity acting on the satellite is given by:
where
G is the gravitational constant
is the Earth's mass
m is the mass of the satellite
r is the distance of the satellite from the Earth's centre
Here we have
m = 700 kg
Substituting into the equation, we find:
<em>Note that the distance mentioned in the problem (2.4 x 10^6 meters) is not realistic, since it is less than the radius of the Earth (6.37 x 10^6 meters).</em>
