<h3>
</h3><h3>Given</h3>
v = 20m\s
a = 3m\s^2
t = 4sec
Firstly we have to find u
a =
3m\s =
12m\s = 20 - u
20 - u = 12m\s
- u = -8
u = 8
Now we can easily find distance by using second equation of motion
s = ut + 1\2 at^2
s = 8(4) + 1\2(3)(16)
s = 32 + 24
s = 56
So distance is 56 m\s hope it helps
Answer:
5. 9GmM/(10R)
Explanation:
m is the mass of the satellite
M is the mass of the earth
W is the energy required to launch the satellite
Energy at earth surface = Potential energy (PE) + W
W = Energy at earth surface - Potential energy (PE)
But PE =
Therefore: W = Energy at earth surface -
Energy at earth surface (E) at an altitude of 5R =
But
Therefore:
W = E - PE
Answer:
C. 590 mph
Explanation:
Given:
- velocity of jet,
- direction of velocity of jet, east relative to the ground
- velocity of Cessna,
- direction of velocity of Cessna, 60° north of west
Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.
Refer the attached schematic.
So,
&
Now the vector of relative velocity of Cessna with respect to jet:
Now the magnitude of this velocity:
is the relative velocity of Cessna with respect to the jet.
Since everything in the circuit is in series .. .
-- The total resistance is (3 + 2) = 5 ohms.
-- The voltage across the 3-ohm resistor is 3/5 of the total voltage.
-- The voltage across the 2-ohm resistor is 2/5 of the total voltage.
(2/5) of (9 volts) = 18/5 = 3.6 volts .