Answer:
a) 112.5 m
b) 15.81s
Explanation:
a)We can use the following equation of motion to calculate the velocity v of the rocket at s = 500 m at a constant acceleration of a = 2.25 m/s2
After the engine failure, the rocket is subjected to a constant deceleration of g = -10 m/s2 until it reaches its maximum height where speed is 0. Again if we use the same equation of motion we can calculate the vertical distance h traveled by the rocket after engine failure
So the maximum height that the rocket could reach is 112.5 + 500 = 612.5 m
b) Using ground as base 0 reference, we have the following equation of motion in term of time when the rocket loses its engine:
t = 15.81 or t = -6.33
Since t can only be positive we will pick t = 15.81s
Answer:
(a)T= M2 × g, (b)T= (M1 + M2)g, (c)T= M2 (a + g) and (d)T=(M1 + M2) (a + g)
Explanation:
M1 is hanged upper and M2 is lower at Rest.
(a) For M2
T2 = Weight of the Body M2= M2 × g
(b) T1 = Weight of the Body M2 + Weight of the Body M2
T1 = M1 g + M2 g = (M1 + M2)g
M1 is hanged upper and M2 is lower at accelerated upwards ( F = T - W)
(c) For M2
⇒T = M2a + M2g = M2 (a + g)
(d) For M1
T = (M1 + M2) a + (M1 + M2) g
⇒ T = (M1 + M2) (a + g)
Answer:
is proved.
Explanation:
The magnetic field in the long current carrying wire is,
Here, I is the current, B is the magnetic field.
Now, by using cylindrical coordinates for the divergence of B.
Put the value of B in above equation.
Hence, it is prove that for a long current I carrying wire magnetic field divergence that is .
During the first phase of acceleration we have:
v o = 4 m/s; t = 8 s; v = 13 m/s, a = ?
v = v o + a * t
13 m/s = 4 m / s + a * 8 s
a * 8 s = 9 m/s
a = 9 m/s : 8 s
a = 1.125 m/s²
The final speed:
v = ?; v o = 13 m/s; a = 1.125 m/s² ; t = 16 s
v = v o + a * t
v = 13 m/s + 1.125 m/s² * 16 s
v = 13 m/s + 18 m/s = 31 m/s
Answer:
A.) a proton and an electron
