Answer:
Explanation:
each grid corresponding 0.1s⁻¹.
0.2grid unit = 0.2×0.1 =0.02s⁻¹
distance of the star from telescope
d = 1/p
d= 1/0.02= 50 par sec
1par sec = 3.26 light year
1 light year = 9.5×10¹²km
3.26ly=3.084×10¹³km
d= 50×3.084×10¹³=1.55×10¹⁵km
Answer:
F = 4000 N
Explanation:
given,
mass of rocket (M)= 5000 Kg
10 Kg gas burns at speed (m)= 4000 m/s
time = 10 s
average force = ?
at the end the rocket is at rest
by conservation of momentum
M v + m v' = 0
5000 x v - 10 x 4000 = 0
5000 v = 40000
v = 8 m/s
speed of rocket = 8 m/s
now,
we know
change in momentum = F x Δ t
F = 4000 N
Hence, the average force applied to the rocket is equal to F = 4000 N
Answer:
The displacement is 20 m northward.
The distance is 44 m
Explanation:
Let northward motion be in positive direction
Let southward motion be in negative direction
Dispalcement is the shortest path covered by the hockey puck at a specified direction.
The dispalcement of the hockey puck will be calculated as;
Displacement = 26 m northward - 12m southward + 6 m northward
Displacement = 20 m in northward direction
Distance is the total path covered by the hockey puck.
Distance = 26 m + 12m + 6 m
Distance = 44 m
Answer:
988.39 N/C
Explanation:
First, we start by finding the magnitude of the electric dipole moment, as it's going to be needed
p = 4.40*10^-9 * 3.10*10^-3
p = 1.364*10^-11 Cm
The charges are in a uniform electric field E whose direction makes an angle of 36.4° with the line connecting the charges.
Magnitude of torque exerted on the dipole,
t = 8.0×10−9 N.m
pEsinO = t, making E subject of formula
E = t / psinO
E = 8.0*10^-9/1.364*10^-11 (sin36.4)
E = 8.0*10^-9/1.364*10^-11 * (0.5934)
E = 8.0*10^-9 / 8.094*10^-12
E = 988.39 N/C
the magnitude of this field is E = 988.39 N/C
They are feeling weightless because there is no external contact force pushing nor pulling upon their body, in each case gravity is the only force acting upon their body.