Answer:
The speed of the car when load is dropped in it is 17.19 m/s.
Explanation:
It is given that,
Mass of the railroad car, m₁ = 16000 kg
Speed of the railroad car, v₁ = 23 m/s
Mass of additional load, m₂ = 5400 kg
The additional load is dropped onto the car. Let v will be its speed. On applying the conservation of momentum as :
v = 17.19 m/s
So, the speed of the car when load is dropped in it is 17.19 m/s. Hence, this is the required solution.
Answer:
1.04μT
Explanation:
Due to both wires have opposite currents, the magnitude of the total magnetic field is given by
I: electric current = 10A
mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2
r1: distance from wire 1 to the point in which B is measured.
r2: distance from wire 2.
The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m
By replacing in the formula you obtain:
hence, the magnitude of the magnetic field is 1.04μT
Answer:
a) h = 593.50 m
b) h₁₁ = 103 m
c) vf = 107.91 m/s
Explanation:
a)
We will use second equation of motion to find the height:
where,
h = height = ?
vi = initial speed = 0 m/s
t = time taken = 11 s
g = 9.81 /s²
Therefore,
<u>h = 593.50 m</u>
b)
For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:
where,
vf = final velocity at tenth second = v₁₀ = ?
t = 10 s
vi = 0 m/s
Therefore,
Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:
where,
h = height covered during last second = h₁₁ = ?
vi = v₁₀ = 98.1 m/s
t = 1 s
Therefore,
<u>h₁₁ = 103 m</u>
c)
Now, we use first equation of motion for complete motion:
where,
vf = final velocity at tenth second = ?
t = 11 s
vi = 0 m/s
Therefore,
<u>vf = 107.91 m/s</u>