Answer:
v_f = 0.87 m/s
Explanation:
We are given;
F_avg = -17700 N (negative because it's backward)
m = 117 kg
Δt = 5.50 × 10^(−2) s
v_i = 7.45 m/s
Now, formula for impulse is given by;
I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s
From impulse momentum theory, we know that;
Change in momentum of particle is equal to impulse.
Thus,
Δp = I = m•v_f - m•v_i
Thus,
-973.5= 117(v_f - 7.45)
Thus,
-973.5/117 = (v_f - 7.45)
-8.3205 + 7.45 = v_f
v_f = - 0.87 m/s
We'll take absolute value as;
v_f = 0.87 m/s
7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J
8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m
9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333
A 50w motor can do 500w in 5 seconds
Answer:
Radius of the solenoid is 0.93 meters.
Explanation:
It is given that,
The magnetic field strength within the solenoid is given by the equation,
, t is time in seconds
The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m
The electric field due to changing magnetic field is given by :
x is the distance from the axis of the solenoid
, r is the radius of the solenoid
r = 0.93 meters
So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.
