Answer:
The acceleration of the car, a = -3.75 m/s²
Explanation:
Given data,
The initial velocity of the airplane, u = 75 m/s
The final velocity of the plane, v = 0 m/s
The time period of motion, t = 20 s
Using the I equations of motion
v = u + at
a = (v - u) / t
= (0 - 75) / 20
= -3.75 m/s²
The negative sign indicates that the plane is decelerating
Hence, the acceleration of the car, a = -3.75 m/s²
Answer:
v = 50.5 m/s
Explanation:
F = (m)(^v/^t)
115N = (0.04551kg)(v/(0.020s))
2,526.917161 m/s² = v/(0.020s)
v = 50.53834322 m/s
v = 50.5 m/s
Answer:
t = 5.56 ms
Explanation:
Given:-
- The current carried in, Iin = 1.000002 C
- The current carried out, Iout = 1.00000 C
- The radius of sphere, r = 10 cm
Find:-
How long would it take for the sphere to increase in potential by 1000 V?
Solution:-
- The net charge held by the isolated conducting sphere after (t) seconds would be:
qnet = (Iin - Iout)*t
qnet = t*(1.000002 - 1.00000) = 0.000002*t
- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:
V = k*qnet / r
Where, k = 8.99*10^9 ..... Coulomb's constant
qnet = V*r / k
t = 1000*0.1 / (8.99*10^9 * 0.000002)
t = 5.56 ms
Answer:
do not worry bro you will know how to use it
Answer:
1m= 3.2 feets
? = 5280 feets
let ? be n
cross multiply
3.2n = 1×5280
n= 5280÷3.2
n= 1650
so, 5280 feets is equal to 1650 metres
