Answer:
a = (v2 - v1) / t
From A to B (8 - 4) m/s / 1 s = 4 m / s^2
From A to D ( 7 - 4) m/s / 5 s = .6 m / s^2
Note these equations hold for "uniform" values
They say nothing about the acceleration at intermediate points - the equation just says that his average speed increased from 4 m/s to 7 m/s during a 5 sec period
Answer:5250 N
Explanation: ig:iihoop.vince
The average power is 
Explanation:
First of all, we calculate the work done to accelerate the car; according to the work-energy theorem, the work done is equal to the change in kinetic energy of the car:
where
:
is the final kinetic energy of the car, with
m = 2000 kg is the mass of the car
v = 60 m/s is the final speed of the car
is the initial kinetic energy of the car, with
u = 30 m/s is initial speed of the car
Soolving:
Now we can find the power required for the acceleration, which is given by

where
t = 9 s is the time elapsed
Solving:

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Answer:
23.5 mV
Explanation:
number of turn coil 'N' =22
radius 'r' =3.00 cm=>
0.03m
resistance = 1.00 Ω
B= 0.0100t + 0.0400t²
Time 't'= 4.60s
Note that Area'A' = πr²
The magnitude of induced EMF is given by,
lƩl =ΔφB/Δt = N (dB/dt)A
=N[d/dt (0.0100t + 0.0400 t²)A
=22(0.0100 + 0.0800(4.60))[π(0.03)²]
=0.0235
=23.5 mV
Thus, the induced emf in the coil at t = 4.60 s is 23.5 mV