Answer:
R=V/I=6/2=3ohm
time =5minutes =5*60=300seconds
I=2A
Heat =I^2Rt=(2)^2*3*300=4*900=3600J
Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,

Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?

a = -1.47 m/
(a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/
(since same friction force is applied)

s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Answer:
Your answer should be Incentives
Explanation:
Gravitational field exists in
the space surrounding a charged particle and exerts a force on other charged
particles. Gravitational waves are ripples of waves travelling outward from the
source. The more massive the orbit of two bodies, the more it emits
gravitational wave. And everything around it that is near within the wave
experiences a ‘pull’ toward the orbiting bodies.