Answer:
Electrostatic Force...or a non-contact force
Explanation:
Answer:
Work done, W = 6 J
Explanation:
It is given that,
Force of gravity acting on the book, weight of the book is 15 N
We need to find the work done in lifting the book straight up for a distance of 0.4 meters.
The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

So, the magnitude of work done in lifting the book is 6 joules.
Answer:
4
Explanation:
In order for the current to continue flowing through the circuit (and for the bulbs to continue shining), there must be a closed path containing the battery where current can flow. Let's see the effect of removing each bulb on the circuit:
- 1: when removing bulb 1 only, the current can still flow through the path battery-bulb 3- bulb 4
- 2: when removing bulb 2 only, the current can still flow through the path battery-bulb 3- bulb 4
- 3: when removing bulb 3 only, the current can still flow through the path battery-bulb 1-bulb 2- bulb 4
- 4: when removing bulb 4 only, the current can no longer flow. In fact, there is no closed path that contains the battery now, so the current will not flow and all the bulbs will stop shining.
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