B would be the correct answer
Answer:
0 J
Explanation:
given,
mass of the ball = 5 kg
radius of the horizontal circle = 0.5 m
tension in the string = 10 N
Work done = ?
Work done by the tension in the circular path will be equal to zero.
This is because body moves in circular path, the centripetal force act along the radius of the circle and motion is right angle to the tension on the string.
so, work done = F s cos θ
θ = 90°,
work done = F s cos 90° ∵ cos 90° = 0
Work done = 0 J
Answer:
, the minus meaning west.
Explanation:
We know that linear momentum must be conserved, so it will be the same before (
) and after (
) the explosion. We will take the east direction as positive.
Before the explosion we have 
.
After the explosion we have pieces 1 and 2, so 
.
These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.
Since we know momentum must be conserved we have:
Which means (since we want 
and 
):
So for our values we have:
Answer:
3°C
Explanation:
We can that heat Q=m
dT
Where m is the mass = specific heat capacity
dT = Temperature difference
here we have given m=625 g =.625 kg
specific heat of granite =0.79 J/(g-K) = 0.79 KJ/(kg-k)
=25°C
we have to find
we have also given Q=10.9 KJ
10.9=0.625×0.79×(25-)
25- =22
=3°C
Answer:
largest lead = 3 m
Explanation:
Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.
So, first we need to calculate the velocities of both the anchorman
given data:
Distance = d = 100 m
Time arrival for A = 9.8 s
Time arrival for B = 10.1 s
Velocity of anchorman A = D / Time arrival for A
=100/ 9.8 = 10.2 m/s
Velocity of anchorman B = D / Time arrival for B
=100/10.1 = 9.9 m/s
As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use
d = vt
= 9.9 x 9.8 = 97 m
So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.
largest lead = 100 - 97 = 3 m
So if his lead no more than 3 m anchorman A win the race.
