Answer:
Explanation:
Wheel completes four revolution.
The linear displacement is zero.
The angular displacement is 4 x 2π = 8π radian.
So, option (c) is correct.
The formula is Ke = 1/2 m v^2
The two of them together have a Ke of mv^2. So you either increase m or v. That's what makes the problem difficult. He can do D or B. We have to choose.
A is no solution. The Ke goes down because Paul loses Ivan's mass.
C is out of the question 3 meters/sec is a big reduction from 5 m/s. So now what do we do about B and D?
The question is what does the third person add. The tandoms I've peddled only allow for 1 or 2 people to add to the motion. So the third person only adds mass. He does not have a v that he is contributing to. To say that he is going 5m/s is true, but he's not contributing anything to that motion.
I pick B, but it is one of those questions that the correctness of it is in the head of the proposer. Be prepared to get it wrong. Argue the point politely if you agree with me, but back off as soon as you have presented your case.
B <<<<====== answer.
Answer:
Explanation:
total weight acting downwards
= 3g + 10g
13 g
volume of lead = 10 / 11.3 = .885 cm³
Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g
Buoyant force on lead = .885 x 1 x g
total buoyant force = vg + .885 g
For floating
vg + .885 g = 13 g
v = 12.115 cm³
total volume of bobber
= 4/3 x 3.14 x 2³
= 33.5 cm³
fraction of volume submerged
= 12.115 / 33.5
= .36
= 36 %
Answer:
The skater's speed after she stops pushing on the wall is 1.745 m/s.
Explanation:
Given that,
The average force exerted on the wall by an ice skater, F = 120 N
Time, t = 0.8 seconds
Mass of the skater, m = 55 kg
It is mentioned that the initial sped of the skater is 0 as it was at rest. The change in momentum of skater is :
The change in momentum is equal to the impulse delivered. So,
So, the skater's speed after she stops pushing on the wall is 1.745 m/s.
Answer:
The heat transferred from water to skin is 6913.5 J.
Explanation:
Given that,
Weight of water = 25.0 g
Suppose that water and steam, initially at 100°C, are cooled down to skin temperature, 34°C, when they come in contact with your skin. Assume that the steam condenses extremely fast. We will further assume a constant specific heat capacity c=4190 J/(kg°K) for both liquid water and steam.
We need to calculate the heat transferred from water to skin
Using formula for stream
Put the value into the formula
Hence, The heat transferred from water to skin is 6913.5 J.
