K.E. = 1/2 mv²
K.E. is directly proportional to v^2
So, when K.E. increase by 2, K.E. increase by root. 2
v' = 1.41v
original v value was 3 so, final would be:
v' = 1.41*3 = 4.23
After round-off to it's tenth value, it will be:
v' = 4.2
So, option B is your answer!
Hope this helps!
Explanation:
C,
.hahxxjdndjdndjgfndkndidjdodnxondos
Answer:
yes
Explanation:
using law of HC(heat capacity), which is
- heat loss=heat gain
- energy H=MCQ
Where M is mass of substance,C is specific heat capacity, and Q is temperature change
In case of two substance
- the H = Mc*Cc*Q+Mw*Cw*Q(provided the initial and final temperature are given)
<span>One end of a uniform meter stick is placed against a vertical wall. The other end is held by a lightweight cord that makes an angle, theta, with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.390. A. what is the maximum value...</span>
Answer: T= 715 N
Explanation:
The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:
T = mv² / r
At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.
So the kinetic energy will be the following:
K = 1/2 m v² = 15. 0 J
Solving for v², and replacing in the expression for T:
T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N
