Answer:
Explanation:
a )
change in the gravitational potential energy of the bear-Earth system during the slide = mgh
= 45 x 9.8 x 11
= 4851 J
b )
kinetic energy of bear just before hitting the ground
= 1/2 m v²
= .5 x 45 x 5.8²
= 756.9 J
c ) If the average frictional force that acts on the sliding bear be F
negative work done by friction
= F x 11 J
then ,
4851 J - F x 11 = 756.9 J
F x 11 = 4851 J - 756.9 J
= 4094.1 J
F = 4094.1 / 11
= 372.2 N
We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"
- it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
From the question we are told
the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?
Generally the equation for the Force is mathematically given as
F=\frac{F}{dx}
Therefore
F=-kdx
k=600Nm^{-1}
now
K.E=0.5x ds^2
K.E=600*(-0.1^2)
K.E=3J
Therefore
the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
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Answer:
14 seconds
Explanation:
Distance covered by sports car is measured with 21*t where t is time.
Distance covered by police is measured with at^2/2(t-time, a-acceleration) as it doesnt have starting velocity. If the distances are equal(police catches sport car)
then we have 21t=3/2*t^2 Solving this equation we get t=14 & t=0;
excluding the starting point we have 14 seconds
Displacement is d
Vf² = Vi² + 2 g d
(-20²) = (+10²) + 2 (-9.8) d
-19.6 d = 300
d = -15.3 m
negative means lower
time is t
d = Vi t + 1/2 g t²
-15.3 = 10 t + (-4.9) t²
4.9 t² - 10 t -15.3 = 0
t = 3.06 s
Hope this helps -John
Answer:
v' = 1.5 m/s
Explanation:
given,
mass of the bullet, m = 10 g
initial speed of the bullet, v = 300 m/s
final speed of the bullet after collision, v' = 300/2 = 150 m/s
Mass of the block, M = 1 Kg
initial speed of the block, u = 0 m/s
velocity of the block after collision, u' = ?
using conservation of momentum
m v + Mu = m v' + M u'
0.01 x 300 + 0 = 0.01 x 150 + 1 x v'
v' = 0.01 x 150
v' = 1.5 m/s
Speed of the block after collision is equal to v' = 1.5 m/s
