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PIT_PIT [208]
3 years ago
10

A 25 kg child is bouncing on a trampoline, which can be treated as an ideal spring with spring constant 2900 N/m. If the trampol

ine compresses to a maximum of 0.38 m during the bouncing, what is the maximum height that she reaches above the uncompressed surface of the trampoline? You may ignore friction.
Physics
1 answer:
White raven [17]3 years ago
5 0

Answer:

Explanation:

Energy store in the compressed trampoline = potential energy of at the maximum height ignoring friction

energy stored in the trampoline = \frac{1}{2} kx^{2} where k is spring constant and x is the distance compressed  = 0.5 × 2900 × 0.38² = 209.38 J

209.38 J = 25 × 9.8 × h

h maximum height attain = 209.38 J / ( 25 × 9.8) = 0.855 m

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A cart moves with negligible friction or air resistance along a roller coaster track. The cart starts from rest at the top of a
lina2011 [118]

Answer:

hinit = 17.5 m

Explanation:

  • Assuming no friction present, the mechanical energy must be conserved, which means that at any point of the trajectory, the sum of the gravitational potential energy and the kinetic energy must keep the same.
  • At the top of the hill, since it starts from rest, all the energy must be potential, and we can express it as follows:

       E_{o} = U_{o} = m*g*h_{init}  (1)

  • When the car arrives to the top of the second hill, as we know that it is lower than the first one, the energy of the car, must be part gravitational potential energy, and part kinetic energy.
  • We can express this final energy as follows:

       E_{f} = U_{f} + K_{f}  = m*g* h_{2} + \frac{1}{2} *m*v_{f} ^{2}  (2)

  • In order to find hinit, we need to make (1) equal to (2), and solve for it.
  • In (2) we have the value of h₂ (10 m), but we still need the value of the speed at the top of the second hill, vf.
  • Now, when the car is at the top of the hill, there are two forces acting on it, in opposite directions: the normal force (upward) and the weight (downward).
  • We know also that there is a force that keeps the car along the circular track, which is the centripetal force.
  • This force is just the net downward force acting on the car (it's vertical at the top), and is just the difference between the weight and the normal force.
  • If the cart just barely loses contact with the track at the top of the second hill, this means that at that point the normal force becomes zero.
  • So, the centripetal force must be equal to the weight.
  • The centripetal force can be expressed as follows:

       F_{c} = m*\frac{v_{f} ^{2}}{R}  (3)

  • We have just said that (3) must be equal to the weight:

       F_{c} = m*\frac{v_{f} ^{2}}{R} = m*g (4)

  • Simplifying, and rearranging, we can solve for vf², as follows:

       v_{f}^{2} = R*g  (5)  

  • Replacing (5) in (2), simplifying and rearranging in (1) and (2) we finally have:

      h_{init} = h_{2} + \frac{1}{2} R = 10m + 7.5 m = 17.5 m (6)

7 0
3 years ago
Physical science tries to a. Survey all of science b. Describe the environment and interaction between living things c. Describe
RideAnS [48]

Answer:

option "c" is correct

Explanation:

Physical science or Physics is the study of matter, energy and their mutual relationship.

7 0
3 years ago
PLEASE HELP
bixtya [17]

Answer:

AS- X 3.42 Y 3. B) X Y c) x Ross TET V V. a 1.71 1.71 LLL LLL 2.42 N al

Explanation:

3 0
3 years ago
Read 2 more answers
A concave mirror has a radius of curvature of 10 cm. Find the location an height of the image if the distance of the object is 8
puteri [66]
R=10
F=R/2
F=10/2=5
F=-5(CONCAVE MIRROR)
U=-8(CONCAVE MIRROR)
HEIGHT OF OBJECT=1.5
V=?
HEIGHT OF IMAGE=?
I/F=1/U+1/V
-I/5=-1/8-1/V
-1/V=-1/5+1/8
-1/V=-8+5/40
-1/V=-3/40
1/V=3/40
V=40/3

HEIGHT OF IMAGE/HEIGHT OF OBJECT =-V/U
HEIGHT OF IMAGE=40/3*1/-8*15/10
                              =-20/8
                              =-2.5
7 0
3 years ago
A race car is one lap behind the lead race car when the lead car has 55 laps to go in a race. If the speed of the lead car is 62
ella [17]

Answer:

The second car must go with a speed of 63.43 m/sec

Explanation:

Speed V of lead car = 62.3 m/sec

Distance S =   55 laps = 55 ×400 meters=22000 m

We know

S = V × t

So,

t= S/V

We put values of S and V here, we get

t=22000/62.3

t= 353.1 sec

So in 353.1 sec the second car which is one lap behind - must go a distance of 55+1=56 laps or 56×400 m = 22400 meters to catch the lead car before it finishes.

i-e for second car

Distance S= 22400m

Time t = 353.1 sec

V= ?

using again

S=Vt

we get

V= S/t

V= 22400/353.1= 63.43 m/sec

7 0
4 years ago
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