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inna [77]
4 years ago
12

At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large, compressed spring. Th

e spring with a force constant 42.0 N/cm and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 68.0 kg are pushed against the other end, compressing the spring 0.390 m. The sled is then released with zero initial velocity.What is the sled's speed when the spring returns toits uncompressed length?m/s
Physics
1 answer:
steposvetlana [31]4 years ago
4 0

The sleds speed when the spring returns toits uncompressed length is v = 0.03 m/s.

<u>Explanation</u>:

Given,

force constant = 42 N/cm = 0.42 N/m,   mass m = 68 kg, spring x = 0.39 m

The potential energy, U, stored in the spring is

                     U = 1/2 kx^2  

                       = 1 / 2 \times 0.42 \times (0.39)^2

                       = 0.032 J

All its potential energy has been converted into kinetic energy since it has a uncompressed length.

                    K = 1/2 mv^2

                     v = sqrt (2K / m)

                       = √(( 2 \times 0.032) / 68)

                    v = 0.03 m/s .

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A cylinder that is 20 cm tall is filled with water. If a hole is made in the side of the cylinder 5 cm below the top level, how
lara31 [8.8K]

Answer:

See description

Explanation:

This is an example where we need Tornicelli's law, which states that the horizontal speed of a fluid that starts falling from an orifice is the same speed that an object acquires from free-falling.

v = \sqrt{2gh}

we are given:

h_{cilinder} = 0.2 [m]\\h = 0.05 [m]\\d=0.15[m]

the horizontal velocity of the water at the start is:

v = \sqrt{2(9.8)(0.05)}=0.989949 [m/s]=1[m/s]

now we need to find the time for the water drops to fall d:

as the gravity is the only force interacting with the water we have:

y(t) = \frac{1}{2} g*t^2

replace for y = d

0.15 = \frac{1}{2} g*t^2=>t=\sqrt{\frac{2*0.15}{9.8}}=0.1749[s]

now that we have t we notice that there are no horizontal forces interacting with the water, so the horizontal position is given by:

x(t)=v*t

Finally, we replace v and t:

x(2.45) = 1*0.1749 = 0.1749 [m]=17.49[cm]

4 0
4 years ago
A car mass 600kg starts from rest moving uniform acceleration 0.2 m/s^2 after 60 seconds collides with stationary pick up van of
galben [10]

Answer:

the phenomenon of the system supports the principle of conservation of momentum.

Explanation:

The law of conservation of momentum says that:

Initial Momentum = Final Momentum

So, first we calculate initial momentum of the system:

Initial Momentum  = m₁u₁ + m₂u₂

where,

m₁ = mass of car = 600 kg

m₂ = mass of van = 400 kg

u₁ = Initial Speed of Car

For initial speed of car, we use:

Vf = Vi + at

Vf = 0 m/s + (0.2 m/s²)(60 s)

Vf = u₁ = 12 m/s

u₂ = Initial Speed of Van = 0 m/s

Therefore,

Initial Momentum  = (600 kg)(12 m/s) + (400 kg)(0 m/s)

Initial Momentum  = 7200 Ns   --------------- equation (1)

Now, for the final momentum:

Final Momentum  = m₁v₁ + m₂v₂

where,

v₁ = v₂ = Final Speed of Car and van (both are locked) = 7.2 m/s

Therefore,

Final Momentum = (600 kg)(7.2 m/s) + (400 kg)(7.2 m/s)

Final Momentum = 7200 Ns   ------------- equation (2)

Comparing equation (1) and (2):

Initial momentum = Final Momentum

<u>Hence, the phenomenon of the system supports the principle of conservation of momentum.</u>

5 0
3 years ago
When chemical bonds are formed, energy is _______the bond? a) Stored in b)Released from c)Not needed in Please help
miss Akunina [59]
A, chemical bondings would yield a positive enthalpy change, thus endothermic.
6 0
3 years ago
What does area under a velocity time graph represent
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Velocity hope that helps
7 0
3 years ago
A shot putter shoots a 7.3kg shot from rest to 14m/s in 1.5s. what was the average power?
sweet [91]

Answer: 477W

Explanation:

Given the following :

Mass (m) = 7.3kg

Initial Velocity (u) = 0

Final velocity (v) = 14m/s

time (t) = 1.5s

Power = workdone (W) / time (t)

The workdone can be calculated as the change in kinetic energy (KE) :

Recall ;

KE = 0.5mv^2

Therefore, change in KE is given by:

0.5mv^2 - 0.5mu^2

Change in KE = 0.5(7.3)(14^2) - 0.5(7.3)(0^2)

Change in KE = 715.4J

Therefore ;

Average power = Workdone / time

Workdone = change in KE = 715.4N

Average power = 715.4 / 1.5

Average power = 476.93333 W

= 477W

5 0
4 years ago
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