A rock is pulled back in a slingshot as shown in the diagram below. The elastic on the slingshot is displaced 0.2 meters from its initial position. The rock is pulled back with a force of 10 newtons.When the rock is released, what is its kinetic energy?

Answers

Answer 1

Answer:

The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.

What will be the speed of the rock?

Initial speed of the rock, u = 40m/s

Final position of the rock s = 0m taking the release point as reference. The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.

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Initial speed of the rock, u = 40m/s

Final position of the rock s = 0m taking the release point as reference

From the second equation of motion:

solving above we get:

t = 0s or t = 8.16s, t =0 seconds is neglected since it represents the initial position which is the same as the final position at t = 8.16s

So, the rock takes 8.16 seconds to return to the release point.

Therefore, The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.

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Answer 2

Answer:

Answer:

Explanation:

i don't know

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As a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 2.43 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 24 m/s. One horsepower equals 746 W.

Answers

Answer:

F=75.53N

Explanation:

To calculate the power we define the equation,

$P=Fv$

Where,

F= Force

V= Velocity,

Here we have that 2.43hp is equal to 1812.78W,

clearing F,

$F=(P)/(v) = (1812)/(24)\nF=75.53N$

A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?

Answers

Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

potential energy of the mass is equal to the energy stored in the spring

m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the balloon is 0.18 kg/m3. The radius of the balloon (when filled) is R = 4.9 m. The total mass of the empty balloon and basket is mb = 121 kg and the total volume is Vb = 0.073 m3. Assume the average person that gets into the balloon has a mass mp = 73 kg and volume Vp = 0.077 m3. 1)What is the volume of helium in the balloon when fully inflated? m3 2)What is the magnitude of the force of gravity on the entire system (but with no people)? Include the mass of the balloon, basket, and helium. N

Answers

Answer:

1) The volume of helium in the ballon when is fully inflated is 492.8070 m³

2) The magnitude of the force of gravity (with no people) is 869.3119 N

Explanation:

Given:

ρair = density of air = 1.28 kg/m³

ρhelium = density of helium = 0.18 kg/m³

R = radius of balloon = 4.9 m

mtotal = 121 kg

Vtotal = 0.073 m³

mp = average mass per person = 73 kg

Vp = 0.077 m³

g = gravity = 9.8 m/s²

Questions:

1) What is the volume of helium in the balloon when fully inflated, Vhelium = ?

2) What is the magnitude of the force of gravity on the entire system (but with no people), Fg = ?

1) The volume of helium in the ballon when is fully inflated

$V_(helium) =(4)/(3) \pi R^(3) =(4)/(3) \pi *4.9^(3) =492.8070m^(3)$

2) First, you need to calculate the mass of helium

$m_(helium) =\rho _(helium) *V_(helium) =0.18*492.8070=88.7053kg$

The magnitude of the force of gravity (with no people)

$F_(g) =m_(helium) *g=88.7053*9.8=869.3119N$

A car with tires of radius 0.25 m come to a stop from 28.78 m/s (100 km/hr) in 50.0 m without any slipping of tires. Find: (a) the angular acceleration of the wheels; (b) number of revolutions made while coming to rest.

Answers

Answer:

The answer is below

Explanation:

a) Using the formula:

$\omega^2=\omega_o^2+2\alpha \theta\n\n\omega=final\ angular\ velocity,\omega_o=initial\ anglular\ velocity,\alpha= angular\ acceleration,\n\theta=angular\ distance\n\nGiven\ that:\n\ninitial\ velocity(u)=28.78m/s,distance(s)=50\ m,radius(r)=0.25\ m,\nfinal/ velocity(v)=0(stop)\n\n\omega=v/r=(28.78m/s)/(0.25m) =115.12\ rad/s,\omega_o=0,\theta=s/r=(50\ m)/(0.25\ m)=200\ rad\n \n\omega^2=\omega_o^2+2\alpha \theta\n\n115.12^2=0^2+2\alpha(200)\n\n2\alpha(200)=13252.6144\n\n\alpha=33.13\ rad/s^2$

$\theta=200\ rad=200\ rad*(1\ rev)/(2\pi\ rad)=31.83\ rev$

A boy and a girl are pulling a heavy crate at the same time with 7 units of firce each. What is the net force acts on the ibject? Is the object balanced or unbalanced?

Answers

Answer:

Net force= 14 units

The object is unbalanced

Explanation:

The net force refers to the sum of all forces applied to an object. However, the direction of force applied determine the net force. In this question, a boy and girl is pulling a heavy crate at the same time.

This means that the force is in the same direction, hence, the net force will be:

F(N) = 7 + 7 = 14 unit

However, since the pull is occuring at the same direction. This means that the object has a net force, therefore, will move in a particular direction. This means that the OBJECT IS UNBALANCED

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