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tatyana61 [14]
1 year ago
11

Bill has a mass of 85 kg and is skating west. He increases his speed from 3 m/s to 5 m/s by applying a force for 3 seconds. What

force did he apply?
Physics
1 answer:
likoan [24]1 year ago
4 0
F = ma
F = 85×(5-3)÷3
F = 85×(2÷3)
F = 85×0.667
F = 56.67N
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Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water wit
melamori03 [73]

Answer:

Explanation:

Given that,

Height of the bridge is 20m

Initial before he throws the rock

The height is hi = 20 m

Then, final height hitting the water

hf = 0 m

Initial speed the rock is throw

Vi = 15m/s

The final speed at which the rock hits the water

Vf = 24.8 m/s

Using conservation of energy given by the question hint

Ki + Ui = Kf + Uf

Where

Ki is initial kinetic energy

Ui is initial potential energy

Kf is final kinetic energy

Uf is final potential energy

Then,

Ki + Ui = Kf + Uf

Where

Ei = Ki + Ui

Where Ei is initial energy

Ei = ½mVi² + m•g•hi

Ei = ½m × 15² + m × 9.8 × 20

Ei = 112.5m + 196m

Ei = 308.5m J

Now,

Ef = Kf + Uf

Ef = ½mVf² + m•g•hf

Ef = ½m × 24.8² + m × 9.8 × 0

Ef = 307.52m + 0

Ef = 307.52m J

Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw

7 0
1 year ago
A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static f
Snezhnost [94]

Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

\mu mg = m\frac{v^2}{r}

where

\mu is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

v=\sqrt{\mu gr}

And by substituting the data of the problem, we find the speed at which the car begins to skid:

v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s

7 0
2 years ago
Read 2 more answers
Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley
kenny6666 [7]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : x ( acceleration of mass 4m ) = \frac{g}{7}

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

Explanation:

Given data:

mass suspended = 4 meters

mass suspended at other end = 3 meters

first we have to express the kinetic and potential energy equations

The general kinetic energy of the system can be written as

T = \frac{4m}{2} x^2  + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2

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U = -4mgx-3mg(-x+y)-mg(-x-y)+constant=-2mgy +constant

The Lagrangian of the problem can now be setup as

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next we will take the Euler-Lagrange equation for the generalized equations :

Euler-Lagrange  equation = 4x-y =0\\-2y+x +g = 0

solving the equations simultaneously

x ( acceleration of mass 4m ) = \frac{g}{7}

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

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Ierofanga [76]

1-125/500)x100=efficiency

1-1/4)x100 =75pc

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