If you start skating down this hill, your potential energy will be converted to kinetic energy. At the bottom of the hill, your kinetic energy will be equal to your potential energy at the top. What will be your speed at the bottom of the hill?

Answers

Answer 1

Answer: Your potential energy at the top of the hill was (mass) x (gravity) x (height) .

Your kinetic energy at the bottom of the hill is (1/2) x (mass) x (speed)² .

If there was no loss of energy on the way down, then your kinetic energy
at the bottom will be equal to your potential energy at the top.

(1/2) x (mass) x (speed)² = (mass) x (gravity) x (height)

Divide each side by 'mass' :

(1/2) x (speed)² = (gravity) x (height) . . . The answer we get
will be the same for every skater, fat or skinny, heavy or light.
The skater's mass doesn't appear in the equation any more.

Multiply each side by 2 :

(speed)² = 2 x (gravity) x (height)

Take the square root of each side:

Speed at the bottom = square root of(2 x gravity x height of the hill)

We could go one step further, since we know the acceleration of gravity on Earth:

Speed at the bottom = 4.43 x square root of (height of the hill)

This is interesting, because it says that a hill twice as high won't give you
twice the speed at the bottom. The final speed is only proportional to the
square root of the height, so in order to double your speed, you need to
find a hill that's 4 times as high.

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Question 17 options:A 71.8 kg man goes from an area where the acceleration due to gravity is 9.79 m/s2 to an area where the acceleration due to gravity is 9.82 m/s2.What is the change (in kilograms), to the mass of the man? Respond to three significant digits expressed as a.bc and remember to record your responses for a, b, and c in that order.(i was thinking the answer would be 0 but im not sure)

Answers

Mass doesn't change, no matter where you take it.
Your first impression of ' 0 ' is totally correct.

A bicyclist steadily speeds up from rest to 11.0m/s in 3.40s. How far did she travel during this time?

Answers

Answer:

To find the distance traveled by the bicyclist during the given time, we can use the formula:

Distance = (Initial Velocity * Time) + (0.5 * Acceleration * Time^2)

Since the bicyclist starts from rest, the initial velocity is 0 m/s.

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 11.0 m/s

Time (t) = 3.40 s

Using the formula, we can calculate the distance traveled:

Distance = (0 * 3.40) + (0.5 * Acceleration * 3.40^2)

To find the acceleration, we can use the equation:

Acceleration = (Final Velocity - Initial Velocity) / Time

Acceleration = (11.0 - 0) / 3.40

Acceleration = 11.0 / 3.40

Now, we substitute the value of acceleration into the distance formula:

Distance = (0 * 3.40) + (0.5 * (11.0 / 3.40) * 3.40^2)

Simplifying further:

Distance = 0 + (0.5 * (11.0 / 3.40) * 11.56)

Distance = (0.5 * (11.0 / 3.40) * 11.56)

Distance = (0.5 * 11.0 * 3.40)

Distance = 0.5 * 37.4

Distance = 18.7 meters

Therefore, the bicyclist traveled a distance of 18.7 meters during the given time of 3.40 seconds.

should your directions to a friend for traveling from one city to another include displacements or distances?Explain

Answers

Displacements because it includes direction since you want to lead your friend to the right place. If you only include distances then it's like which way should they go. They don't know if it's north, east, south or west. They also don't know if they should travel in a straight line or do they have to make a lot of turns

You observe a distant galaxy. you find that a spectral line normally found in the visible part of the spectrum is shifted toward the infrared. what do you conclude?

Answers

The galaxy is moving away from you.

A graph depicts force versus position. What represents the work done by the force over the given displacement?A. The work done is equal to the product of the maximum force times the maximum position.
B. The work done is equal to the area under the curve.
C. Work cannot be determined from this type of graph.
D. The work done is equal to length of the curve.
E. The work done is equal to the slope of the curve.

Answers

Work = (force) x (distance)

With a force/distance graph, the work is the area under the graph.

If the graph is curved, broken, or some other screwy shape that makes it difficult to
get the area in any simple way, then work is the integral of the function represented
by the graph.

Final answer:

The work done by a force over a given displacement, as represented in a force versus position graph, is equal to the area under the curve.

Explanation:

In a force versus position graph, the work done by the force over the given displacement is represented by the area under the curve. The work done is the integral of the force with respect to displacement which, in a graphical representation, translates to the area under the curve of the force versus position graph. For example, if the force is constant, the graph will be a rectangle, and the work done will be the product of force (height of the rectangle) and displacement (width of the rectangle). If the force is variable, the area under the curve might need to be calculated by dividing it into small sections and summing up their areas.

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A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b and outer radius c. The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length λCalculate the electric field
(a) At any point between the cylinders a distance r from the axis and
(b) At any point outside the outer cylinder.
(c) Graph the magnitude of the electric field as a function of the distance r from the axis of the cable, from r = 0 to r = 2c.
(d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

Answers

Answer:

Part a)

$E = (\lambda)/(2\pi \epsilon_0 r)$

Part b)

$E = (\lambda)/(2\pi \epsilon_0 r)$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

$\int E. dA = (q)/(\epsilon_0)$

$E. 2\pi rL = (\lambda L)/(\epsilon_0)$

$E = (\lambda)/(2\pi \epsilon_0 r)$

Part b)

Outside the outer cylinder we will again use Guass law

$\int E. dA = (q)/(\epsilon_0)$

$E. 2\pi rL = (\lambda L)/(\epsilon_0)$

$E = (\lambda)/(2\pi \epsilon_0 r)$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Final answer:

The electric field between the cylinders is given by E = λ / (2πε₀r). The electric field outside the outer cylinder is zero due to the absence of net charge. Graph the electric field magnitude using the equation E = λ / (2πε₀r). The inner surface charge of the outer cylinder is -λ and the outer surface charge is 0.

Explanation:

To calculate the electric field between the cylinders at a distance r from the axis, you can use Gauss's Law. Since the charging is uniform, the electric field will also be uniform. Therefore, the electric field at any point between the cylinders is given by E = λ / (2πε₀r), where ε₀ is the permittivity of free space.

To calculate the electric field at any point outside the outer cylinder, you can use the principle of superposition. The electric field due to the outer cylinder is zero because it has no net charge. The electric field due to the inner cylinder can be calculated using the same formula as before.

To graph the magnitude of the electric field as a function of the distance r from the axis, you can plot the equation E = λ / (2πε₀r) for values of r ranging from 0 to 2c.

The charge per unit length on the inner surface of the outer cylinder is -λ, while the charge per unit length on the outer surface of the outer cylinder is 0. This is because the outer cylinder has no net charge and the inner cylinder has a uniform positive charge per unit length λ.

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