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postnew [5]
3 years ago
7

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?
Physics
1 answer:
baherus [9]3 years ago
6 0
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:

<u>Speed at the bottom = square root of(2 x gravity x height of the hill)</u>

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
<em>square root </em>of the height, so in order to double your speed, you need to
find a hill that's <em>4 times</em> as high.






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8 0
3 years ago
How long does it take a car to cross a 20m bridge if it starts from rest and accelerates at 5 m/s^2?
polet [3.4K]

The correct answer is 2.8s

5 0
4 years ago
brainly a child on a swing set swings back and forth with a period of 3.3 s and an amplitude of 25°. what is the maximum speed o
beks73 [17]

Maximum speed of the child as she swings is  2.23 m/s.

<h3>Step by Step Calculation:</h3>

T=3.3 s is the oscillation's time period.

The swing's greatest angle is 25° (max).

The swing's bottom will have the following kinetic energy:

k=12mv2...........(1)

The mass in this situation is m, and the speed is v.

The potential energy change is expressed as,

∆U=mgL1-cosθmax...............(2)

Here, L is a string's length and g is the acceleration caused by gravity. L is given as,

L=gT24π2

Combine equation (1) with (2)

12mv2=mgL1-cos, maxv=2g, maxv=gT24, maxv=g2T22, maxv=9.8 m/s

22.33 m/s, 22.31 s22.31 cos25°

Therefore, the child's top speed is 2.23 m/s.

<h3>What is Oscillation ?</h3>
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6 0
1 year ago
With the switch open, roughly what must be the resistance of the resistor on the right for the current out of the battery to be
Delvig [45]

Answer:

The resistance must be 6.67\Omega

Solution:

Resistance, R_{1} = 20\Omega

Resistance, R_{2} = 10\Omega

For the current to be the same when the switch is open or closed, the resistances must be connected in parallel as current is distributed in parallel with the same voltage across the circuit:

Thus in parallel:

\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}

\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{10}

R_{eq} = 6.67\ \Omega

4 0
3 years ago
A 70 ft rope hangs from a helicopter above this room. The rope has a mass per unit length of 2 lb/ft. In order to be rescued fro
Mrac [35]

Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope  = 2 lb/ft x 70 ft

                                         = 140 lbs.

Total mass to be pulled to the helicopter, M = 120 lbs  + 140 lbs  

                                                                       = 260 lbs

The work done is calculated from work-energy theorem as follows;

W = Mgh

where;

g is acceleration due gravity = 32.17 ft/s²

h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s²  x 70 ft

W = 585494 lb.ft²/s²

1 lb.ft²/s² = 0.0421 J

W = 585494 lb.ft²/s²  = 2.47 X 10⁴ J.

Therefore, the work done to get you safely away from the test is  2.47 X 10⁴ J.

4 0
3 years ago
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