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MariettaO [177]

1 year ago

9

You are testing a new amusement park roller coaster with an empty car with a mass of 130 kg. One part of the track is a vertical

loop with a radius of 12.0 m. At the bottom of the loop (point A) the car has a speed of 25.0 m/s and at the top of the loop (point B) it has speed of 8.00 m/s.

1 answer:

vlada-n [284]1 year ago

6 0

Answer:

Work done by friction along the motion is given as

$W_f = -5857.8 J$

Explanation:

As per work energy theorem we can say

Work done by all forces = change in kinetic energy of the system

so here car is moving from bottom to top

so here the change in kinetic energy is total work done on the car

so here we will have

$W_f + W_g = \frac{1}{2}m(v_f^2 - v_i^2)$

$W_f - mgH = \frac{1}{2}m(v_f^2 - v_i^2)$

now plug in all data in it

$W_f - (130)(9.81)(2\times 12) = \frac{1}{2}(130)(8^2 - 25^2)$

$W_f = 30607.2 - 36465$

$W_f = -5857.8 J$

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Sort the phrases based on whether they represent kinetic energy or potential energy.

alexandr1967 [171]

Kinetic energy:
*the energy of a moving body*
rising water vapor.
the wings of flying hummingbird.
rolling marble.

Potential energy:
*the energy that is stored in a body so that any small change in position or state of the body, will result in this body movement by transforming all this potential energy into kinetic energy*
stone resting.
disconnected battery (the potential energy is what called voltage here)
stretched rubber band.

Hope this helps.

8 0

1 year ago

Read 2 more answers

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P

tatiyna

Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites $r_{sat1} = 48000 \times 10^{3}$ m

Distance of second satellites $r _{sat2} = 64000 \times 10^{3}$ m

Distance of charon $r_{c} = 19600 \times 10^{3}$ m

Time period of charon $T_{c} = 6.39$ days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

$T^{2} = r^{3}$

$\frac{T}{r^{\frac{3}{2} } } = constant$

For first satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} } }$

${T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat1} = 24.46$ days

For second satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} } }$

${T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat2} = 37.67$ days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

8 0

2 years ago

A capacitor, C1, consists of two parallel circular plates with radius R and separation of d. A second capacitor, C2, consists of

qaws [65]

Answer:

$\frac{E_1}{E_2}= 4$

Explanation:

Capacitance C is given by

$C= \frac{\epsilon_0A}{d}$

A= area of capacitor cross section

d= distance

therefore,

$C_1= \frac{\epsilon_0A_1}{d_1}$

A_1= πR^2

d_1= d

$C_2= \frac{\epsilon_0A_2}{d_2}$

A_= π(2R)^2

d_2 = 2d

$q= \frac{\epsilon_0A_1}{d_1}V_1$

threfore

$V_1= \frac{qd_1}{\epsilon_0A_1}$

and

$V_2= \frac{qd_2}{\epsilon_0A_2}$

also we know that E= V/d

⇒ $\frac{E_1}{E_2}= \frac{V_1}{V_2}\times\frac{d_2}{d_1}$

⇒ $\frac{E_1}{E_2}= \frac{qd_1}{\epsilon_0A_1}\times\frac{\epsilon_0A_2}{qd_2}\times\frac{d_2}{d_1}$

= A_1/A_2= $\frac{4R^2}{R^2}$ =4

therefore,

$\frac{E_1}{E_2}= 4$

5 0

1 year ago

An air-track glider undergoes a perfectly inelastic collision with an identical glider that is initially at rest. what fraction

DiKsa [7]

Refer to the diagram shown below.

The initial KE (kinetic energy) of the system is
KE₁ = (1/2)mu²

After an inelastic collision, the two masses stick together.
Conservation of momentum requires that
m*u = 2m*v
Therefore
v = u/2

The final KE is
KE₂ = (1/2)(2m)v²
= m(u/2)²
= (1/4)mu²
= (1/2) KE₁

The loss in KE is
KE₁ - KE₂ = (1/2) KE₁.

Conservation of energy requires that the loss in KE be accounted for as thermal energy.

Answer: 1/2

5 0

2 years ago

Read 2 more answers

A gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position. Why?

spin [16.1K]

Answer:

The body's rotational inertia is greater in layout position than in tucked position. Because the body remains airborne for roughly the same time interval in either position, the gymnast must have much greater kinetic energy in layout position to complete the backflip.

Explanation:

A gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position.

When the body is straight , its moment of rotational inertia is more than the case when he folds his body round. Hence rotational inertia ( moment of inertia x angular velocity ) is also greater. To achieve that inertia , there is need of greater imput of energy in the form of kinetic energy which requires greater effort.

So a gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position.

6 0

2 years ago