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Delicious77 [7]
3 years ago
11

Apply the law of conservation of energy and describe the energy transformations that occur as you coast down a long hill on a bi

cycle and then apply the brakes to make the bike stop at the bottom.
A roller coaster is at the top of a hill and rolls to the top of a lower hill. If mechanical energy is constant, then on the top of which hill is the kinetic energy from the roller coaster`s motion greater?
Physics
1 answer:
m_a_m_a [10]3 years ago
4 0
As you coast down a long hill on your bicycle, potential energy from your height is converted to kinetic energy as you and your bike are pulled downward by gravity along the slope of the hill. While there is air resistance and friction slowing you down by a little bit, your speed increases gradually until you apply the brakes, causing enough friction to slow yourself and the bike to a stop at the bottom.

A roller coaster will have higher kinetic energy at the lower hill because it will have already been moving as opposed to the initial hill. But I'm not one hundred percent certain. You can always google this stuff, but I do know for sure that at the first hill, the roller coaster will have higher potential energy.
Hope this helps!
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A CD has a mass of 17 g and a radius of 6.0 cm. When inserted into a player, the CD starts from rest and accelerates to an angul
Stells [14]

Answer:

τ =9.41 * 10⁻⁴ N*m

Explanation:

Kinematics of the CD

The CD rotates with constant angular acceleration and its angular acceleration is calculated as follows:

\alpha = \frac{\omega_{f}- \omega_{i}}{t}   Formula (1)

Where:

α : angular acceleration. (rad/s²)

ωf: final angular velocity  (rad/s)

ωi : initial angular velocity  (rad/s)

t = time interval (s)

Data

ωf= 20 rad/s

ωi =0

t = 0.65 s.

Calculating of the angular acceleration of the CD

We replace data in the formula (1)

\alpha = \frac{20 -0}{0.65}

α  = 30.77 rad/s²

Newton's second law  in rotation:

F = ma has the equivalent for rotation:

τ = I * α   Formula  (2)

where:

τ : It is the net torque applied to the body.  (N*m)

I :  it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Calculating of the moment of inertia  of the CD

The moment of inertia of a disk with respect to an axis perpendicular to the plane  and passing through its center is calculated by the following formula:

I = (1/2) M*R² Formula (3)

Data

M= 17 g = 17/1000 kg = 0.017 kg  : CD mass

R= 6.0 cm= 6/100 m = 0.06 m : CD  radius

We replace data in the formula (3) :

I = (1/2) ( 0.017 kg)*(0.06 m)² = 3.06 * 10⁻⁵ kg*m²

Calculating of the  net torque acting on the CD

Data

α  = 30.77 rad/s²

I =  3.06 * 10⁻⁵ kg*m²

We replace data in the formula (2) :

τ = I * α

τ =( 3.06 * 10⁻⁵ kg*m²) * (30.77 rad/s²)

τ =9.41 * 10⁻⁴ N*m

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