Furthermore, the AEC said that the Joint Committee has made its position clear that it would no longer authorize any such subsidies. Yet, Mr. Chairman, we find.

8 0

3 years ago

A skier moving at 5.71 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220

emmainna [20.7K]

Answer:

She travels 7.56 m before stopping.

Explanation:

Hi there!

According to the work-energy theorem, the magnitude of the work done by a force on a moving object to bring it to stop will be equal to the kinetic energy of the object:

W = KE

Where:

W = work

KE = kinetic energy

In this case, the force that stops the skier is the friction force. Then, the work done by friction will be:

W = Fr · d

Where:

Fr = friction force.

d = traveled distance.

The friction force is calculated as follows:

Fr = N · μ

Where:

N = normal force.

μ = coefficient of kinetic friction.

The forces acting in the vertical direction are the weight of the skier (w, downward) and the normal force (N, upward). Since the skier is not being accelerated in the vertical direction, then the sum of vertical forces is equal to zero:

∑Fy = N - w = 0 ⇒N = w

The weight is calculated as follows:

w = m · g

Where m is the mass of the skier and g is the acceleration due to gravity.

Then, the work done by friction can be expressed as follows:

W = Fr · d

W = N · μ · d

Since N = w = m · g

W = m · g · μ · d

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

Where v is the speed of the skier.

Appliyng work-energy theorem:

W = KE

m · g · μ · d = 1/2 · m · v²

Solving for d:

d = 1/2 · v² / g · μ

d = 1/2 · (5.71 m/s)² / (9.8 m/s² · 0.220)

d = 7.56 m

She travels 7.56 m before stopping.

3 0

4 years ago

Monochromatic light with a wavelength of 6.4 E -7 meter passes through two narrow slits, producing an interference pattern on a

seropon [69]

Answer:

The distance between the slits is given by 1.3 × $10^{-4}$ m

Given:

$\lambda = 6.4 \times 10^{-7} m$

D = 4 m

y = $2 \times 10^{-2} m$

m = 1

To find:

distance between slits, d = ?

Formula used:

y = $\frac{m \times \lambda \times D}{d}$

y = distance of first bright band from central maxima

D = distance between screen and source

d = distance between slits

$\lambda$ = wavelength

Solution:

distance of first bright band from central maxima is given by,

y = $\frac{m \times \lambda \times D}{d}$

y = distance of first bright band from central maxima

D = distance between screen and source

d = distance between slits

$\lambda$ = wavelength

Thus,

d = $\frac{m \times \lambda \times D}{y}$

d = $\frac{1 \times 6.4 \times 10^{-7} \times 4 }{2 \times 10^{-2} }$

d = 1.28 × $10^{-4}$

d = 1.3 × $10^{-4}$ m

The distance between the slits is given by 1.3 × $10^{-4}$ m

8 0

4 years ago

A stone is thrown at an angle of 30 degrees above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A

natta225 [31]

v = initial velocity of launch of the stone = 12 m/s

θ = angle of the velocity from the horizontal = 30

Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.

v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s

a = acceleration of the stone = - 9.8 m/s²

t = time of travel = 4.8 s

Y = vertical displacement of stone = vertical height of the cliff = ?

using the kinematics equation

Y = v₀ t + (0.5) a t²

inserting the values

Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²

Y = - 84.1 m

hence the height of the cliff comes out to be 84.1 m

5 0

3 years ago