Ask question

Ask question

All categories

I am Lyosha [343]

4 years ago

15

A shuffleboard disk is accelerated to a speed of 5.8 m/s and released. If the coefficient of kinetic friction between the disk a

nd the concrete court is 0.31, how far does the disk go before it comes to a stop?

1 answer:

MatroZZZ [7]4 years ago

3 0

Answer:

5.53 m

Explanation:

$v_{o}$ = initial speed of shuffleboard disk = $5.8 ms^{-1}$

$v_{f}$ = final speed of shuffleboard disk = $0 ms^{-1}$

$\mu$ = Coefficient of kinetic friction between the disk and concrete court = 0.31

acceleration due to friction is given as

$a = - \mu g\\a = - (0.31) (9.8)\\a = - 3.04 ms^{-2}$

$d$ = distance traveled by disk before it stops

using the kinematics equation that fits the above the data, we have

$v_{f}^{2} = v_{o}^{2} + 2 a d\\0^{2} = 5.8^{2} + 2 (- 3.04) d\\d = 5.53 m$

You might be interested in

In making water safe to drink, which process removes impurities by forcing water through sand or charcoal? A. aeration B. coagul

spin [16.1K]

The answer is C! I knew this way before taking the test

4 0

4 years ago

A wheelbarrow is pushed with a force of 40 N. If 6,000 J of work is

stepladder [879]

Answer:

Distance = 150 meters

Explanation:

Given the following data;

Work done = 6,000 Joules

Force = 40 Newton

To find the total distance covered by the wheelbarrow;

Workdone = force * distance

Substituting into the formula, we have;

6000 = 40 * distance

Distance = 6000/40

Distance = 150 meters

Therefore, the total distance the wheelbarrow was pushed is 150 meters.

5 0

3 years ago

a skier starts from rest and skis down a 82 meter tall hill labeled h1, into a valley and staught back up another 35 meter hill(

horrorfan [7]

Answer:

She is going at 30.4 m/s at the top of the 35-meter hill.

Explanation:

We can find the velocity of the skier by energy conservation:

$E_{1} = E_{2}$

On the top of the hill 1 (h₁), she has only potential energy since she starts from rest. Now, on the top of the hill 2 (h₂), she has potential energy and kinetic energy.

$mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2}^{2}$ (1)

Where:

m: is the mass of the skier

h₁: is the height 1 = 82 m

h₂: is the height 2 = 35 m

g: is the acceleration due to gravity = 9.81 m/s²

v₂: is the speed of the skier at the top of h₂ =?

Now, by solving equation (1) for v₂ we have:

$v_{2}^{2} = \frac{2mg(h_{1} - h_{2})}{m}$

$v_{2} = \sqrt{2g(h_{1} - h_{2})} = \sqrt{2*9.81 m/s^{2}*(82 m - 35 m)} = 30.4 m/s$

Therefore, she is going at 30.4 m/s at the top of the 35-meter hill.

I hope it helps you!

6 0

2 years ago

A force parallel to the x-axis acts on a particle moving along the x-axis. This force produces potential energy U(x) given by U(

noname [10]

Answer:

Explanation:

U(x)=αx⁴

Force F = - dU / dx

= 4αx³

= 4 x .75 x³

= 3 x³

Force when x = .8 m

F = 3 (0.8)³

= 1.54 N

8 0

3 years ago

A thin beam of light of wavelength 50 μm (in the infrared portion of the em spectrum) goes through

Brut [27]

Solution :

Given :

Wavelength of the thin beam of light, λ = 50 μm

Distance of the screen from the slit, D = 3.00 m

Width of the fringe, Δy = ±8.24 mm

Therefore, width of the slit is given by :

$d=\frac{n \lambda D}{\Delta y}$

$=\frac{2 \times 50 \times 10^{-9} \times 3}{2 \times 8.24 \times 10^{-3}}$

= 0.000018203 m

= 0.0182 mm

= 0.018 mm

The intensity of light is given by :

$I=I_0\left(\frac{\sin \beta /2}{\beta/ 2}\right)^2$ , where $\beta=\frac{2 \pi D \sin \theta}{\lambda}$

$I=I_0\left(\frac{\sin \frac{\pi d \sin \theta}{\lambda}}{\frac{\pi d \sin \theta}{\lambda}}\right)^2$

$I=I_0\left(\frac{\sin \frac{\pi d y}{D\lambda}}{\frac{\pi d y}{D\lambda}}\right)^2$

Now, $\frac{dy}{D \lambda} = \frac{8.24 \times 10^{-3}\times 0.018 \times 10^{-3}}{4 \times 50\times 10^{-9}\times 4}$

= 0.1854

≈ 0.18

$I=I_0\left(\frac{\sin 0.56}{0.56}\right)^2$

$=I_0 \times 0.81$

= 2 x0.81

$= 1.62 \ W/m^2$

6 0

3 years ago