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Anni [7]
3 years ago
7

) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffi

cient of kinetic friction between the stone and the surface
Physics
1 answer:
natali 33 [55]3 years ago
4 0

Answer:

-0.3

Explanation:

F' = μmg ........... Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma ............ Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g............... Equation 3

From the equation of motion,

v² = u²+2as................. Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

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Current flows from High Potential (Positive) to Low potential (Negative)

So, option D is your answer!

Hope this helps!
3 0
3 years ago
Read 2 more answers
A solenoid of radius 2.0 mm contains 100 turns of wire uniformly distributed over a length of 5.0 cm. It is located in air and c
tankabanditka [31]

Answer:

The magnetic field strength inside the solenoid is 5.026\times10^{-3}\ T.

Explanation:

Given that,

Radius = 2.0 mm

Length = 5.0 cm

Current = 2.0 A

Number of turns = 100

(a). We need to calculate the magnetic field strength inside the solenoid

Using formula of the magnetic field strength

Using Ampere's Law

B=\dfrac{\mu_{0}NI}{l}

Where, N = Number of turns

I = current

l = length

Put the value into the formula

B=\dfrac{4\pi\times10^{-7}\times100\times2.0}{5.0\times10^{-2}}

B=0.005026=5.026\times10^{-3}\ T

(b). We draw the diagram

Hence, The magnetic field strength inside the solenoid is 5.026\times10^{-3}\ T.

4 0
3 years ago
A 40W lamp wastes 34 J of energy every second by heating its surroundings.
Artemon [7]

Answer:

15\%.

Explanation:

The efficiency of a machine is the percentage of energy input that was turned into useful energy.

The power rating of this lamp is 40\; \rm W (same as 40\; \rm J \cdot s^{-1},) meaning that 40\; \rm J of energy is supplied to this lamp every second.

The question states that 34\; \rm J out of that 40\; \rm J of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the (40\; \rm J - 34\; \rm J) = 6\; \rm J of energy supplied to this lamp would be turned into useful energy output.

Thus, every second, this lamp would receive 40\; \rm J of energy input and would outputs 6\; \rm J of useful work. The efficiency of this lamp would be:

\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}.

4 0
2 years ago
A diffraction grating with 1000 lines per mm is used in a spectrometer to measure the wavelengths of light emitted from a gas di
frez [133]

Answer:

= 9.8°

Explanation:

Width of one slit (a₁ ) = 1 / 1000 mm=0.001 mm = 10⁻⁶ m.

width of one slit in case 2 (a₂ ) = 1/500 =2 x 10⁻⁶ m

angular position of fringe, Sinθ  = n λ /a

n is order of fringe , λ is wave length of light and a  is slit aperture

So Sinθ  ∝ 1 / a

Sin θ₁ /Sin θ₂ = a₂/a₁ ;

Sin20°/sinθ₂ = 2 / 1

sinθ₂ = Sin 20° / 2 = .342/2 = .171

θ₂ = 9.8 °

4 0
3 years ago
Richard is driving home to visit his parents. 150 mi of the trip are on the interstate highway where the speed limit is 65 mph .
Serggg [28]

Answer:

t = 25.5 min

Explanation:

To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.

t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h

Next, you calculate the difference between both times t1 and t2:

\Delta t=t_1-t_2=2.30h-1.875h=0.425h

This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:

0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s

hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph

3 0
3 years ago
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