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dusya [7]
4 years ago
7

1. A mass is on a level plane, it has a weight of 20N. What is the coefficient of kinetic friction if an applied force

Physics
1 answer:
Arturiano [62]4 years ago
4 0

Answer:

0.4

Explanation:

F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.

Since the mass is moving with a constant velocity, there's no acceleration hence

F=Fr=\mu N where N is the weight of object and \mu is coefficient of kinetic friction.

F=\mu N and making [tex]\mu the subject

\mu=\frac {F}{N}

Substituting F for 8 N and N for 20 N

\mu=\frac {8}{20}=0.4

Therefore, coefficient of kinetic friction is 0.4

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In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo
Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of 32^{\circ} (when light enters from the air.)

Explanation:

Let v_{1} denote the speed of light in the first medium. Let v_{\text{air}} denote the speed of light in the air. Assume that the light entered the boundary at an angle of \theta_{1} to the normal and exited with an angle of \theta_{\text{air}}. By Snell's Law, the sine of \theta_{1}\! and \theta_{\text{air}}\! would be proportional to the speed of light in the corresponding medium. In other words:

\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}.

When light enters a boundary at the critical angle \theta_{c}, total internal reflection would happen. It would appear as if the angle of refraction is now 90^{\circ}. (in this case, \theta_{\text{air}} = 90^{\circ}.)

Substitute this value into the Snell's Law equation:

\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}.

Rearrange to obtain an expression for the speed of light in the first medium:

v_{1} = v_{\text{air}} \cdot \sin(\theta_{1}).

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For 0 < \theta < 90^{\circ}, \sin(\theta) is monotonically increasing with respect to \theta. In other words, for \!\theta in that range, the value of \sin(\theta)\! increases as the value of \theta\! increases.

Therefore, compared to the medium in this question with \theta_{c} = 27^{\circ}, the medium with the larger critical angle \theta_{c} = 32^{\circ} would have a larger \sin(\theta_{c}). such that light would travel faster in that medium.

4 0
3 years ago
Carbon-14 is used to determine the age of ancient objects. If a sample today contains 0.060 g of carbon-14, how much carbon-14 m
MakcuM [25]

Answer: 86.47 g of carbon-14 must have been present in the sample 11,430 years ago.

Explanation:

Half-life of sample of carbon -14= 5,730 days

\lambda=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{5,730 days}=0.00012 day^{-1}

Let the sample present 11,430 years(t) ago = N_o

Sample left till today ,N= 0.060 g

N=N_o\times e^{-\lambda t}

ln[N]=ln[N]_o-\lambda t

\log[0.060 g]=\log[N_o]-2.303\times 0.00012 day^{-1}\times 11,430 days

\log[N_o]=1.9369

N_o=86.47 g

86.47 g of carbon-14 must have been present in the sample 11,430 years ago.

5 0
3 years ago
Read 2 more answers
The isotope 23893Np has a half-life of 2.0 days. If 4.00 grams are produced at noon on Monday, what will be the mass of neptuniu
viva [34]

Answer:

0.25 gram of neptunium is remaining

Explanation:

First we calculate the no. of half lives passed. For that we have formula:

n = t/T

where,

n = no. of half lives passed = ?

t = total time passed = 8 days (From Monday noon to Tuesday noon of following week)

T = Half Life Period = 2 days

Therefore,

n = 8 days/2 days

n = 4

Now, for the remaining mass of neptunium, we use the formula:

m = (mi)/(2)^n

where,

mi = initial mass of neptunium = 4.00 grams

m = remaining mass of neptunium = ?

Therefore,

m = 4 grams/2⁴

<u>m = 0.25 gram</u>

3 0
3 years ago
Which of the following is not a physical change?
Alekssandra [29.7K]
C because of galvination is sized
5 0
4 years ago
A world class sprinter is travelling with speed 12.0 m/s at the end of a 100 meter race. Suppose he decelerates at the rate of 2
Solnce55 [7]

Answer:

after 6 second it will stop

he travel 36 m to stop

Explanation:

given data

speed = 12 m/s

distance = 100 m

decelerates rate = 2.00 m/s²

so acceleration a = - 2.00 m/s²

to find out

how long does it take to stop and how far does he travel

solution

we will apply here first equation of motion that is

v = u + at   ......1

here u is speed 12 and v is 0 because we stop finally

put here all value in equation 1

0 = 12 + (-2) t

t = 6 s

so after 6 second it will stop

and

for distance we apply equation of motion

v²-u² = 2×a×s  ..........2

here v is 0 u is 12 and a is -2 and find distance s

put all value in equation 2

0-12² = 2×(-2)×s

s = 36 m

so  he travel 36 m to stop

3 0
3 years ago
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