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kirill [66]
2 years ago
13

A cart for hauling ore out of a gold mine has a mass of 435 kg, including its load. the cart runs along a straight stretch of tr

ack that is sloped 4.77° from the horizontal. a donkey, trudging along and to the side of the track, has the unenviable job of pulling the cart up the slope with a 438-n force for a distance of 193 m by means of a rope that is parallel to the ground and makes an angle of 13.3° with the track. the coefficient of friction for the cart\'s wheels on the track is 0.0167. use g = 9.81 m/s2. find the work that the donkey performs on the cart during this process.
Physics
1 answer:
marta [7]2 years ago
4 0
What are the answer choices 

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A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w
frosja888 [35]

Answer:

(a) n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) n_{2} = 1.349

(c) v_{1} = 2.04\times 10^{8}\ m/s

(d) v_{2} = 2.22\times 10^{8}\ m/s

Solution:

As per the question:

Refractive index of medium 1, n_{1} = 1.47

Angle of refraction for medium 1, \theta_{1} = 59^{\circ}

Angle of refraction for medium 2, \theta_{1} = 69^{\circ}

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

n_{1}sin\theta_{1} = n_{2}sin\theta_{2}

where

n_{2} = Refractive Index of medium 2

Now,

n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}

n_{2} = 1.349

(c) To calculate the velocity of light in medium 1:

We know that:

Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}

Thus for medium 1

n_{1} = \frac{c}{v_{1}

v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s

(d) To calculate the velocity of light in medium 2:

For medium 2:

n_{2} = \frac{c}{v_{2}

v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s

5 0
2 years ago
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The sphere that refers to Earth's water is called what?
WARRIOR [948]

Answer:

Its called the hydrosphere UwU!

Explanation:

6 0
2 years ago
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Can I write a narrative writing how the first time I fell in love with blue eyes?
STatiana [176]

Yes you can. Fiction or real. You can almost write any topic you want...
4 0
3 years ago
In a 30.0-s interval, 500 hailstones strike a glass window with an area of 0.600m2 at an angle of 45.0°to the window surface. Ea
aliina [53]

or one hailstone we have;
Force = Mass X acceleration = 0.005kg x 9.8.} This is when the hailstone is not inclined at an angle.
When the hailstone is inclined at an angle of 45, then the component of force along the glass window will be F =0.005kg x 9.8 x sin45= 0.005kg x 9.8 x 0.707= 0.0346N.
Therefore, total force for the 500 hailstones would be 500x0.0346N=17.32N
This force is acting on an area equal to 0.600m2
Pressure = Force per unit area = 17.32N/0.600m2 = 28.9Pa

5 0
3 years ago
1. A car accelerates uniformly from 12 m/s to 39 m/s in 12 seconds. What is the car's average acceleration
Alinara [238K]

Answer:

(1) 2.25m/s^2

(2) 45.6m

Explanation:

(1) A car accelerates uniformly from 12m/s to 39m/s in 12 seconds

Therefore the average acceleration can be calculated as follows

a = 39-12/12

a = 27/12

a= 2.25m/s^2

(2) A butterfly is flying at 4m/s , it accelerates uniformly at 1.2 m/s for 6 seconds

u= 4

a= 1.2

t= 6

Therefore the distance can be calculated as follows

S= ut + 1/2at^2

= 4×6 + 1/2 × 1.2 × 6^2

= 24 + 1/2 × 1.2 × 36

= 24 + 1/2 × 43.2

= 24 + 21.6

S = 45.6m

Hence the butterfly travels at 45.6m

5 0
2 years ago
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