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Nesterboy [21]
3 years ago
8

A 40 kg skier starts at the top of a 12 meter high slope. At the bottom, she is traveling 10 miles. How much energy does she los

e to friction?
Physics
2 answers:
givi [52]3 years ago
8 0
Use conversion of energy. mgh+.5mv^2 initial = final. at the top there's no kinetic energy, at the bottom theres no potential. so mgh at the top=.5mv^2 at the bottom.
mgh at the top = 40 (9.8)(12)
not sure what you mean by 10 miles, do you mean 10 mph or 10 meters/second. if it isnt in SI units, do the proper conversions and put it in m/s
.5mv^2=.5 (40)(final velocity)
DaniilM [7]3 years ago
7 0
2,704 J i believe. Not sure
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A layer of ethyl alcohol (n = 1.361) is on top of water (n = 1.333). To the nearest degree, at what angle relative to the normal
AleksAgata [21]

Answer:

a. 78 degree

Explanation:

According to Snell's Law, we have:

(ni)(Sin θi) = (nr)(Sin θr)

where,

ni = Refractive index of medium on which light is incident

ni = Refractive index of ethyl alcohol = 1.361

nr = Refractive index of medium from which light is refracted

nr = Refractive index of ethyl alcohol = 1.333

θi = Angle of Incidence

θr = Angle of refraction

So, the Angle of Incidence is know as the Critical Angle (θc), when the refracted angle becomes 90°. This is the case of total internal reflection. That is:

θi = θc

when, θr = 90°

Therefore, Snell's Law becomes:

(1.361)(Sin θc) = (1.333)(Sin 90°)

Sin θc = 1.333/1.361

θc = Sin⁻¹ (0.9794)

θc = 78.35° = 78° (Approximately)

Therefore, correct answer will be:

a. <u>78 degree</u>

8 0
3 years ago
In which direction does the electric field point at a position directly north of a positive charge?
navik [9.2K]

Answer:

B. South

Explanation:

An electric field can be defined as the amount of electric force per unit charge. The direction of the electric field can be determined by the motion of a positive test charge under the electric force.

The direction of electric field is radially outward for a positive charge and radially inward for a negative charge. Thus, for the electric field points toward SOUTH at a position directly south of a positive charge.

5 0
3 years ago
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If the block is subjected to the force of F = 500 N, determine its velocity at s = 0.5 m. When s = 0, the block is at rest and t
STatiana [176]

Answer:

The velocity is  4.6 m/s^2

Explanation:

Given:

Force = 500N

Distance  s= 0

To find :

Its velocity at s = 0.5 m

Solution:

\sum F_{x}=m a

F\left(\frac{4}{5}\right)-F_{S}=13 a

500\left(\frac{4}{5}\right)-\left(k_{s}\right)=13 a

400-(500 s)=13 a

a = \frac{400 -(500s)}{13}

a = (30.77 -38.46s) m/s^2

Using the relation,

a=\frac{d v}{d t}=\frac{d v}{d s} \times \frac{d s}{d t}

a=v \frac{d v}{d s}

v d v=a d s

Now integrating on both sides

\int_{0}^{v} v d v=\int_{0}^{0.5} a d s

\int_{0}^{v} v d v=\int_{0}^{0.5}(30.77-38.46 s) d s

\left[\frac{v^{2}}{2}\right]_{0}^{2}=\left[\left(30.77 s-19.23 s^{2}\right)\right]_{0}^{0.5}

\left[\frac{v^{2}}{2}\right]=\left[\left(30.77(0.5)-19.23(0.5)^{2}\right)\right]

\left[\frac{v^{2}}{2}\right]=[15.385-4.807]

\left[\frac{v^{2}}{2}\right]=10.578

v^{2}=10.578 \times 2

v^{2}=21.15

v = \sqrt{21.15}

v = 4.6 m/s^2

8 0
3 years ago
Which statement will be true if you increase the frequency of a periodic wave?
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3 years ago
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lara [203]
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Next, you can use the rate the Jetson's car is going (180km/h) and divide the 38,955.75 by it to see how many hours it would take at that constant speed.

38,955.75 / 180 = 216.42 hours

Then you can divide that by 24 to get how many days
6 0
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