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sukhopar [10]
4 years ago
14

A 60-kg skier starts at the top of a 10-meter high slope. At the bottom, she is traveling 10 m/s. How much energy does she lose

to friction?
Physics
1 answer:
Karo-lina-s [1.5K]4 years ago
4 0
The potential energy of the skier at the top is
60 kg (9.81 m/s2) (10 m) = 5886 J

Her kinetic energy at the bottom is
(1/2) (60 kg) (10 m/s)^2 = 3000 J

The energy lost to friction is
5886 J - 3000 J = 2886 J
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A tuning fork generates sound waves with a frequency of 240 Hz. The waves travel in opposite directions along a hallway, are ref
Bingel [31]

Answer:

The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

Explanation:

Given that,

Frequency of sound wave = 240 Hz

Distance = 46.0 m

Distance of fork = 14 .0 m

We need to calculate the path difference

Using formula of path difference

\Delta x=2(L_{2}-L_{1})

Put the value into the formula

\Delta x =2((46.0-14.0)-14.0)

\Delta x=36\ m

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{v}{f}

Put the value into the formula

\lambda=\dfrac{343}{240}

\lambda=1.42\ m

We need to calculate the phase difference

Using formula of the phase difference

\phi=\dfrac{2\pi}{\lambda}\times \delta x

Put the value into the formula

\phi=\dfrac{2\pi}{1.42}\times36

\phi=159.29\ rad

\phi\approx 68.2^{\circ}

Hence, The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

7 0
3 years ago
Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 wi
Flauer [41]

Answer:

a. Fnet =37.67N

b. The direction = 133.4 from the x axis counter clockwise.

c. Option 2

Explanation:

Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.

Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..

Given that F1 is 71N at 20°, then it is in the first quadrant.

a. Fnet= F1+F2+F3

Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j

Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j

Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

Magnitude of the vector

Fnet= √((-2.04)^2+(37.62)^2)

Fnet= 37.67N

Fnet is approximately 38N.

b. Direction of the Fnet.

Angle=arctan(y/x)

Angle=arctan(-37.61/2.04)

Angle= -43.37°

The angle is in the negative x axis and positive y axis.

Then the direction becomes 180-43.37

Therefore, the direction of the net force is 133.37°.

c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.

Fnet=ma

Fnet= mv/t

So the velocity is in the direction of the Fnet.

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3 years ago
Which are the three types of mutations?
WITCHER [35]
Replication, Multiplication, and Substitution.
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A child standing on a bus remains still when the bus is at rest. When the bus moves forward and then slows down, the child conti
pochemuha
B. Newton's First Law, I'm pretty sure. The first states that an object in motion stays in motion, and an object at rest stays at rest until an outside force is applied, and that seems pretty relevant.
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4 years ago
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The statement about "<span>efficiency compared the output work to the output force" is false. Efficiency can be compared from the input work to the output work.</span>
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