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masha68 [24]
3 years ago
12

A skier of mass 103 kg comes down a slopeof constant angle 32◦with the horizontal.What is the force on the skier parallel tothe

slope? Neglect friction. The accelerationof gravity is 9.8 m/s2.Answer in units of N.
Physics
1 answer:
LuckyWell [14K]3 years ago
3 0

Answer:

534.9 N

Explanation:

The skier weight is his mass times gravitational acceleration g

W = mg = 103 * 9.8 = 1009.4 N

This weight can be divided into 2 components, one perpendicular and the other parallel to the 32-degree slope. The parallel component would equal to

Wsin(32^0) = 1009.4sin(32^0) = 534.9 N

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<h3>What is the cylinder's total surface area?</h3>

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5 0
1 year ago
If a rock climber accidentally drops a 52.5-g piton from a height of 325 meters, what would its speed be just before striking th
alex41 [277]
Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.  
So we have 1/2 MV^2 = MGH 
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V = âš( 2 * 9.8 * 325)  
V = âš 6370
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6 0
3 years ago
A 2-kg book falls off a shelf. It hits a student traveling 2 m/s. How much kinetic energy does the book have?
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4 J

1/2 mv^2 is ke formula

3 0
3 years ago
Earth's gravity attempts to change the velocity of all objects on Earth's surface toward _____ at a rate of 9.8 meters per secon
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8 0
3 years ago
Read 2 more answers
A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7
aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

f'=(\frac{v+v_r}{v+v_s})f

where

f is the original frequency

f is the apparent frequency

v is the velocity of the wave

v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

v_s is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

v_s = v_A = -28.7 m/s (surface A is the source, which is moving towards the receiver)

v_r = +62.2 m/s (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz

(b) 0.232 m

The wavelength of a wave is given by

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m

(c) 1481.2 Hz

Again, we can use the same formula

f'=(\frac{v+v_r}{v+v_s})f

In the source frame (= on surface A), we have

v_s = v_B = -62.2 m/s (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

v_r = +28.7 m/s (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz

(d) 0.225 m

The wavelength of the wave is given by

\lambda=\frac{v}{f}

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m

8 0
3 years ago
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