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inn [45]
3 years ago
5

A hiker climbs a mountain. Starting at the base of the mountain, he first moved up 520m at a 32.0 degree angle. What is the fina

l displacement of the hiker?
Physics
1 answer:
balu736 [363]3 years ago
5 0

Answer:

\displaystyle \vec{d}=

Explanation:

<u>Displacement Vector</u>

Suppose an object is located at a position  

\displaystyle P_1(x_1,y_1)

and then moves at another position at

\displaystyle P_2(x_2,y_2)

The displacement vector is directed from the first to the second position and can be found as

\displaystyle \vec{d}=

If the position is given as magnitude-angle data ( z , α), we can compute its rectangular components as

\displaystyle x=z\ cos\alpha

\displaystyle y=z\ sin\alpha

The question describes the situation where the initial point is the base of the mountain, where both components are zero

\displaystyle P_1(0,0)

The final point is given as a 520 m distance and a 32-degree angle, so  

\displaystyle x_2=520\ cos32^o= 440.99\ m

\displaystyle y_2=520\ sin32^o=275.6\ m

The displacement is

\displaystyle \vec{d}=

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starting from rest , a formula one car accelerates uniformly at 25m\s2 for 30secs. what distance does it cover in the last one s
Anestetic [448]

The distance covered in the last second of motion is 737.5 m

Explanation:

The motion of the car is a uniformly accelerated motion, so we can use the suvat equations.

First of all, we have to find the velocity of the car when the last second of motion starts, that is the velocity of the car after t = 29 s. We can use the equation:

v = u + at

where

u = 0 is the initial velocity

a=25 m/s^2 is the acceleration

Substituting t = 29 s,

v=0+(25)(29)=725 m/s

Now we can find the distance covered in the last second of motion by using

s=ut+\frac{1}{2}at^2

where

u = 725 m/s is the velocity at the beginning of the last second

t = 1 s is the time interval considered

a=25 m/s^2 is the acceleration

Substituting,

s=(725)(1)+\frac{1}{2}(25)(1)^2=737.5 m

Note that the acceleration of 25 m/s^2 is not realistic for a car, but I still have used the data of the problem.

Learn more about acceleration:

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#LearnwithBrainly

6 0
3 years ago
Look at the picture and help me plz
Elenna [48]

Answer:

AI

Explanation:

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3 years ago
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Sound waves that start in air can move into water. This is because it is __________ that is transferred and not .
ikadub [295]
The correct answer is B
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3 years ago
At which point will the riders experience centripetal acceleration?
Korvikt [17]

the answer would be (X)

8 0
3 years ago
Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note
Sophie [7]

(a) 2 Hz

The frequency of the nth-harmonic is given by

f_n = n f_1

where

f_1 is the fundamental frequency

Therefore, the frequency of the third harmonic of the A (f_1 = 440 Hz) is

f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz

while the frequency of the second harmonic of the E (f_1 = 659 Hz) is

f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz

So the frequency difference is

\Delta f = 1320 Hz - 1318 Hz = 2 Hz

(b) 2 Hz

The beat frequency between two harmonics of different frequencies f, f' is given by

f_B = |f'-f|

In this case, when the strings are properly tuned, we have

- Frequency of the 3rd harmonic of A-note: 1320 Hz

- Frequency of the 2nd harmonic of E-note: 1318 Hz

So, the beat frequency should be (if the strings are properly tuned)

f_B = |1320 Hz - 1318 Hz|=2 Hz

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

f_1 \propto \sqrt{T}

this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.

In this situation, the beat frequency is 4 Hz (four beats per second):

f_B = 4 Hz

And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,

f_B = |f_3-f_2|

and f_3 = 1320 Hz, we have two possible solutions for f_2:

f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz

However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be

1324 Hz

8 0
3 years ago
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