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sergey [27]
3 years ago
5

A man goes for a walk, starting from the origin of an xyz coordinate system, with the xy plane horizontal and the x axis eastwar

d. Carrying a bad penny, he walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high
(a) In unit-vector notation, what is the displacement of the penny from start to its landing point?
(b) When the man returns to the origin, what is the magnitude of his displacement for the return trip?
Physics
2 answers:
Mazyrski [523]3 years ago
7 0

Answer:

Explanation:

d1 = 1300 m east

d2 = 2400 m north

d3 = 640 m downward

(a)

The displacement of penny is given by

\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}

\overrightarrow{d}=1300\widehat{i}+2400 \widehat{j}-640\widehat{k}

(b) For the return journey of man, the displacement is given by

\overrightarrow{d}=-\overrightarrow{d_{1}}-\overrightarrow{d_{2}}

\overrightarrow{d}=-1300\widehat{i}-2400 \widehat{j}

The magnitude of the displacement is given by

d=\sqrt{1300^{2}+2400^{2}}=2729.47 m

tekilochka [14]3 years ago
3 0
<h2>a) Displacement of penny = 1300 i + 2400 j - 640 k</h2><h2>b) Magnitude of his displacement = 2729.47 m</h2>

Explanation:

a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.

1300 m east = 1300 i

2400 m north = 2400 j

Drops the penny from a cliff 640 m high = -640 k

Displacement of penny = 1300 i + 2400 j - 640 k

b) Displacement of man for return trip = -1300 i - 2400 j

    \texttt{Magnitude = }\sqrt{(-1300)^2+(-2400)^2}=2729.47m

    Magnitude of his displacement = 2729.47 m

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A 6- F capacitor is charged to 90 V and is then connected across a 700- resistor. What is the initial charge on the capacitor
Lorico [155]

Answer:

540C.

Explanation:

A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;

Q = CV          ----------(i)

From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.

Where;

C = 6F

V = 90V

Substitute these values into equation (i) as follows;

Q = 6 x 90

Q = 540 C

Therefore, the initial charge on the capacitor is 540C.

7 0
3 years ago
A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr
mihalych1998 [28]

(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

f' = (\frac{v}{v\pm v_s})f

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

v_s is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)

f is the original frequency of the sound

Here we have

f = 350 Hz

When the train is approaching, we have

v_s = -40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s - 40.4 m/s})(350 Hz)=396.7 Hz

When the train has passed the platform, we have

v_s = +40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s + 40.4 m/s})(350 Hz)=313.1 Hz

Therefore the overall shift in frequency is

\Delta f = 313.1 Hz - 396.7 Hz = -83.6 Hz

And the negative sign means the frequency has decreased.

(b) 0.865 m

The wavelength and the frequency of a wave are related by the equation

v=\lambda f

where

v is the speed of the wave

\lambda is the wavelength

f is the frequency

When the train is approaching the platform, we have

v = 343 m/s (speed of sound)

f = f' = 396.7 Hz (apparent frequency)

Therefore the wavelength detected by a person on the platform is

\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m

5 0
3 years ago
A 1.80-m string of weight 0.0126 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
Veseljchak [2.6K]

Answer:

W = 0.135 N

Explanation:

Given:

- y (x, t) = 8.50*cos(172*x -2730*t)

- Weight of string m*g = 0.0126 N

- Attached weight = W

Find:

The attached weight W given that Tension and W are equal.

Solution:

The general form of standing mechanical waves is given by:

                            y (x, t) = A*cos(k*x -w*t)  

Where k = stiffness and w = angular frequency

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- Calculate wave speed V:

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- Tension in the string T:

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where Y: is the mass per unit length of the string.

- The tension T and weight attached W are equal:

                           T = W = Y*V^2 = (w/L*g)*V^2

                            W = (0.0126 / 1.8*9.81)*(13.78)^2

                            W = 0.135 N

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