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Dahasolnce [82]
3 years ago
10

Dante is leading a parade across the main street in front of city hall. Starting at city hall, he marches the parade 4 blocks ea

st, then 3 blocks south. From there, the parade marches 1 block west and 9 blocks north and finally stops. What is the vector displacement and direction of the the parade, starting from the city hall and the stopping point? (1 point)

Physics
2 answers:
REY [17]3 years ago
6 0

Answer:

 R = 6.7  blocks and    θ = 63.4

Explanation:

The displacement vector is the vector sum of each individual displacement, the easiest way to do this is to find the magnitude of the displacement.

X axis

         x₁ = 4

         x₂ = -1

         x = 4 - 1 = 3

Y Axis  

      y₁ = -3

      y₂ = 9

      y = -3 + 9 = 6

We use the Pythagorean theorem

     R = √ (x² + y²)

     R =√ (3² + 6²)

     R = 6.7

We look for the displacement angle with trigonometry

        tan θ = y / x

       tan θ = 6/3

       θ = tan⁻¹ 2

       θ = 63.4

Alisiya [41]3 years ago
4 0

Answer:

The displacement between the city hall and the stopping point is 6.7 blocks.

The direction is 63.43°

Explanation:

The diagram explains better.

From the diagram, point O is the city hall and point D is the stopping point.

The displacement between O and D is OD.

Using Pythagoras theorem, we can find this:

OD² = OX² + DX²

From the diagram:

OX = 4 - 1 = 3 blocks

DX = 9 - 3 = 6 blocks

=> OD² = 3² + 6² = 9 + 36

OD² = 45

=> OD = 6.7 blocks

To get the direction, θ, we use SOHCAHTOA:

tanθ = DX/OX

tanθ = 6/3 = 2

=> θ = 63.43°

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A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i
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8 0
3 years ago
A 50.0 Watt stereo emits sound waves isotropically at a wavelength of 0.700 meters. This stereo is stationary, but a person in a
photoshop1234 [79]

Answer:

a) f' = 432 Hz

b) I = 8.12*10^-4 W/m^2

Explanation:

a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:

f'=f(\frac{v-v_o}{v+v_s})    (1)

where

f: frequency of the source = ?

v: speed of sound = 343 m/s

vo: speed of the observer = 40.0 m/s

vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

v=\lambda f\\\\

v: speed of sound

λ: wavelength = 0.700 m

f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz

Next, you replace the values of all parameters in the equation (1):

f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz

hence, the frequency perceived by the car is 432 Hz

b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:

I=\frac{P}{4\pi r^2}

P: power of the source = 50.0 W

r: distance to the source = 70.0 m

I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}

hence, the intensity is 8.12*10^⁻4 W/m^2

3 0
3 years ago
A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

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P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0
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