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2 years ago

14

A 2600-m-high mountain is located on the equator. how much faster does a climber on top of the mountain move than a surfer at a

nearby beach? the earth's radius is 6400 km.

1 answer:

puteri [66]2 years ago

5 0

The climber move 0.19 m/s faster than surfer on the nearby beach.

Since both the person are on the earth, and moves with the constant angular velocity of earth, however there linear velocity is different.

Number of seconds in a day, t=24*60*60=86400 sec

The linear speed on the beach is calculated as

V1= $\frac{2πr}{t}$

Here, t is the time

Plugging the values in the above equation

V1= $\frac{2π*6.4*10^6}{86400}$ =465.421 m/s

Velocity on the mountain is calculated as

V2= $\frac{2π(r+h)}{t}$

Plugging the values in the above equation

V2= $\frac{2π(6.4*10^6+2600}{86400}$ =465.61 m/s

Therefore person on the mountain moves faster than the person on the beach by 465.61-465.421=0.19 m/s

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In two separate double slit experiments, an interference pattern is observed on a screen. In the first experiment, violet light

sashaice [31]

Answer:

$\lambda_3 = 4.72*10^{-7} m$

Explanation:

given data:

wavelength \lambda = 708nm = 708*10^{-9} m

using the following relation:

$y = \frac{mL\lambda}{d}$

according to the given information

second and third dark fringe is at same location. so

$y_2 = y_3$

$\frac{m_2L\lambda_2}{d} = \frac{m_3L\lambda_3}{d}$

$m_2\lambda_2 = m_3\lambda_3$

$2*708*10^{-9} = 3*\lambda_3$

$\lambda_3 = \frac{2*708*10^{-9}}{3}$

$\lambda_3 = 4.72*10^{-7} m$

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2 years ago

A boy jumps into an indoor swimming pool. He notices that the water appears to get colder as he

Viefleur [7K]

Answer:

cold air is more dense than warm water so it sinks to the bottom of the pool

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2 years ago

A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the

likoan [24]

Answer:

The separation between the first two minima on either side is 0.63 degrees.

Explanation:

A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:

$a\sin \theta_n=n\lambda$

with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:

for the first minimum

$a\sin \theta_1=(1)\lambda$

solving for θ1:

$\theta_1=\arcsin (\frac{\lambda}{a})=\arcsin (\frac{550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_1=0.63 degrees$

for the second minimum:

$a\sin \theta_2=(2)\lambda$

$\theta_2=\arcsin (\frac{2\lambda}{a})=\arcsin (\frac{2*550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_2=1.26 degrees$

So, the angular separation between them is the rest:

$\Delta \theta =1.26-0.63$

$\Delta \theta=0.63$

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2 years ago

A person is working on a steel structure while standing on the ground. An accident occurred where 5 A pass through the structure

matrenka [14]

Answer:

35mA

Explanation:

Hello!

To solve this problem we must use the following steps

1. Find the electrical resistance of the metal rod using the following equation

$R=\alpha \frac{l}{a}$

WHERE

α=

metal rod resistivity=2x10^-4 Ωm

l=leght=2m

A= Cross-sectional area

$A=\frac{\pi }{4} d^2=\frac{\pi }{4} (0.06)^2=0.00283$

solving

$R=(2x10^-4)\frac{2}{0.00283} =0.14$

2. Now we model the system as a circuit with parallel resistors, where we will call 1 the metal rod and 2 the man(see attached image)

3.we know that the sum of the currents in 1 and 2 must be equal to 5A, by the law of conservation of energy

I1+I2=5

4.as the voltage on both nodes is the same we can use ohm's law in resitance 1 and 2 (V=IR)

V1=V2

(0.14I1)=2000(i2)

solving for i1

I1=14285.7i2

5.Now we use the equation found in step 3

14285.7i2+i2=5

$i2=\frac{5}{14285.7+1} =3.5x10^-4A=35mA$

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2 years ago

A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which

goldfiish [28.3K]

Answer:

160 Hz , 240 Hz , 400 Hz

Explanation:

Given that

Frequency of forth harmonic is 320 Hz.

Lets take fundamental frequency = f₁

$f_1=\dfrac{320}{4}\ Hz$

f₁=80 Hz

Frequency of first harmonic = f₂

f₂=2 f₁

f₂ =2 x 80 = 160 Hz

Frequency of second harmonic = f₃

f₃= 3 f₁=3 x 80 = 240 Hz

Frequency of fifth harmonic = f₅

f₅= 5 f₁= 5 x 80 = 400 Hz

Three frequencies are as follows

160 Hz , 240 Hz , 400 Hz

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2 years ago