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Ray Of Light [21]

3 years ago

6

Let A be the last two digits and let B be the sum of the last three digits of your 8-digit student ID. (Example: For 20245347, A

= 47 and B = 14) In a remote civilization, distance is measured in urks and an hour is divided into 125 time units named dorts. The length conversion is 1 urk = 58.0 m. Consider a speed of (25.0 + A + B) urks/dort. Convert this speed to meters per second (m/s). Round your final answer to 3 significant figures.

1 answer:

zheka24 [161]3 years ago

7 0

Answer:

The speed is 173 m/s.

Explanation:

Given that,

A = 47

B = 14

Length 1 urk = 58.0 m

An hour is divided into 125 time units named dorts.

3600 s = 125 dots

dorts = 28.8 s

Speed v= (25.0+A+B) urks/dort

We need to convert the speed into meters per second

Put the value of A and B into the speed

$v=25.0+47+14$

$v =86\ urk s/dort$

$v=86\times\dfrac{58.0}{28.8}$

$v=173.19\ m/s$

Hence, The speed is 173 m/s.

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What is the gravitational force on a 70kg that is 6.38x10^6m above the earths surface

NARA [144]

Answer:

171.5 N

Explanation:

The gravitational force on an object due to the Earth is given by

$F=mg$

where

m is the mass of the object

g is the acceleration due to gravity

The acceleration due to gravity at a certain height h above the Earth is given by

$g=\frac{GM}{(R+h)^2}$

where:

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

$R=6.37\cdot 10^6 m$ is the Earth's radius

Here,

$h=6.38\cdot 10^6 m$

So the acceleration due to gravity is

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})}{(6.37\cdot 10^6 + 6.38\cdot 10^6)^2}=2.45 m/s^2$

We know that the mass of the object is

m = 70 kg

So, the gravitational force on it is

$F=mg=(70)(2.45)=171.5 N$

5 0

3 years ago

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

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3 years ago

A horizontal line above the time axis of a speed vs. time graph means an object is ___.

miv72 [106K]

A horizontal line on a speed/time graph means a constant speed.

6 0

3 years ago

A circular loop of wire lies flat on a level table top in a region where the magnetic field vector points straight upward. The m

jarptica [38.1K]

The magnetic field direction and direction of induced current in a wire are related by the right hand grip rule. Since the magnetic field was upwards, the thumb points upwards and the fingers curl around it. When viewed from above, it is seen as a current flowing in the counter clockwise direction.

4 0

3 years ago

How much total energy is dissipated in 10. seconds

noname [10]

Answer : Total energy dissipated is 10 J

Explanation :

It is given that,

Time. t = 10 s

Resistance of the resistors, R = 4-ohm

Current, I = 0.5 A

Power used is given by :

$P=\dfrac{E}{t}$

Where

E is the energy dissipated.

So, E = P t.............(1)

Since, $P=I^2R$

So equation (1) becomes :

$E=I^2Rt$

$E=(0.5\ A)^2\times 4\Omega \times 10\ s$

$E=10\ J$

So, the correct option is (3)

Hence, this is the required solution.

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3 years ago

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