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3 years ago

A car traveling at 100km/hr .how many hours will it take to cover a distance of 750 km

Physics

2 answers:

grin007 [14]3 years ago

4 0

Answer:

7.5 right?

Explanation: if im wrong shoot me

DedPeter [7]3 years ago

4 0

Answer:

7.5 hours( 7 hours 30 minutes)

Explanation:

Speed = distance/time

time = distance/speed

750 km divided by 100 km/hr = 7.5 hours

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What distance does electromagnetic radiation travel in 55.0 μs ?

Komok [63]

Electromagnetic radiation is an energy that is known as light. so electromagnetic radiation will have the same speed as the speed of light which is 3 x 10^8 m/s. so the distance it travel at 55 x 10^-6 s is:
D = ( 3 x 10^8 m/s ) ( 55 x 10^-6 s )
D = 16500 m

4 0

3 years ago

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

3 0

3 years ago

1. Calculate the average velocity of the following trip. You walk to Pershing Square 58

Gre4nikov [31]

Explanation:

Velocity = displacement / time

v = √((58 m)² + (135 m)²) / (12 min × 60 s/min)

v = 0.20 m/s

7 0

3 years ago

Two electric charges, held a distance, dd, apart experience an electric force of magnitude, FF, between them. If one of the char

lorasvet [3.4K]

Answer:

$F'=2F$

Explanation:

The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:

$F=\frac{kq_1q_2}{d^2}$

In this case, we have $q_1'=2q_1$ :

$F'=\frac{kq'_1q_2}{d^2}\\F'=\frac{k(2q_1)q_2}{d^2}\\F'=2\frac{kq_1q_2}{d^2}\\F'=2F$

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3 years ago

A car is traveling in a straight line at a constant speed. The car's engine is exerting a constant force on the car. Explain why

kherson [118]

Answer:

because its not going down a long hill instead its going on a leveled street

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3 years ago