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2 years ago

To reach a distance of 2 km, a car has to cover a path of 2.4 km in 10 minutes. Calculate its average speed and velocity

Physics

1 answer:

nexus9112 [7]2 years ago

5 0

Answer:

Average speed = 0.24 km/min.

Average velocity = 0.2 km/min.

Explanation:

Finding average velocity :

Least possible distance / Time taken
2 km / 10 min
0.2 km/min.

Finding average speed :

Path distance / Time taken
2.4 km / 10 min
0.24 km/min.

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What is the final velocity of a car that starts at 22 m/s and accelerates at 3.78 m/s for distance of 45 m

Pepsi [2]

v^2 = v0^2 +2ad v^2 = 22^2 + 2*3.78*45 = 824.2 v= √824.2 = 28.7 m/s

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3 years ago

For a 50 kV anode voltage, what is the maximum photon energy of the x-ray radiation?

V125BC [204]

Answer:

The energy of photon, $E=8\times 10^{-15}\ J$

Explanation:

It is given that,

Voltage of anode, $V=50\ kV=50\times 10^3\ V=5\times 10^4\ V$

We need to find the maximum energy of the photon of the x- ray radiation. The energy required to raise an electron through one volt is called electron volt.

$E=eV$

e is charge of electron

$E=1.6\times 10^{-19}\times 5\times 10^4$

$E=8\times 10^{-15}\ J$

So, the maximum energy of the x- ray radiation is $8\times 10^{-15}\ J$ . Hence, this is the required solution.

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2 years ago

An ideal gas is in a sealed rigid container. the average kinetic energy of the gas molecules depends most on

finlep [7]

It depends most on the temperature of the gas.

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3 years ago

The red light from a helium-neon laser has a wavelength of 721.4 nm in air. Find the speed, wavelength, and frequency of helium-

saveliy_v [14]

Answer:

(a) the speed of helium-neon laser light in air is 3 x 10⁸ m/s

the wavelength of helium-neon laser light in air is 721.4 nm

the frequency of helium-neon laser light in air is 415.86 THz

(b) the speed of helium-neon laser light in water is 2.26 x 10⁸ m/s

the wavelength of helium-neon laser light in water is 542.4nm

the frequency of helium-neon laser light in water is 416.67THz

the wavelength of helium-neon laser light in glass is 480.9nm

the frequency of helium-neon laser light in glass is 415.88THz

From the results above, it can be seen that speed of the light is directly proportional to its wavelength, while the frequency of the light remained fairly constant for the different media.

Explanation:

Part (a) the speed, wavelength, and frequency of helium-neon laser light in air

Given;

wavelength of helium-neon laser light in air, λ = 721.4 nm

speed of light in air, v = 3 x 10⁸ m/s

v = f λ

where;

f is the frequency of helium-neon laser light in air

$f = \frac{v}{\lambda} = \frac{3*10^8}{721.4 *10^{-9}} =4.1586*10^{14} \ Hz$

f = 415.86 THz

Part (b) the speed, wavelength, and frequency of helium-neon laser light in water

refractive index of water = 1.33

$Refractive \ index \ of \ water =\frac{speed \ of \ light \ in \ air}{speed \ of \ light \ in \ water} = \frac{wavelength \ of \ light \ in \ air}{wavelength \ of \ light \ in \ water}$

$speed \ of \ light \ in \ water = \frac{speed \ of \ light \ in \ air}{Refractive \ index \ of \ water} \\\\speed \ of \ light \ in \ water = \frac{3*10^8}{1.33} = 2.26 *10^8 \ m/s$

Again;

$wavelength \ of \ light \ in \ water = \frac{wavelength \ of \ light \ in \ air}{Refractive \ index \ of \ water} \\\\wavelength \ of \ light \ in \ water = \frac{721.4 \ nm}{1.33} = 542.4 \ nm$

$f = \frac{v}{\lambda} = \frac{2.26*10^8}{542.4 *10^{-9}} =4.1667*10^{14} \ Hz$

f = 416.67 THz

Part (c) the speed, wavelength, and frequency of helium-neon laser light in glass

Refractive index of glass = 1.5

$speed \ of \ light \ in \ glass = \frac{speed \ of \ light \ in \ air}{Refractive \ index \ of \ glass} \\\\speed \ of \ light \ in \ glass = \frac{3*10^8}{1.5} = 2 *10^8 \ m/s$

Also;

$wavelength \ of \ light \ in \ glass = \frac{wavelength \ of \ light \ in \ air}{Refractive \ index \ of \ glass} \\\\wavelength \ of \ light \ in \ glass = \frac{721.4 \ nm }{1.5} = 480.9 \ nm$

$f = \frac{v}{\lambda} = \frac{2*10^8}{480.9 *10^{-9}} =4.1588*10^{14} \ Hz$

f = 415.88 THz

5 0

3 years ago

Read 2 more answers

Isolating a variable in two equations is easiest when one of them has a coefficient of 1. Let's say we have the two equations 3A

xxTIMURxx [149]

Answer:

$B=3A-5$

Explanation:

Variable Isolation

It's a common practice when dealing with equations that we have to isolate one variable in terms of other variables and/or constants. The isolation of the variable usually implies adding, subtracting, multiplying or dividing by constants. The following example shows how to isolate the A:

$2A+3B=-4\\\\2A=-4-3B\\\\\displaystyle A=\frac{-4-3B}{2}$

We are required to find the equation where the variable has a coefficient of 1 and isolate it. The following equation fits into the description:

$3A-B=5$

Isolating B:

$B=3A-5$

3 0

3 years ago