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3 years ago

A car travels 200km in 3.0 hours. Determine the average velocity of the car

Physics

1 answer:

Masja [62]3 years ago

4 0

Answer:

Explanation:

The average velocity of the car can be found by using the formula

$a = \frac{d}{t } \\$

d is the distance

t is the time taken

From the question we have

$a = \frac{200}{3} \\ = 66.66666...$

We have the final answer as

Hope this helps you

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The photographer realizes that with the lens she is currently using, she can't fit the entire landscape she is trying to photogr

dexar [7]

She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.

What is focal length?

The focal length is a measure of how strongly the system converges or diverges light.

A positive focal length indicates that a system converges light, while a negative focal length indicates that the system diverges light.

For a standard rectilinear lens,

FOV = 2 arctan (x/2f)

FOV ∝ 1 / f

where x is the diagonal of the film.

Focal length (f) and field of view (FOV) of a lens are inversely proportional.

From the equation we can say that,

A shorter focal length gives you a wide angle of view which allows more view to fit in the frame.

Hence,

She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.

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2 years ago

What is the equation for determining the value of a specific gas constant using the universal gas constants

maria [59]

Answer:

Physically, the gas constant is the constant of proportionality that relates the energy scale in physics to the temperature scale, when a mole of particles at the stated temperature is being considered. Thus, the value of the gas constant ultimately derives from historical decisions and accidents in the setting of the energy and temperature scales, plus similar historical setting of the value of the molar scale used for the counting of particles.

Explanation:

Pa follow

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3 years ago

In a region of space where gravitational forces can be neglected, a sphere is accelerated by a uniform light beam of intensity 8

BlackZzzverrR [31]

Answer:

The correct answer is B

Explanation:

To calculate the acceleration we must use Newton's second law

F = m a

a = F / m

To calculate the force we use the defined pressure and the radiation pressure for an absorbent surface

P = I / c absorbent surface

P = F / A

F / A = I / c

F = I A / c

The area of area of a circle is

A = π r²

We replace

F = I π r² / c

Let's calculate

F = 8.0 10⁻³ π (1.0 10⁻⁶)²/3 10⁸

F = 8.375 10⁻²³ N

Density is

ρ = m / V

m = ρ V

m = ρ (4/3 π r³)

m = 4500 (4/3 π (1 10⁻⁶)³)

m = 1,885 10⁻¹⁴ kg

Let's calculate the acceleration

a = 8.375 10⁻²³ / 1.885 10⁻¹⁴

a = 4.44 10⁻⁹ m/s² absorbent surface

The correct answer is B

4 0

3 years ago

Calculate the potential difference across a 25-Ohm. resistor if a 0.3-A current is flowing through it.

nikdorinn [45]

Answer:7.5V

Explanation:

Ohm's law, V=IR

so, V=0.3×25

V=7.5V

4 0

4 years ago

Read 2 more answers

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

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7 0

3 years ago