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2 years ago

Answer pls . ......

Physics

1 answer:

zhannawk [14.2K]2 years ago

3 0

Answer:

yeah its distance

Explanation:

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2. A car traveled 240 km in 3 hours. What was its average speed in km per hour?

vekshin1

Answer:

80 km

Explanation:

Since 240 divided by three is 80, the velocity the car is traveling at is 80 kilometers per hour.

8 0

2 years ago

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Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

3 years ago

A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a

Stella [2.4K]

Answer:

y = 67.6 feet, y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

tan θ = y / L

For the small triangle. Projector up to the person

tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

y₀ / (L -d) = y / L

y = y₀ L / (L-d)

The distance from the screen (d), we look for it with kinematics

v = d / t

d = v t

we replace

y = y₀ L / (L - v t)

y = 5.2 22 / (22 - 3 t)

y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

y = 114.4 / (22-13)

y = 67.6 feet

7 0

3 years ago

Daisy made the graph shown to display the data recorded during an experiment. Daisy is studying the effect of soil on the growth

andreev551 [17]

im in flvs too if thats what this is but anyway im doing it right now and i believe it is sunlight was not kept constant

3 0

3 years ago

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Please help me with this

Firlakuza [10]

Answer:

can't see anything sorry can't help

7 0

2 years ago

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Answer pls . ......​

Answer pls . ......