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3 years ago

Which of these statements suggests that the orbital period of Venus is shorter than that of Saturn?

Physics

1 answer:

Snowcat [4.5K]3 years ago

5 0

Answer:

C) Venus is closer to the Sun than Saturn is.

Explanation:

This is the right answer.

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How much heat would be absorbed by 75.20 g of iron when heated from 22 C to 28 C

Bumek [7]

Mass x SH x °C (or K) ΔT

= 75g x 0.45J/g/K x 6.0 ΔT

= 202.5 Joules of heat absorbed.

(202.5J / 4.184J/cal = 48.4 calories).

I guess that is the answer

8 0

3 years ago

Read 2 more answers

A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on

n200080 [17]

Answer:

6858.5712 m/s

Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

V = 6858.5712 m/s

6 0

2 years ago

The old rubber boot has two leaks. The top of the boot is 0.3 m higher than the leaks. What is the velocity of the water coming

Otrada [13]

Answer:

Leak 1 = 3.43 m/s

Leak 2 = 2.42 m/s

Explanation:

Given that the top of the boot is 0.3 m higher than the leaks.

Let height H = 0.3m and the acceleration due to gravity g = 9.8 m/s^2

From the figure, the angle of the leak 1 will be approximately equal to 45 degrees. While the leak two can be at 90 degrees.

Using the third equation of motion under gravity, we can calculate the velocity of leak 1 and 2

Find the attached files for the solution and figure

7 0

2 years ago

A closed system:

adoni [48]

The answer is b i just did the test

3 0

1 year ago

Consider a double slit experiment in the air. The wavelength of light is 500nm light, the screen is 1m away and two adjacent bri

Bess [88]

Answer: $0.75\ cm$

Explanation:

Given

Wavelength of light $\lambda=500\ nm$

Screen is $D=1\ m$ away

Distance between two adjacent bright fringe is $\Delta y=\dfrac{\lambda D}{d}$

When same experiment done in water, wavelength reduce to $\dfrac{\lambda }{\mu}$

So, the distance between the two adjacent bright fringe is $\Delta y'=\dfrac{\lambda D}{\mu d}$

Keeping other factor same, distance becomes

$\Rightarrow \dfrac{1}{\frac{4}{3}}=\dfrac{3}{4}\quad \text{Refractive index of water is }\dfrac{4}{3}\\\\\Rightarrow 0.75\ cm$

3 0

2 years ago