Which event occurs when Earth is in a position between the moon and the sun so that Earth's shadow falls on the moon?

1 answer:

6 0

The answer is A a solar eclipse is when the moon covers the sun for a short period of time while a lunar eclipse is when the moon is being covered by Earth
i hope this helped

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A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

2.87 = -1.442 + 1.8 u

u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0

4 years ago

Calculate the ratio of the drag force on a jet flying at 950 km/h at an altitude of 10 km to the drag force on a prop-driven tra

Black_prince [1.1K]

Answer:

$\frac{F_1}{F_2}=3.55$

Explanation:

F = Force

C = Drag coefficient equal for both aircrafts

ρ = Density of air

A = Surface area equal for both aircrafts

v = Velocity

$v_2=\frac{2}{5}v_1$

$F_1=\frac{1}{2}\rho_1 CAv_1^2$

$F_2=\frac{1}{2}\rho_2 CAv_2^2\\\Rightarrow F_2=\frac{1}{2}\rho_2 CA\left(\frac{2}{5}v_1\right)^2$

Dividing the above two equations we get

$\frac{F_1}{F_2}=\frac{\frac{1}{2}\rho_1 CAv_1^2}{\frac{1}{2}\rho_2 CA\left(\frac{2}{5}v_1\right)^2}\\\Rightarrow \frac{F_1}{F_2}=\frac{\rho_1}{\rho_2\frac{4}{25}}\\\Rightarrow \frac{F_1}{F_2}=\frac{0.38}{0.67\frac{4}{25}}\\\Rightarrow \frac{F_1}{F_2}=3.55$