Ask question

Ask question

All categories

3 years ago

8

If you were an Astronaut in the middle of the near side of the moon during a full moon, how would the ground around you look? Ho

w would the earth high in the sky look? Describe what is in sunlight and what is in darkness

1 answer:

Alenkasestr [34]3 years ago

6 0

Dark light a round ball and stuff

You might be interested in

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00

Nutka1998 [239]

Answer:

$6.99535\times 10^{-6}\ V/m$

Explanation:

P = Power Output = 1000 W

r = Radius = 35000000 m

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

Intensity of Electric radiation is given by

$I=\dfrac{P}{A}\\\Rightarrow I=\dfrac{P}{4\pi r^2}\\\Rightarrow I=\dfrac{1000}{4\pi\times 35000000^2}\ W/m^2$

Intensity of Electric radiation is given by

$I=\dfrac{1}{2}c\epsilon_0E_0\\\Rightarrow E_0=\sqrt{\dfrac{2I}{c\epsilon_0}}\\\Rightarrow E_0=\sqrt{\dfrac{2\times \dfrac{1000}{4\pi\times 35000000^2}}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E_0=6.99535\times 10^{-6}\ V/m$

The amplitude of the electric field vector is $6.99535\times 10^{-6}\ V/m$

6 0

4 years ago

What happens to the volume of a solid when it is moved to a larger container?

Goryan [66]

Remains unchanged (as it's a solid)

6 0

3 years ago

A sealed syringe contains 10 × 10−6 m3 of air at 1 × 105 Pa. The plunger is pushed until the volume of trapped air is 4 × 10−6 m

Scilla [17]

Answer: $P_2=2.5\times 10^5\ Pa$

Explanation:

Given

Initial volume $V_1=10\times 10^{-6}\ m^3$

Initial Pressure $P_1=10^5\ Pa$

Final trapped air $V_2=4\times 10^{-6}\ m^3$

If there is no change in temperature then We can write

$P_1V_1=P_2V_2\quad \text{(From ideal gas equation)}$

$P_2=\dfrac{P_1V_1}{V_2}$

$P_2=10^5\times \frac{10\times 10^{-6}}{4\times 10^{-6}}$

$P_2=2.5\times 10^5\ Pa$

6 0

3 years ago

Which statement is a hypothesis?

solong [7]

B.

This is because a hypothesis is an explanation explaining things

7 0

4 years ago

You are on the Pirates of the Caribbean attraction in the Magic Kingdom at Disney World. Your boat rides through a pirate battle

maxonik [38]

Answer:

a) v₀ₓ = 62.76 m / s, b) θ₁ = 17.6º, θ₂ = 67.0º

Explanation:

We can solve this exercise using the projectile launch ratios

a) Let's find the time it takes for the bullet to reach the water level

y = y₀ + v_{oy} t - ½ g t²

when it reaches the water its height is zero y = 0, as the bullet is fired horizontally its initial vertical velocity is zero

0 = y₀ + 0 - ½ g t²

t = $\sqrt{2y_o/g}$

t = $\sqrt{2 \ 7 /9.8}$

t = 1,195 s

now we can calculate the speed with the horizontal movement

x = v₀ₓ t

v₀ₓ = x / t

v₀ₓ = 75.0 / 1.195

v₀ₓ = 62.76 m / s

b) if the speed of the bullets is half of that found

v₀ = 62.76 / 2 = 31.38 m / s

let's write the expressions for the distance

x = v₀ cos θ t

y = y₀ + v_{oy} sin θ t - ½ g t²

t = $\frac{x}{v_o \ cos \theta}$

we substitute

$0 = y_o + v_o sin \theta \ \frac{x}{v_o \cos \thetay} - 1/2 g \ (\frac{x}{v_o \ cos \theta})^2$

$0 = y_o + x tan \theta - \frac{1}{2} g \ \frac{x^2}{ v_o^2 \ cos^2 \theta}$

let's use the identified trigonometry

sec² θ = 1 + tan² θ

sec θ = 1 / cos θ

we substitute

$0 = y_o + x tan \theta - \frac{g x^2}{2 v_o^2} ( 1 + tan^2 \theta)$

$\frac{g x^2}{2v_o^2} tan^2 \theta - x tan \theta + \frac{gx^2}{2v_o^2} - y_o = 0$

we change variable

tan θ = H

$\frac{gx^2}{2 v_o^2 } H^2 - x H + \frac{gx^2}{2v_o^2}-y_o =0$

we subtitle the values

$\frac{9.8 \ 75^2}{2 \ 31.38^2} H^2 - 75 H + \frac{9.8 \ 75^2}{2 \ 31.38^2}-7 =0$

27.99 H² - 75 H + 20.99 = 0

H² - 2.679 H + 0.75 = 0

we solve the quadratic equation

H = [2.679 ± $\sqrt{2.679^2 - 4 0.75}$ ] / 2

H = [2,679 ± 2,044] / 2

H₁ = 0.3175

H₂ = 2.3615

now we can find the angles

H₁ = tan θ₁

θ₁ = tan⁻¹ H₁

θ₁ = tan⁻¹ 0.3175

θ₁ = 17.6º

θ₂ = 67.0º

for these two angles the bullet hits the boat

3 0

3 years ago