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3 years ago

Explain the qualifications that must be met for any vehicle to go in space.

Physics

2 answers:

Lubov Fominskaja [6]3 years ago

8 0

Answer: It must be able to handle a lot of pressure and heat

Explanation:

Shkiper50 [21]3 years ago

5 0

Ais 1+1is2 so you put 9x9 in 2+2 what the answer is your answer hope it help

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Even though you praise your dog for sitting inside on his bed and chewing the bone you gave him, your dog insists on going outsi

qwelly [4]

Answer:

dog

Explanation:

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2 years ago

How many days does it take for a free to grow?

vovikov84 [41]

Idk what is growing but if it’s a free than c

7 0

3 years ago

A wave with a frequency of 80 Hz travels through rubber with a wavelength of 7.0 m. What is the velocity of the wave?

Scorpion4ik [409]

Answer:

560 m/s

Explanation:

Given,

Frequency ( f ) = 80 hz

Wavelength ( λ ) = 7.0 m = 7m

To find : Velocity ( v )

Formula : -

v = f λ

v = 80 x 7

v = 560 m/s

Hence, the velocity of the wave is 560 m/s.

5 0

2 years ago

What type of force is keeping this person from falling through the wall

Softa [21]

Answer:

Friction

Explanation:

Friction keeps things from sliding into things according to newton's laws

5 0

2 years ago

We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes

Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

$1.56421\times 10^{-6}\ T$

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = $\frac{d}{2}=\frac{7}{2}=3.5\ cm$

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

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

The intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2$

5% of this energy goes to the visible light so the intensity is

$I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2$

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

$u=\frac{1}{2}\epsilon_0E^2$

Energy density is also given by

$\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m$

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

$B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T$

The amplitude of the magnetic field at this surface is $1.56421\times 10^{-6}\ T$

7 0

3 years ago