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3 years ago

Suppose an astronaut were to visit a planet where the force of gravity is half that of Earth. His mass on that planet would be:

1 answer:

8 0

"Mass" is what YOU ARE. It doesn't matter whether you're on Earth, on the moon, on an asteroid, on a meteoroid, on the sun, swimming in chicken soup on Saturn, or in a space capsule falling from one solar system body to another one. Your mass DOESN'T change.

Whatever mass the astronaut had after he ate that steak dinner just before he was launched, THAT's the mass he has while he's on the way to another planet, and THAT's the same mass he'll have when he gets there. It makes no difference what the gravity is like on the other planet. His mass will not change. (choice-B)

Only his WEIGHT changes when he's in different places.

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You have to do the math of each and see which one adds up to 66.5

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A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 3300 kcal o

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Answer

given,

heat added to the gas,Q = 3300 kcal

initial volume, V₁ = 13.7 m³

final volume, V₂ = 19.7 m³

atmospheric pressure, P = 1.013 x 10⁵ Pa

a) Work done by the gas

W = P Δ V

W = 1.013 x 10⁵ x (19.7 - 13.7)

W = 6.029 x 10⁵ J

b) internal energy of the gas = ?

now,

change in internal energy

Δ U = Q - W

Q = 3300 x 10³ cal

Q = 3300 x 10³ x 4.186 J

Q = 1.38 x 10⁷ J

now,

Δ U = 1.38 x 10⁷ - 6.029 x 10⁵

Δ U = 1.32 x 10⁷ J

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3 years ago

The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

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3 years ago

To balance a chemical equation we use______. -coefficients -subscripts

Vitek1552 [10]

Answer:

Coefficients

Explanation:

Coefficients are used to balance chemical equations.
According to the law of conservation of mass, the mass of the reactants in a chemical equation should be equivalent to the mass of the products.
The conservation of mass is achieved in chemical equations is achieved through balancing chemical equations.
Balancing chemical equations is a try and error of putting appropriate coefficients to the reactants or products to ensure that the number of atoms of each element are equal on the side of the reactants and that of products.

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3 years ago

9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Sku

Sedaia [141]

Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately 33.23 m/s in a direction from the North of

approximately 9.18°.

Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

v₁ = -( $\frac{\sqrt{2} }{2}$ × 3.5)·i + ( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

v₂ = ( $\frac{\sqrt{2} }{2}$ × 4.0)·i - ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-( $\frac{\sqrt{2} }{2}$ × 3.5) - ( $\frac{\sqrt{2} }{2}$ × 4.0))·i + (( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5) + ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15))·j

v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

Red Skull's velocity relative to Captain America, v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

Red Skull appears to be moving West at 5.3 m/s and North at 32.8 m/s

The direction is $arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}$

Therefore;

Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

$|v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23$ ·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ 33.23 m/s

Learn more here:

brainly.com/question/24430414

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2 years ago