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3 years ago

What would be the weight of a 59.1-kg astronaut on a planet with the same density as Earth and having twice Earth's radius?

1 answer:

5 0

You could always directly use F = G M m / r², but that takes forever.

It is best to know that, on Earth's surface, the astronaut would weigh 
F = m g = 579.2 N 
where m = 59.1 kg and g = 9.8 m/s². 

You then notice from F = G M m / r² that doubling the mass of the earth would double his weight and doubling the radius would give 1/4 his original weight. So... 
F = 1/4 * 2 * 579.2 N = 290 N

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Which of these stars has the hottest core?A)a blue main-sequnce star. B)a red supergiant. C)a red main-sequence star.

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The question you have asked is a

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3 years ago

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Which equation can be used to calculate the normal force on an object if you know the speed of the object, the coefficient of ki

tino4ka555 [31]

Answer:

Explanation:

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2 years ago

Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis

ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × $10^{-9}$ m

wavelength λ = 700 nm = 700 × $10^{-9}$ m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ ...........1

here m is 1 for single slit and d is = $\frac{1}{900*10^3 m}$

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 380 × $10^{-9}$

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 700 × $10^{-9}$

θ = 0.681 rad

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4 years ago

What is the displacement of the student from the bus stop to the mailbox?

Reika [66]

Answer:C

Explanation:

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3 years ago

Each plate of a parallel‑plate capacitor is a square of side 3.63 cm, and the plates are separated by 0.473 mm. The capacitor is

NARA [144]

Answer:

E= 55.53 x 10³ V/m

Explanation:

Given that

a= 3.63 cm

Area ,A= a²

distance ,d= 0.473 mm

Stored energy ,U = 8.49 nJ

Value of capacitor given as

$C=\dfrac{\varepsilon _oA}{d}$

By putting the values

$C=\dfrac{8.85\times 10^{-12}\times 3.63^2\times 10^{-4}}{0.473\times 10^{-3}}$

C=2.46 x 10⁻¹¹ F

$U=\dfrac{1}{2}CV^2$

V=Voltage difference

$V=\sqrt{\dfrac{2U}{C}}$

$V=\sqrt{\dfrac{2\times 8.49\times 10^{-9}}{2.46\times 10^{-11}}}$

V=26.27 V

V= E d

E=Electric filed

26.27 = E x 0.473 x 10⁻³

E= 55.53 x 10³ V/m

7 0

3 years ago