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3 years ago

How does Mercury's close proximity to the sun and thin atmosphere affect its ability to maintain liquid water

Physics

1 answer:

alekssr [168]3 years ago

5 0

Answer

A thin atmosphere does not supply much oxygen, and the heat from the sun would evaporate it, because mercury is close to the sun.

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only a relatively small part of the electromagnetic spectrum is visible. what determines which bands of the electromagnetic spec

swat32

Answer:

hsnzbssj

Explanation:

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3 0

3 years ago

A solid cylinder has a mass of 5 kg and radius of 2 m and is fixed so that it is able to rotate freely around its center without

kari74 [83]

Answer:

0.893 rad/s in the clockwise direction

Explanation:

From the law of conservation of angular momentum,

angular momentum before impact = angular momentum after impact

L₁ = L₂

L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)

L₂ = angular momentum of cylinder and angular momentum of bullet after collision.

L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision

So,

L₁ = L₂

L₁ = (I₁ + I₂)ω

ω = L₁/(I₁ + I₂)

ω = L₁/(1/2MR² + mR²)

ω = L₁/(1/2M + m)R²

substituting the values of the variables into the equation, we have

ω = L₁/(1/2M + m)R²

ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²

ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)

ω = + 9 kgm²/s/(2.52 kg)(4 m²)

ω = +9 kgm²/s/10.08 kgm²

ω = + 0.893 rad/s

The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.

7 0

3 years ago

A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightl

USPshnik [31]

Given Information:

Wavelength = λ = 39.1 cm = 0.391 m

speed of sound = v = 344 m/s

linear density = μ = 0.660 g/m = 0.00066 kg/m

tension = T = 160 N

Required Information:

Length of the vibrating string = L = ?

Answer:

Length of the vibrating string = 0.28 m

Explanation:

The frequency of beautiful note is

f = v/λ

f = 344/0.391

f = 879.79 Hz

As we know, the speed of the wave is

v = √T/μ

v = √160/0.00066

v = 492.36 m/s

The wavelength of the string is

λ = v/f

λ = 492.36/879.79

λ = 0.5596 m

and finally the length of the vibrating string is

λ = 2L

L = λ/2

L = 0.5596/2

L = 0.28 m

Therefore, the vibrating section of the violin string is 0.28 m long.

3 0

3 years ago

당신을 결코 뒤지지 않는 사람을 사랑하는 방법과 당신에게 맞는 사람을 찾는 방법

Fiesta28 [93]

Explanation:

내가 좋은 사람이 필요하고 내가 믿을 수 있는 사람이 필요하기 때문에 친구가 없다고 말하는 사람은 거의 없지만 대부분은 가짜이고 한국어를 모릅니다.

7 0

2 years ago

ASTM A229 oil-tempered carbon steel is used for a helical coil spring. The spring is wound with D = 50 mm d = 10.0 mm, and a pit

horrorfan [7]

Answer

given,

D = 50 mm = 0.05 m

d = 10 mm = 0.01 m

Force to compress the spring

$F = \dfrac{d^4G\delta}{8D^3N}$

$\dfrac{\delta}{N} = p - d = 14 - 10 = 4 mm$

$F = \dfrac{d^4G}{8D^3}\times 0.004$

$F = \dfrac{0.1^4\times 79\times 10^9}{8\times 0.05^3}\times 0.004$

F = 3160 N

stress correction factor from stress correction curve is equal to 1.1

now, calculation of corrected stress

$\tau = \dfrac{8FDk_s}{\pi d^3}$

$\tau = \dfrac{8\times 3160 \times 0.05 \times 1.1}{\pi \times 0.01^3}$

= 442.6 Mpa

The tensile strength of the steel material of ASTM A229 is equal to 1300 Mpa

now,

$\tau_s \leq 0.45 S_u$

$\tau_s \leq 0.45 \times 1300$

$\tau_s \leq 585\ Mpa$

since corrected stress is less than the $\tau_s$

hence, spring will return to its original shape.

6 0

3 years ago

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