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kirill [66]
3 years ago
7

The radius of Mercury (from the center to just above the atmosphere) is 2440 km (2440 × 103 m), and its mass is 3 × 1023 kg. An

object is launched straight up from just above the atmosphere of Mercury.
a. What initial speed is needed so that when the object is far from Mercury its final speed is 2000 m/s?
b. What initial speed is needed so that when the object is far from Mercury its final speed is 0 m/s?
Physics
1 answer:
KonstantinChe [14]3 years ago
5 0

Answer:

(a)4676m/s

(b) 4227m/s

Explanation:

Let the mass of mercury be Mm = 3×10²³ kg and the mass of the object be M

Also let the distance from the center of mercury be R1 = 2.240×10⁶m let the distance the object is launched to be R

From the energy conservation equation

K1 + U1 = K2 + U2

K1, K2 = initial and final kinetic energies.

U1, U2 = initial and final potential energies.

1/2MV1² + (-GMm×M/R1) = 1/2MV2² + (-GMm×M/R)

Assuming that the very far R approaches infinity then the second term to the right side of the equation above approaches zero

So dividing through by M since M is common to both sides we have that

1/2V1² + (-G×Mm/R1) = 1/2V2²

Rearranging,

V1² = 2×( 1/2V2²+ GMm/R1 )

Given V2 = 2000m/s and also the constant value of G = 6.67×10‐¹¹ N•m²/kg²

Substituting,

V1² =2×( 1/2×2000² + 6.67×10-¹¹ ×3×10²³ /(2.24×10⁶))

V1² = 21866071.43

V1 = 4676m/s

(b) this is still similar to the case described above but this time V2 = 0m/s

V1² = 2×( 1/2V2²+ GMm/R1 ) = 2×(1/2×0+ GMm/R1) = 2GMm/R1

V1² = 2×6.67×10-¹¹ × 3×10²³/(2.24×10⁶)

= 17866071.43

V1 = 4227m/s

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zavuch27 [327]
'  +4 m/s² ' means that the pigeon's speed is  4 m/s greater every second.

Starting from zero speed, after 10 seconds, its speed is 

                           (10 x 4m/s)  =  40 m/s.

We can't say anything about its velocity, because we have
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3 years ago
Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 mi
zepelin [54]

Answer:

D = 1.8677 miles , θ = 24.28º at South of West

Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

     cos ( 360-30) = cos (-30) = x₁ / d

     sin (-30) = y₁ / d

     x₁ = d cos (-30)

     y₁ = d sin (-30)

     x₁ = 1.2 cos (-30) = 1,039 miles

     y₁ = 1.2 sin (-30) = -0.6 miles

second shift. d = 2.0 miles to 20º West of South

       cos (270-20) = x₂ / d

       cos (250) = y₂ / d

       x₂ = 2.0 cos 250 = -0.684 miles

       y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

       cos (180 + 30) = x₃ / d

       sin (210) = y₃ / d

       x₃ = 1.6 cos 210 = -1.3856 miles

       y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

       cos (90 + 15) = x₄ / d

       sin (105) = y₄ / d

       x₄ = 2.6 cos 105 = -0.6729 miles

       y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis  (West-East direction)

       X = x₁ + x₂ + x₃ + x₄

       X = 1.039 -0.684 - 1.3846 - 0.6729

       X = -1.7025 miles

   

       Y = y₁ + y₂ + y₃ + y₄

       Y = -0.6 -1.879 -0.8 +2.511

       Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

      D = √ (X² + Y²)

      D = √ (1.7025² + 0.768²)

      D = 1.8677 miles

let's use trigonometry to find the direction

       tan θ = Y / X

       θ = tan⁻¹ Y / x

       θ = tan⁻¹ (0.768 / 1.7025)

       θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

         θ = 24.28º at South of West

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3 years ago
A 438kg car is accelerating east at 2.55m/s^2. What is the total force acting east on the car
lisabon 2012 [21]

Answer:

<h2>1116.9 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 438 × 2.55

We have the final answer as

<h3>1116.9 N</h3>

Hope this helps you

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The glandular epithelial tissues make up the liver.

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Answer:

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