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Zarrin [17]
2 years ago
15

The moon has a mass of 7.4 × 1022 kg and completes an orbit of radius 3.8×108 m about every 28 days. The Earth has a mass of 6 ×

1024 kg and completes an orbit of radius 1.5 × 1011 m every year. What is the speed of the Moon in its orbit? Answer in units of m/s.
Physics
1 answer:
matrenka [14]2 years ago
5 0

To solve this problem it is necessary to apply the kinematic equations of movement description, where the speed of a body is defined as the distance traveled in a given time, that is to say

v = \frac{x}{t}

where,

x = Displacement

t = time

Since the moon is making a path equal to that of a circle, we know by geometry that the perimeter of a circle is given by

x = 2\pi r

x = 2\pi (3.8*10^8)

x=2.39*10^9m

At the same time we have that time is equal to

t = 28days(\frac{86400s}{1day})

t = 2419200s

Using the equation for velocity we have finally that

v= \frac{x}{t}

v = \frac{2.39*10^9}{2419200}

v = 987.92 m/s

Therefore the speed of the Moon in its orbit is 987.92m/s

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marysya [2.9K]

Answer:

From the initial height h

Explanation:

When a material or substance is drop from a height h, it possesses potential energy, immediately it is dropped from that height, the potential energy is gradually converted to kinetic energy, it gets to a point where the potential energy equals the kinetic energy, as the material touches the ground, all potential energy has been converted to kinetic energy already

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2 years ago
An object at rest is suddenly broken apart into two fragments by an explosion one fragment acquires twice the kinetic energy of
satela [25.4K]
<span>First, we use the kinetic energy equation to create a formula: Ka = 2Kb 1/2(ma*Va^2) = 2(1/2(mb*Vb^2)) The 1/2 of the right gets cancelled by the 2 left of the bracket so: 1/2(ma*Va^2) = mb*Vb^2 (1) By the definiton of momentum we can say: ma*Va = mb*Vb And with some algebra: Vb = (ma*Va)/mb (2) Substituting (2) into (1), we have: 1/2(ma*Va^2) = mb*((ma*Va)/mb)^2 Then: 1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2 We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1: 1/2(ma) = (ma^2)/mb Cancel the ma of the left, leaving the right one with exponent 1: 1/2 = ma/mb And finally we have that: mb/2 = ma mb = 2ma</span>
8 0
2 years ago
A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the
lions [1.4K]

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

3 0
2 years ago
An object moving on the x axis with a constant acceleration increases its x coordinate by 82.9 m in a time of 2.51 s and has a v
Aneli [31]

We are given: Final velocity (v_f)=20 m/s .

Time t= 2.51 s and

distance s = 82.9 m.

We know, equation of motion

v_f = v_i + at.

Let us plug values of final velocity, and time in above equation.

20=v_i+a(2.51)

20=v_i+2.51a

Subtracting 2.51a from both sides, we get

20-2.51a=v_i  -----------equation(1)

Using another equation of motion

v_f-v_i=2as

Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.

We get,

20-(20-2.51a)=2*a(82.90)

Now, we need to solve it for a.

20-20+2.51a=165.8a.

-163.29a=0

a=0.

So, the acceleration would be 0 m/s^2.


5 0
2 years ago
A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly abov
Rudik [331]

Answer:

Apparent depth (Da) = 60.15 cm (Approx)

Explanation:

Given:

Distance from fish (D) = 80 cm

Find:

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Computation:

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So,

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Apparent depth (Da) = 60.15 cm (Approx)

5 0
2 years ago
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