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3 years ago

Calculate the average speed of spacecraft orbiting mars

1 answer:

4 0

That completely depends on the size of the orbit. Similarly at Earth, a TV satellite takes 1 day for an orbit, but the Moon takes 27.3 days.

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You have $20 and your boss gives you $10

balandron [24]

You start out with $20 then the next second you get $10 then the next second you get another $10 and the next second you got another $10 which equals up to $50

3 0

4 years ago

Read 2 more answers

A circus tightrope walker weighing 800 N is standing in the middle of a 15 meter long cable stretched between two posts. The cab

Lorico [155]

Answer:

T = 10010 N

Explanation:

To solve this problem we must use the translational equilibrium relation, let's set a reference frame

X axis

Fₓ-Fₓ = 0

Fₓ = Fₓ

whereby the horizontal components of the tension in the cable cancel

Y Axis

$F_{y} + F_{y} - W =0$

2 $F_{y}$ = W

let's use trigonometry to find the angles

tan θ = y / x

θ = tan⁻¹ (0.30 / 0.50 L)

θ = tan⁻¹ (0.30 / 0.50 15)

θ = 2.29º

the components of stress are

F_{y} = T sin θ

we substitute

2 T sin θ = W

T = W / 2sin θ

T = $\frac{ 800}{ 2sin 2.29}$

T = 10010 N

4 0

3 years ago

The figure below shows an acceleration-versus-force graph for three objects pulled by rubber bands. The mass of object 2 is 0.20

lawyer [7]

Such a great question
answer:
object 1 = 0.1kg
object 2 = 0.5kg

if you can read this

4 0

3 years ago

Read 2 more answers

For a particular pipe in a pipe-organ, it has been determined that the frequencies 576 Hz and 648 Hz are two adjacent natural fr

N76 [4]

Answer:

(A) Fo = 72 Hz

(B) The pipe is open at both ends

This problem involves the application of the knowledge of standing waves in pipes.

Explanation:

The full solution can be found in the attachment below.

For pipes open at both ends the frequency of the pipe is given by

F = nFo = nv/2L where n = 1, 2, 3, 4.....

For pipes closed at one end the frequency of the pipe is given by

F = nFo = nv/4L where n = 1, 3, 5, 7...

The full solution can be found in the attachment below.

3 0

3 years ago

A uniform 1.3-kg rod that is 0.67 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both

cestrela7 [59]

Answer:

α = 5.75°

Explanation:

In this case, the problem states that both springs have identical lenghts and we also have theri constant. We want to know the angle of the rod with the horizontal. This can be found with the following expression:

sinα = Δx/L

α = sin⁻¹ (Δx/L) (1)

However, we do not have Δx. This can be found when half of the weight of the rod is balanced. In this way:

F₁ = k₁*x₁ ----> x₁ = F₁ / k₁ (2)

And the force is the weight in half so: F₁ = mg/2

Replacing in (2) we have:

x₁ = (1.3 * 9.8) / (2 * 58) = 0.1098 m

Doing the same thing with the other spring, we have:

x₂ = (1.3 * 9.8) / (2 * 36) = 0.1769 m

Now the difference will be Δx:

Δx = 0.1769 - 0.1098 = 0.0671 m

Finally, we can calculate the angle α, from (1):

α = sin⁻¹(0.0671 / 0.67)

Hope this helps

3 0

4 years ago