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hodyreva [135]
2 years ago
14

In concept simulation 10.2 you can explore the concepts that are important in this problem. astronauts on a distant planet set u

p a simple pendulum of length 1.2 m. the pendulum executes simple harmonic motion and makes 100 complete oscillations in 360 s. what is the magnitude of the acceleration due to gravity on this planet
Physics
1 answer:
Margarita [4]2 years ago
5 0
Period of a simple pendulum = 2π √(L/G)

(360s/100) = 2π √(1.2m/G)

1.8s / π = √1.2m / √G

√G · (1.8s/π) = √1.2m

√G = (π · √1.2m) / 1.8s

Square each side:

G = π² · 1.2m / 3.24 s²

G = (1.2 · π² / 3.24)  m/s²

G = 3.66 m/s²

So I just went and looked up Mars gravity. Floogle says it's 3.711 m/s² there.
That seems awfully close ... only 1.4% greater than our astronauts measured. 
You don't suppose . . . . .
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Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire
Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

F = ma \rightarrow a = \frac{F}{m}

Replacing,

a =\frac{42N}{83000kg}

a =5.06*10^{-4}m/s^2

The total speed change

\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity } we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}

We have that,

t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }

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7 0
2 years ago
A passenger compartment of a rotating amusement park ride contains a bench on which a book of mass
Basile [38]

a) 120 s

b) v = 0.052R [m/s]

Explanation:

a)

The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).

The graph of the problem is missing, find it in attachment.

To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.

The first point we take is t = 0, when the position of the book is x = 0.

Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.

Therefore, the period is

T = 120 s - 0 s = 120 s

b)

The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.

The perimeter of the wheel is:

L=2\pi R

where R is the radius of the wheel.

The period of revolution is:

T=120 s

Therefore, the tangential speed of the book is:

v=\frac{L}{T}=\frac{2\pi R}{120}=0.052R

8 0
3 years ago
Several light bulbs, each of resistance 1.5 Ω, are connected in a series across a 120 V source of emf. If the current through th
Leni [432]
<h3><u>Answer;</u></h3>

40 light bulbs

<h3><u>Explanation</u>;</h3>

The total resistance of components or bulbs in series is given as the sum of resistance of all the components.

Thus; if there are bulbs in series each with a resistance of 1.5 Ω, the the total resistance will be; 1.5nΩ

From the ohms law;

V = IR , where V is the voltage, I is the current and R is the resistor.

Thus; R = V/i

         R = 120/2

            = 60 Ω

But, there are n bulbs each with 1.5 Ω; thus there are;

n = 60/1.5

<u>  = 40 Bulbs </u>

7 0
3 years ago
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