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vodomira [7]
3 years ago
14

The gravitational acceleration on Mars is 3.71 m/s2. If the pendulums were set in motion on the red planet, how would that affec

t their periods?
Physics
1 answer:
Elanso [62]3 years ago
6 0
On Earth, the period of a pendulum is given by:
T_{earth}=2\pi  \sqrt{ \frac{L}{g_{earth} }
where L is the length of the pendulum and g_{earth}=9.81~m/s^2 is the gravitational acceleration on Earth.
Similarly, the period of the same pendulum on Mars will be
T_{mars}=2\pi \sqrt{ \frac{L}{g_{mars} }
where g_{mars}=3.71~m/s^2 is the gravitational acceleration on Mars.
Therefore, if we want to see how does the period of the pendulum on Mars change compared to the one on Earth, we can do the ratio between the two of them:
\frac{T_{mars}}{T_{earth}}= \sqrt{ \frac{g_{earth}}{g_{mars}} }  =  \sqrt{ \frac{9.81~m/s^s}{3.71~m/s^2} }=1.63
Therefore, the period of the pendulum on Mars will be 1.63 times the period on Earth.
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If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in
Sindrei [870]

Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

Moment of inertia I = 0.33 kg⋅m2

Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

The negative sign implies it's a negative acceleration.

Frictional torque that brought it to rest must be equal to the original torque.

Torqu = I x §

T = 0.33 x 0.33 = 0.11 Nm

5 0
3 years ago
Which graph BEST represents the relationship between the potential energy and kinetic energy of a cannon ball as it flies over t
nexus9112 [7]

Answer:C

Explanation:I think sorry if it’s wrong

6 0
3 years ago
Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?
frosja888 [35]

Answer:

The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Explanation:

From the question given above, the following data were obtained:

Height to which the target is located = 50 m

Initial velocity (u) = 20 m/s

To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 20² – (2 × 10 × h)

0 = 400 – 20h

Collect like terms

0 – 400 = – 20h

– 400 = – 20h

Divide both side by – 20

h = – 400 / – 20

h = 20 m

Thus, the the maximum height to which the cannon ball attained is 20 m.

From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

3 0
3 years ago
Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another
ioda

Answer:

6.1875\times 10^{-8}

Explanation:

Assuming uniform spread of sound with no significant reflections or absorption. We know that sound intensity varies I=\frac {k}{r^{2}} where r is the distance

Since intensity is given then when at 3 m

1.1\times 10^{-7}= \frac {k}{3^{2}}

k=3^{2}\times 1.1\times 10^{-7}= 9.9\times 10^{-7}

Since we have the constant then at 4m

Intensity, I= \frac {9.9\times 10^{-7}}{4^{2}}=6.1875\times 10^{-8}

8 0
3 years ago
Para abrir una puerta se comprime un resorte y, por defecto de este, la puerta se cierra automáticamente. ¿Qué forma de energía
harkovskaia [24]

Answer:

El resorte al comprimirse adquiere energía potencial elástica.

Explanation:

Por conservación de la energía, y si no hay agentes externos (como pérdida de energía por rozamiento), la energía cinética que proviene del movimiento de la puerta al abrirse se transfiere al resorte en forma de energía potencial elástica cuando el resorte se comprime.

En el proceso de descompresión ocurre lo contrario, es decir, la energía potencial elástica del resorte se transforma en energía cinética que es transferida a la puerta para cerrarse.      

Por lo tanto, el resorte al comprimirse adquiere energía potencial elástica.

Espero que te sea de utilidad!            

4 0
2 years ago
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