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Lelu
4 days ago
12

Two balls of unequal mass are hung from two springs that are not identical. The springs stretch the same distance as the two sys

tems reach equilibrium. Then both springs are compressed and released. Which one oscillates faster? a. Springs oscillate with the same frequency. b. The spring with the heavier ball. c. The spring with the lighter ball. d. It is impossible do determine without additional data.
Physics
1 answer:
Ostrovityanka [942]4 days ago
3 0
a. Springs oscillate at an identical frequency.
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A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i
Sav [1095]

Respuesta:

11.4 m/s

Explicación:

La fórmula para la aceleración centrípeta es:

a=\frac{v^2}{R}

donde, a es la aceleración, v la velocidad alrededor de la circunferencia y R el radio del círculo.

En este problema,

a = g = aceleración debida a la gravedad en la cima = 9.81\ m/s^2

v = ?

R = 13.2 m

Por lo tanto,

9.81=\frac{v^2}{13.2}

v^2=9.81\times {13.2}

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8 0
13 days ago
Compressed air is used to fire a 60 g ball vertically upward from a 0.70-m-tall tube. The air exerts an upward force of 3.0 N on
Yuliya22 [1153]

Answer:

2.87 m

Explanation:

Given parameters:

Mass of the ball (m) = 60 g = 0.06 kg

Height of the tube (h) = 0.70 m

Force applied on the ball by compressed air (F) = 3.0 N

Initial velocity of the ball (u) = 0 m/s (Assumed)

Final velocity of the ball at the tube's exit (v) =?

Acceleration of the ball (a) =?

The ball's weight is derived from multiplying mass and gravity. Therefore,

Weight (W) = mg=0.06\times 9.8=0.588\ N

Thus, the total force acting on the ball equals the net of upward force minus the weight.

Net force = Air force - Weight

F_{net}=F-mg\\F_{net}=3.0-0.588 = 2.412\ N

According to Newton's second law, net force equals the mass multiplied by acceleration.

F_{net}=ma\\\\a=\frac{F_{net}}{m}=\frac{2.412\ N}{0.06\ kg}=40.2\ m/s^2

Acceleration (a) is calculated as 40.2 m/s².

Using the motion equation, we find:

v^2=u^2+2ah\\\\v^2=0+2\times 40.2\times 0.7\\\\v=\sqrt{56.28}=7.5\ m/s

Let’s denote the maximum height achieved as 'H'.

Next, we apply the principle of energy conservation from the pipe's peak to the maximum height.

A decrease in kinetic energy equals an increase in potential energy.

\frac{1}{2}mv^2=mgH\\\\H=\frac{v^2}{2g}

Substituting the values, we solve for 'H', yielding:

H=\frac{56.28}{2\times 9.8}\\\\H=\frac{56.28}{19.6}=2.87\ m

Hence, the ball ascends to a height of 2.87 m above the top of the tube.

6 0
4 days ago
If an instalment plan quotes a monthly interest rate of 4%, the effective annual/yearly interest rate would be _____________. 4%
inna [987]

Answer:

More than 48%

Explanation:

If the interest is calculated monthly based on the outstanding balance, it leads to an effective annual rate of...

(1 + 4%)^12 - 1 = 60.1%... more than 48%

4 0
9 days ago
What is the sources of error and suggestion on how to overcome it in the hooke's law experiment?
Yuliya22 [1153]
In the study of physics, Hooke's law can be expressed as:

F = kx

This law indicates that the spring force F is proportional to the extension x, with k being the spring constant.

In experiments, this is often examined using the setup illustrated in the included figure. The spring is tested, and a known weight is applied underneath it. This weight exerts a gravitational pull, essentially its weight, on the spring. While the spring elongates, the displacement can be measured using a ruler.

Several potential errors can arise during this experiment. Firstly, the person's measurement reading may be faulty. Digital scales offer greater accuracy as they reduce human error, while ruler readings can be subjective, especially if not viewed at eye level. Additionally, the object's weight may be inaccurately measured if the scale is untrustworthy. Lastly, the measuring equipment may not be correctly calibrated.

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serg [1189]

Answer:

wow, still in Canada, by the way, my answer is ayy

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