I need to know how to do it and what the answer is

What is the centripetal acceleration of a ball on a string that is whirled horizontally over a boy’s head? The radius of the cir

Leya [2.2K]

Well the centripetal acceleration would be 9.82 squared. That is the correct answer. I hope this helped you. Have a great day! :)

4 0

2 years ago

Read 2 more answers

Emotions refer to the act or process of knowing. True or False

vaieri [72.5K]

Answer:

True

Explanation:

Because they are central to the human experience and are related to everything we do .

4 0

3 years ago

A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po

Umnica [9.8K]

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + $0.5mV1^{2}$ = mgh2 + $0.5mV2^{2}$

where

speed at highest point = V2

speed at lowest point = V1

mass of the girl and swing = m

at the highest point, the speed is minimum (V1 = 0)

at the lowest point the speed is maximum (V2 is the maximum speed)

therefore the equation becomes mgh1 + $0.5mV1^{2}$ = mgh2

m(gh1 + $0.5V1^{2}$ ) = m(gh2)

gh1 + $0.5V1^{2}$ = gh2

V1 = $\sqrt{\frac{gh2 - gh1}{0.5}}$

now we can substitute all required values into the equation above.

V1 = $\sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}$

V1 = $\sqrt{\frac{32.34}{0.5}}$

V1 =8.1 m/s

8 0

2 years ago

An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a

Morgarella [4.7K]

Answer:

0.37 m

Explanation:

The angular frequency, ω, of a loaded spring is related to the period, T, by

$\omega = \dfrac{2\pi}{T}$

The maximum velocity of the oscillation occurs at the equilibrium point and is given by

$v = \omega A$

A is the amplitude or maximum displacement from the equilibrium.

$v = \dfrac{2\pi A}{T}$

From the the question, T = 0.58 and A = 25 cm = 0.25 m. Taking π as 3.142,

$v = \dfrac{2\times3.142\times0.25\text{ m}}{0.58\text{ s}} = 2.71 \text{ m/s}$

To determine the height we reached, we consider the beginning of the vertical motion as the equilibrium point with velocity, v. Since it is against gravity, acceleration of gravity is negative. At maximum height, the final velocity is 0 m/s. We use the equation

$v_f^2 = v_i^2+2ah$

$v_f$ is the final velocity, $v_i$ is the initial velocity (same as v above), a is acceleration of gravity and h is the height.

$h = \dfrac{v_f^2 - v_i^2}{2a}$

$h = \dfrac{0^2 - 2.71^2}{2\times-9.81} = 0.37 \text{ m}$

3 0

2 years ago

A package of aluminum foil contains 50. Ft2 of foil, which weighs approximately 7.0 oz . Aluminum has a density of 2.70 g/cm3. W

kumpel [21]

We know density = Mass / Volume

So Volume = Mass/Density

Volume = Area * Thickness

$Area = 50 ft^2 = 4.645152 m^2\\ \\Mass = 7 oz = 0.198447 kg\\ \\Density = 2.70 g/cm^3 = 2700 kg/m^3\\ \\Substituting \\ \\4.645152 *t = \frac{0.198447}{2700} \\ \\t = 1.58*10^{-5}m = 1.58*10^{-2}mm$

So the approximate thickness of the foil in millimeters = $1.58*10^{-2}mm$

6 0

2 years ago