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solmaris [256]
1 year ago
7

A truck is carrying a refrigerator as shown in the figure. The height of the refrigerator is 158.0 cm, the width is 60.0 cm. The

center of gravity of the refrigerator is 67.0 cm above the midpoint of the base.
What is the maximum acceleration the truck can have so that the refrigerator doesn't tip over? (The static friction between the truck and the refrigerator is very large.)

Physics
1 answer:
Arada [10]1 year ago
3 0

The maximum acceleration the truck can have so that the refrigerator does not tip over is 4.15 m/s².

<h3>What will be the maximum acceleration of the truck to avoid tipping over?</h3>

The maximum acceleration is obtained by taking clockwise moments about the tipping point of rotation.

Clockwise moment = Anticlockwise moment

Ft * 1.58 m = F * 0.67 m

where

  • Ft is tipping force = mass * acceleration, a
  • F is weight = mass * acceleration due to gravity, g

m * a * 1.58 = m * 9.81 * 0.67

a = 4.15 m/s²

The maximum acceleration the truck can have so that the refrigerator does not tip over is 4.15 m/s².

In conclusion, the acceleration of the truck is found by taking moments about the tipping point.

Learn more about moments of forces at: brainly.com/question/27282169

#SPJ1

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2 years ago
determine the loudness (in decibels) of the sound at a rock concert if the intensity of the sound is 1 x 10–1 w/m2. remember, th
EleoNora [17]

The loudness of the sound at the rock concert, where the intensity of the sound is1 x 10⁻¹ Wm⁻² is  110 dB.

Here we are dealing with loudness which is the perception of the Intensity of the sound.

The formula  to refer to in order  to  find the value of the loudness of a sound is ,

  db= 10log(I/I₀)

As we are provided with the current intensity which is  1 x 10⁻¹ Wm⁻². and the initial intensity which is  1 x 10⁻¹² Wm⁻².

So, by substituting the required values in the formula we get

db= 10 * log( 1 x 10⁻¹ /1 x 10⁻¹²)

 = 10 * 11 log(10)

 = 110

So, the result is 110 dB.

To know more about the intensity of sound refer to the link brainly.com/question/9323731?referrer=searchResults.

To know more about questions related to loudness refer to the link brainly.com/question/21094511?referrer=searchResults.

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4 0
2 years ago
A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.
NemiM [27]

Answer:

\dfrac{K_t}{K_r}=\dfrac{5}{2}

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, \omega=2.8\ rad/s

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

K_t=\dfrac{1}{2}mv^2

K_t=\dfrac{1}{2}m(r\omega)^2

K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2

The rotational kinetic energy of the ball is :

K_r=\dfrac{1}{2}I \omega^2

K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2

K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2

Ratio of translational to the rotational kinetic energy as :

\dfrac{K_t}{K_r}=\dfrac{5}{2}

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

4 0
3 years ago
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