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solmaris [256]
2 years 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]2 years 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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Jim is driving a 2268-kg pickup truck at 22 m/s and releases his foot from the accelerator pedal. The car eventually stops due t
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Answer:

610 meters.

Explanation:

Because Jim released the accelerator, the truck started to slow down, so the friction force will eventually stop the truck.

the kinetic energy of the truck just after Jim released the pedal is:

E_k=\frac{1}{2}*m*v^2\\E_k=\frac{1}{2}*2268*(22)^2=548856J

The work done by the friction force is given by:

W_f=F_s*d\\\\d=\frac{548856J}{900N}\\\\d=610m

6 0
3 years ago
How many atoms are represented by one formula unit of aluminum sulfite, al2(so3)3?
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4 0
3 years ago
A refrigerator is used to cool water from 23°C to 5°C in a continuous manner. The heat rejected in the condenser is 570 kJ/min a
GenaCL600 [577]

Answer: Q=5.46 L/s

COP=2.58

Explanation:

Given that

Cp = 4.18 kJ/(kg.C

density  = 1 kg/L

Heat rejected Qr= 570 kJ/min

Power in put W= 2.65 KW

From first law of thermodynamics

U = W+ q

q = Heat absorbed

U = internal energy

W = workdone

U = 570 kJ/min  = 9.5 KW

9.5 = 2.65 + q

q = 6.85 KW

COP = q/W

COP = 6.58 / 2.65

COP=2.58

Lets take volume flow rate is Q

So mass flow rate of water m = ρ Q

q = m Cp ΔT

6.85 = 1 x Q x 4.18 ( 23-5)

Q=0.091 L/min

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7 0
3 years ago
A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th
Stolb23 [73]

Answer:

The current is  I_b  =  400 \ A

Explanation:

From the question we are told that

    The  length of the segment is  l  =  2.50  \  m

     The current is  I_a  =  1000 \ A

     The force felt is  F  =  4.0 \  N

        The distance of the second wire is  d =  5.0 \ cm  = 0.05 \  m

Generally the current on the second wire is mathematically represented as

        I_b  =  \frac{2 \pi * r * F }{ l *  \mu_o  *  I_a }

Here  \mu_o is the permeability of free space with value  \mu_o =  4 \pi * 10^{-7} \ N/A^2

=>      I_b  =  \frac{2 * 3.142  *  0.05 *  4 }{ 2.50  *  4\pi *10^{-7}  * 1000 }

=>      I_b  =  400 \ A

4 0
3 years ago
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