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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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damaskus [11]

Answer:

22.5J

Explanation:

Here the force is given. Also, the displacement is given as 30cm.

First we should check if all the values are in their standard form.

Here 30cm should be converted to metre by dividing it with 100.

Which would give us 0.3m

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The bullet's horizontal velocity and vertical velocity vectors do not affect each other and are known as
lisabon 2012 [21]
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8 0
2 years ago
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Determine the length of the object shown <br> 1.2 cm <br> 1.3 cm <br> 1.25 cm <br> 1.250 cm
stiks02 [169]

The correct answer is 1.25 because it is 1/2 of 1 1/2 and that is 1.25.

6 0
3 years ago
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A.) A bullet accelerates at 8000 m/s2
lukranit [14]

Answer:

0.02 s

160 m/s

Explanation:

Given:

Δx = 1.6 m

v₀ = 0 m/s

a = 8000 m/s²

A) Find t.

Δx = v₀ t + ½ at²

1.6 m = (0 m/s) t + ½ (8000 m/s²) t²

t = 0.02 s

B) Find v.

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (8000 m/s²) (1.6 m)

v = 160 m/s

6 0
2 years ago
The body falls in a free fall 11s.Calculate: a) from what height the body of the paddle) what path did it fall during the last s
JulsSmile [24]

Answer:

a) h = 593.50 m

b) h₁₁ = 103 m

c) vf = 107.91 m/s

Explanation:

a)

We will use second equation of motion to find the height:

h = v_{i}t + (\frac{1}{2})gt^2

where,

h = height = ?

vi = initial speed = 0 m/s

t = time taken = 11 s

g = 9.81 /s²

Therefore,

h = (0\ m/s)(11\ s) + (\frac{1}{2})(9.8\ m/s^2)(11\ s)^2\\\\

<u>h = 593.50 m</u>

b)

For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:

v_{f} = v_{i} + gt

where,

vf = final velocity at tenth second = v₁₀ = ?

t = 10 s

vi = 0 m/s

Therefore,

v_{10} = 0\ m/s + (9.81\ m/s^2)(10\ s)\\\\v_{10} = 98.1\ m/s

Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:

h = v_{i}t + (\frac{1}{2})gt^2

where,

h = height covered during last second = h₁₁ =  ?

vi = v₁₀ = 98.1 m/s

t = 1 s

Therefore,

h_{11} = (98.1\ m/s)(1\ s) + (\frac{1}{2})(9.8\ m/s^2)(1\ s)^2\\\\

<u>h₁₁ = 103 m</u>

c)

Now, we use first equation of motion for complete motion:

v_{f} = v_{i} + gt

where,

vf = final velocity at tenth second = ?

t = 11 s

vi = 0 m/s

Therefore,

v_{f} = 0\ m/s + (9.81\ m/s^2)(11\ s)

<u>vf = 107.91 m/s</u>

8 0
3 years ago
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