The figure shows a rear view of a loaded two-wheeled wheelbarrow on a horizontal surface. It has balloon tires and a weight W = 684 N, which is uniformly distributed. The left tire has a contact area with the ground of AL = 6.20 × 10-4 m2, whereas the right tire is underinflated and has a contact area of AR = 9.20 × 10-4 m2. Find (a) the force from the left tire, (b) the pressure from the left tire, (c) the force from the right tire, (d) the pressure from the right tire that each tire applies to the ground.

The Figure Shows A Rear View Of A Loaded Two-wheeled Wheelbarrow On A Horizontal Surface. It Has Balloon

Answers

Answer 1

Answer:

Explanation:

Summing moments about the CG to zero will show that the two normal forces are equal and have a value of FL = FR = 684/2 = 342 N

Pressure on the left

PL = 342 / 6.20e-4 = 551,612.9032... = 5.5e5 Pa

Pressure on the right

PR = 342 / 9.20e-4 = 371,739.1304... = 3.7e5 Pa

Answer 2

A) The force from the left tire is; FL = 342 N

B) The pressure from the left tire is; PL = 551613 N/m²

C) The force from the right tire is; FR = 342 N

D) The pressure from the right tire is; PR = 371739 N/m²

We see that;

FL and FR are upward forces

W is the downward force.

We know that in equilibrium;

Sum of upward forces = sum of downward forces

Thus;

FL + FR = W

We are given W = 684 N

Since W is at the center, it means that FL = FR. Thus;

FL = FR = 684/2

FL = FR = 342 N

We are given;

Contact area of left tire; AL = 6.2 × 10⁻⁴ m²

Contact area of right tire; AR = 9.2 × 10⁻⁴ m²

Formula for pressure is;

Pressure = Force/Area

Pressure on the left tire;

PL = FL/AL

PL = 342/(6.2 × 10⁻⁴)

PL = 551613 N/m²

Pressure on the t right tire;

PR = FR/AR

PR = 342/(9.2 × 10⁻⁴)

PR = 371739 N/m²

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- BRAINLIEST answerer ❤️

Answer:

9.8

Explanation:

we know ,

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A 400 g ball swings in a vertical circle at the end of
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Answer:

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Answer:

The First 8 Presidents

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How does a balanced chemical equation demonstrate the Law of Conservation of Mass? it shows that only physical changes follow the Law of Conservation of Mass it shows that only physical changes follow the Law of Conservation of Mass it shows that the properties of the elements stay the same after the reaction it shows that the properties of the elements stay the same after the reaction it shows that all compounds remain bonded after the reaction it shows that all compounds remain bonded after the reaction it shows that no atoms have been gained or lost during the reaction

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Answer:

it shows that the properties of the elements stay the same after the reaction

it shows that the properties of the elements stay the same after the reaction

it shows that all compounds remain bonded after the reaction

it shows that all compounds remain bonded after the reaction

it shows that only physical changes follow the Law of Conservation of Mass

it shows that only physical changes follow the Law of Conservation of Mass

it shows that no atoms have been gained or lost during the reaction

it shows that no atoms have been gained or lost during the reaction

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Answer:

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Explanation:

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Answer:

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Answer:

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A wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope that is wound on the wheel and attached to it (see figure). The wheel is released from rest and the block descends 1.5 m in 2.00 s without any slipping of the rope. The tension in the rope during the descent of the block is 20 N. What is the moment of inertia of the wheel?

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Hi there!

We can begin by calculating the acceleration of the block and the wheel using the following equation:

d = vit + 1/2at², where initial velocity = 0 m/s

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Now, we can do a summation of torques:

∑τ =  rT

Rewrite using Newton's 2nd Law for rotation:

Iα = rT

Convert α to a using the relationship α = a/r:

I(a/r) = rT

Ia = r²T

I = r²T/a

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[tex]r = 1.29×10^8\:\text{m}[/tex]

Explanation:

According to Newton's law of universal gravitation, the gravitational force between Uranus and Miranda is

[tex]F_G = G\dfrac{M_UM_M}{r^2}[/tex]

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[tex]r^2 = G\dfrac{M_UM_M}{F_G}[/tex]

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[tex]r = \sqrt{G\dfrac{M_UM_M}{F_G}}[/tex]

Plugging in the values, we get

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[tex]\:\:\:\:\:=1.29×10^8\:\text{m}[/tex]

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Answer:

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We calculate.

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In conclusion using the expression for the speed of waves in materials we can find the result for the speed of sound in aluminum is:

     v = 5050 m / s

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Answer:

Quiet, is the adjective in this sentence

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She had a quiet voice. This statement uses the adjective quiet. Her voice can be characterized as quiet.

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A word used to characterize a person, place, or thing is called an adjective. Adjectives are used to modify nouns or pronouns.

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Answers

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The answer would be B.Crazy

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Hi there!

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In paragraph 1 of this passage, which clues help you figure out the meaning of the word vital?
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Answer:

Question 1 is "need them to survive"

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Answer:

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Answer:

heating prosedure takes place the opposite of condensation

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Answer:

15[m].

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Answers

Answer:

The acceleration of the car is 8.10 m/s²

Explanation:

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a = ?

d = Vᵢ × t + 0.5 × a × t²

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Thus, The acceleration of the car is 8.10 m/s²

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Answer:

True

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Answers

Answer:

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Kinetic friction is defined as a force that acts between moving surfaces. A body moving on the surface experiences a force in the opposite direction of its movement.

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When the mass is not moving, the object experiences static friction. The friction increases as the applied force increases until the block moves. After the block moves, it experiences kinetic friction, which is less than the maximum static friction.

What does it generate?

When surfaces in contact move relative to each other, the friction between the two surfaces converts kinetic energy into thermal energy (that is, it converts work to heat). This property can have dramatic consequences, as illustrated by the use of friction created by rubbing pieces of wood together to start a fire.

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Answers

Answer:

Explanation:

An example of an intense aerobic activity would be running/ sprinting sprinting targets six specific muscle groups: hamstrings, quadriceps, glutes, hips, abdominals and calves. Sprinting is a total body workout featuring short, high-intensity repetitions and long, easy recoveries.

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