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gizmo_the_mogwai

3 days ago

11

A child wants to pump up a bicycle tire so that its pressure is 1.2 × 105 pa above that of atmospheric pressure. if the child us

es a pump with a circular piston 0.035 m in diameter, what force must the child exert?

2 answers:

kicyunya [1K]3 days ago

7 0

The general formula is;
Pressure = Force/Area
Where,
Pressure = Required pressure + Atmospheric pressure = (1.2*10^5) + (101325) = 221325 Pa = 221325 N/m^2

Area = πD²/4 = π*0.035²/4 = 9.621*10^-4 m²

Thus,
Force, F = Pressure*Area = 221325*9.621*10^-4 = 212.94 N

Yuliya22 [1.1K]3 days ago

6 0

Answer:

The force required from the child amounts to 460.8 N.

Explanation:

We are given that pressure $P=1.2\times 10^5pa$

Radius of the circular piston r = 0.035 m

Thus, area $A=\pi r^2=3.14\times 0.035^2=0.003846m^2$

Next, we need to calculate the force achieved by the child.

We know force can be expressed as

$Force=pressure\times area$

Therefore, force will be represented as $Force=1.2\times 10^5\times 0.00384=460.8N$

So the force that the child must exert is 460.8 N.

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The magnitude of the electrical force acting between a +2.4 × 10–8 C charge and a +1.8 × 10–6 C charge that are separated by 0.0

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To solve this problem, Coulomb's law will be applied as follows:
F = k*q1*q2 / r^2 where:
F indicates the force magnitude between the charges
k is a constant = 9.00 * 10^9 N.m^2/C^2
q1 = <span>+2.4 × 10–8 C
q2 = </span><span>+1.8 × 10–6 C
r represents the distance separating the charges = </span><span>0.008 m

By substituting these values, we derive:
F = (9*10^9)(2.4*10^-8)(1.8*10^-6) / (0.008)^2 = 6.075, which rounds to 6.1 Newtons

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2 days ago

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A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

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a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The car's position over time t can be described by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

To find the average velocity, we divide the displacement by the elapsed time:

$v=\frac{\Delta x}{\Delta t}$

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 2.00 s, the position is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

This leads us to the displacement of

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The duration for this interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

Therefore, the average velocity during this period is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 4.00 s, the position is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

Thus, the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

This yields an average velocity of

$v=\frac{20.8}{4.0}=5.2 m/s$

c)

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

And at t = 4 s it is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

This gives us a displacement of

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the resulting average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

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In a given city, the permissible limit of CO (carbon monoxide) in the air is 100 parts per million (ppm). The city monitors the

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Density is defined as the mass divided by the volume.

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10 days ago

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tas watches as his uncle changes a flat tire on a car. his uncle raises the car using a machine called a jack. each time his unc

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The answer is

-Small f and large D.

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Machines are essential for people to amplify their strength; without them, lifting a car would be impossible.

Employing leverage or hydraulic principles, machines enhance your exerted force.

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