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3 years ago

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Rick is moving a wheelbarrow full of bricks out to the curb. The bricks in the wheelbarrow weigh more than Rick is able to carry

, yet he is able to move the bricks. How is this possible?

1 answer:

USPshnik [31]3 years ago

3 0

Answer is given below

Explanation:

This is happen because here when Rick walks with full loaded wheelbarriow of bricks, he able to move it because Rick lifts the wheelbarrow handle
So, most of the weight of full loaded wheelbarrow's load goes on that's wheel and due to friction force between wheel and surface it can easy to move
He uses force to rotate the wheel, much more than the force applied to the rim of the wheel on the axis of rotation or torque

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A monkey has a bit of a heavy for on the gas pedal. As soon as the light turns green the monkey pushes the gas pedal to the floo

Andrei [34K]

Answer:

$s=6.86m/s^2$

Explanation:

Hello,

In this case, considering that the acceleration is computed as follows:

$a=\frac{v_{final}-v_{initial}}{t}$

Whereas the final velocity is 28.82 m/s, the initial one is 0 m/s and the time is 4.2 s. Thus, the acceleration turns out:

$a=\frac{28.82m/s-0m/s}{4.2s}\\ \\s=6.86m/s^2$

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3 years ago

The picture shows an object resting on a balance.

Maslowich

Answer:

4.90kgm^-2

Explanation:

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2 years ago

A car is going 8 meters per second on an access road into a highway

TiliK225 [7]

Answer:

20.96 m/s^2 (or 21)

Explanation:

Using the formula (final velocity - initial velocity)/time = acceleration, we can plug in values and manipulate the problem to give us the answer.

At first, we know a car is going 8 m/s, that is its initial velocity.

Then, we know the acceleration, which is 1.8 m/s/s

We also know the time, 7.2 second.

Plugging all of these values in shows us that we need to solve for final velocity. We can do so by manipulating the formula.

(final velocity - initial velocity) = time * acceleration

final velocity = time*acceleration + initial velocity

After plugging the found values in, we get 20.96 m/s/s, or 21 m/s

3 0

3 years ago

A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electr

slega [8]

<h2>Answer: 1.011 eV</h2>

Explanation:

The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a <u>kinetic energy. </u>

This is what Einstein proposed:

Light behaves like a stream of particles called photons with an energy $E$ :

$E=h.f$ (1)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the kinetic energy $K$ of the photoelectron:

$E=\Phi+K$ (2)

Where $\Phi$ is the <u>minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and </u><u>its value depends on the metal. </u>

In this case $\Phi=2eV$ and $K_{1}=4eV$

So, for the first light source of wavelength $\lambda_{1}$ , and applying equation (2) we have:

$E_{1}=2eV+4eV$ (3)

$E_{1}=6eV$ (4)

Now, substituting (1) in (4):

$h.f=6eV$ (5)

Where:

$h=4.136(10)^{-15}eV.s$ is the Planck constant

$f$ is the frequency

Now, the <u>frequency has an inverse relation with the wavelength </u>

$\lambda_{1}$ :

$f=\frac{c}{\lambda_{1}}$ (6)

Where $c=3(10)^{8}m/s$ is the speed of light in vacuum

Substituting (6) in (5):

$\frac{hc}{\lambda_{1}}=6eV$ (7)

Then finding $\lambda_{1}$ :

$\lambda_{1}=\frac{hc}{6eV }$ (8)

$\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}$

We obtain the wavelength of the first light suorce $\lambda_{1}$ :

$\lambda_{1}=2.06(10)^{-7}m$ (9)

Now, we are told the second light source $\lambda_{2}$ has the double the wavelength of the first:

$\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)$ (10)

Then: $\lambda_{2}=4.12(10)^{-7}m$ (11)

Knowing this value we can find $E_{2}$ :

$E_{2}=\frac{hc}{\lambda_{2}}$ (12)

$E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}$ (12)

$E_{2}=3.011eV$ (13)

Knowing the value of $E_{2}$ and $\lambda_{2}$ , and knowing we are working with the same work function, we can finally find the maximum kinetic energy $K_{2}$ for this wavelength:

$E_{2}=\Phi+K_{2}$ (14)

$K_{2}=E_{2}-\Phi$ (15)

$K_{2}=3.011eV-2eV$

$K_{2}=1.011 eV$ This is the maximum kinetic energy for the second light source

7 0

3 years ago

In which one of the following circumstances could mechanical energy not possibly be conserved, even if friction and air resistan

qwelly [4]

Answer:

A car moves up a hill at a constant velocity

Explanation:

Since the velocity is constant, the speed is also constant and so is the kinetic energy. However, total mechanical energy is sum of gravitational potential energy and kinetic energy, and the car is moving up the hill so its potential energy rises.

Thus, in the circumstances described the mechanical energy cannot be conserved.

The correct answer is A car moving up the hill with constant velocity.

5 0

3 years ago