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IrinaVladis [17]

4 years ago

14

A truck pulls a block 8 meters across a level surface at a force of 216 N over the course of 12 seconds. How much power did the

truck use

1 answer:

iren [92.7K]4 years ago

4 0

Answer:

work = 1728

Power = 134

Explaination:

by using the formula,

Work(W)= Force(F)×Distance(D)

<h2> and</h2>

Power(P)= Work(W)/Time taken(T)

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How many photons will be required to raise the temperature of 1.8 g of water by 2.5 k ?'?

tatyana61 [14]

Missing part in the text of the problem:
"<span>Water is exposed to infrared radiation of wavelength 3.0×10^−6 m"</span>

First we can calculate the amount of energy needed to raise the temperature of the water, which is given by
$Q=m C_s \Delta T$
where
m=1.8 g is the mass of the water
$C_s = 4.18 J/(g K)$ is the specific heat capacity of the water
$\Delta T=2.5 K$ is the increase in temperature.

Substituting the data, we find
$Q=(1.8 g)(4.18 J/(gK))(2.5 K)=18.8 J=E$

We know that each photon carries an energy of
$E_1 = hf$
where h is the Planck constant and f the frequency of the photon. Using the wavelength, we can find the photon frequency:
$\lambda = \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{3 \cdot 10^{-6} m}=1 \cdot 10^{14}Hz$

So, the energy of a single photon of this frequency is
$E_1 = hf =(6.6 \cdot 10^{-34} J)(1 \cdot 10^{14} Hz)=6.6 \cdot 10^{-20} J$

and the number of photons needed is the total energy needed divided by the energy of a single photon:
$N= \frac{E}{E_1}= \frac{18.8 J}{6.6 \cdot 10^{-20} J} =2.84 \cdot 10^{20} photons$

4 0

3 years ago

PLEASE HELP!!!!I IM GIVING BRAINLIEST!! If you answer this correctly ill answer some of your questions you have posted! (30pts)

maksim [4K]

<em>Kinetic Energy</em>

=><em><u>It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.</u></em>

<em>Potential</em><em> </em><em>Energy</em><em> </em>

<u><em>=</em><em>></em><em>potential energy is the energy held by </em></u><em><u>an</u></em>

<em><u> object because of its position relative to </u></em><em><u>other</u></em>

<em><u> objects, stresses within itself, its </u></em><em><u>electric</u></em>

<em><u> charge, or other factors.</u></em>

<h2>Difference:</h2>

=>Potential energy is a <u>stored</u> energy on the other hand kinetic energy is the energy of an object or a system's particle in <em><u>Motion</u></em>.

6 0

3 years ago

Which of the following is a result of reduced levels of red blood cells and elevated levels of white blood cells in the body?

Tju [1.3M]

The cells that circulate in the bloodstream are generally divided into three types, white blood cells.

8 0

3 years ago

What occurs as a ray of light passes from<br> al inilo water?

iVinArrow [24]

Answer:

this may be wrong but I am not sure

3 0

3 years ago

Two small, identical conducting spheres repel each other with a force of 0.030 N when they are 0.65 m apart. After a conducting

solong [7]

Note that the methods applied in solving this question is the appropriate method. Check the parameters you gave in the question if you did not expect a complex number for the charges. Thanks

Answer:

$q_1 = 0.00000119 + j0.00000145 C \\q_2 = 0.00000119 - j0.00000145 C$

Explanation:

Note: When a conducting wire was connected between the spheres, the same charge will flow through the two spheres.

The two charges were 0.65 m apart. i.e. d = 0.65 m

Force, F = 0.030 N

The force or repulsion between the two charges can be calculated using the formula:

$F = \frac{kq^2}{d^2} \\\\0.030 = \frac{9 * 10^9 * q^2}{0.65^2}\\\\q = 1.19 * 10^{-6} C$

Due to the wire connected between the two spheres, $q_1 = q_2 = 1.19 * 10^{-6} C$

The sum of the charges on the two spheres = $q_1 + q_2 = 2.38 * 10^{-6} C$

Note: When the conducting wire is removed, the two spheres will no longer contain similar charges but will rather share the total charge unequally

Let charge in the first sphere = $q_1$

Charge in the second sphere, q₂ = $2.38 * 10^{-6} - q_1$

Force, F = 0.075 N

$F = \frac{k q_1 q_2}{r^2} \\\\0.075 = \frac{9*10^9 * q_1 * (2.38*10^{-6} -q_1 )}{0.65^2}\\\\3.52 * 10^{-12} = q_1 * (2.38*10^{-6} -q_1 )\\\\3.52 * 10^{-12} = 2.38*10^{-6} q_1 - q_1^2\\\\q_1^2 - (2.38*10^{-6}) q_1 + (3.52 * 10^{-12}) = 0\\$

$q_1 = 0.00000119 + j0.00000145 C \\q_2 = 0.00000119 - j0.00000145 C$

6 0

4 years ago