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2 years ago

Physics laws of motion

Physics

1 answer:

expeople1 [14]2 years ago

8 0

The acceleration is $0.095 m/s^2$

Explanation:

We can solve the problem by applying Newton's second law of motion: in fact, the net force acting on an object is equal to the product between the mass of the object and its acceleration. Therefore we can write:

$\sum F = ma$

where:

$\sum F$ is the resultant force acting on the object

m is its mass

a is its acceleration

In this problem, we have the following forces acting on the system:

$F_1 = 2500 N$ (forward)

$F_2 = -2400 N$ (backward)

So, Newton's second law can be rewritten as:

$F_1 -F_2 = ma$

where:

m = 1050 kg is the mass of all the students

Solving the formula for a, we find the acceleration of the system:

$a=\frac{F_1-F_2}{m}=\frac{2500+(-2400)}{1050}=0.095 m/s^2$

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Economy?

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2 years ago

Top-down processing is often used when one encounters an unfamiliar stimulus. Please select the best answer from the choices pro

Naddik [55]

The answer for this question would be B) False or the second option because top-down processing is NOT often used when one encounters an unfamiliar stimulus.

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3 years ago

Read 2 more answers

Why are most of the world’s deserts located between latitudes 10°n to 30°n and 10°s to 30°s.

RoseWind [281]

The bulk of the world's deserts are located at 30 degrees north latitude and 30 degrees south latitude, when the warm equatorial air begins to descend. The heavy, warm, descending air vaporises large amounts of water from the ground's surface. As a result, the environment is rather dry.

<h3>Why are the majority of the desert regions on Earth located between 20 and 30 degrees latitude?</h3>

The zones of falling air are those between 20 and 30 latitudes on the western borders of continents (high pressure and dry weather). As a result, the moisture continues to decrease as the air is compressed and warmed as it falls.

Where the scorching equatorial air starts to descend, the majority of the world's deserts are found between 30 degrees north and 30 degrees south latitude. Large volumes of water are vaporised off the surface of the ground by the thick, warming, falling air. As a result, the climate is extremely dry.

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1 year ago

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$

Now we can find the radiation intensity with the equation

$I=\frac{P}{A}$

Where the area is

$A=4\pi r'^2 = 4\pi(2.4\cdot 10^{11})^2=7.24\cdot 10^{23} m^2$

And substituting,

$I=\frac{3.910\cdot 10^{26}}{7.24\cdot 10^{23}}=540 W/m^2$

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2 years ago

How do you find the force of gravity?

devlian [24]

1<span>Define the equation for the force of gravity that attracts an object, <span>Fgrav = (Gm1m2)/d2</span>
2. </span>Use the proper metric units.
3. Determine the mass of the object in question.
4. <span>Measure the distance between the two objects
5. </span><span>Solve the equation
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3 years ago