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Jobisdone [24]
3 years ago
8

A 15.5 kg box is pushed across the lunch table. The acceleration of the box is 24.2 m/S. What is the net force applied to the bo

x? ​
Physics
1 answer:
MissTica3 years ago
3 0

Answer:

<h2>375.1 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 15.5 × 24.2

We have the final answer as

<h3>375.1 N</h3>

Hope this helps you

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A 80kg astronaut is training in human centrifuge to prepare for a launch. The astronaut uses the centrifuge to practice having a
inessss [21]

The answers on the model of the human centrifuge ready for the launch to each question of the statement are listed below:

a) A force of 2479.210 newtons is acting on the astronaut's back.

b) A <em>net centripetal</em> force of 2479.210 newtons is acting on the astronaut.

c) The <em>centripetal</em> acceleration of the astronaut is 30.990 meters per square second.

d) The astronaut has a <em>linear</em> speed of approximately 19.284 meters per second.

e) The <em>angular</em> speed of the astronaut is 1.607 radians per second (15.346 revolutions per minute).

<h3>How to apply Newton's laws to analyze a process in a human centrifuge training</h3>

The human centrifuge experiments a <em>centripetal</em> acceleration when it reaches a <em>peak</em> angular speed. In this question we must apply Newton's laws of motion and concepts of <em>centripete</em> and <em>centrifugal</em> forces to answer the questions. Now we proceed to answer the questions:

<h3>How much force is acting on the astronaut's back?</h3>

By the third Newton's law the astronaut experiments a <em>rection</em> force (<em>F</em>), in newtons, which has the same magnitude to <em>centrifugal</em> force but opposed to that force. The magnitude of the force acting on the back of the astronaut is equal to:

F = 3.16\cdot (80\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

F = 2479.210\,N

A force of 2479.210 newtons is acting on the astronaut's back. \blacksquare

<h3>What is the net centripetal force on the astronaut?</h3>

By the second and third Newton's laws we know that the <em>net centripetal</em> force on the astronaut is equal to the magnitude of the force found in the previous question. Thus, a <em>net centripetal</em> force of 2479.210 newtons is acting on the astronaut. \blacksquare

<h3>What is the astronaut's centripetal acceleration?</h3>

The centripetal acceleration of the astronaut (<em>a</em>), in meters per square second, is found by dividing the result of the previous question by the mass of the astronaut (<em>m</em>), in kilograms:

a = \frac{F}{m}   (1)

If we know that F = 2479.210 newtons and m = 80 kilograms, then the centripetal acceleration of the astronaut is:

a = \frac{2479.210\,N}{80\,kg}

a = 30.990\,\frac{m}{s^{2}}

The <em>centripetal</em> acceleration of the astronaut is 30.990 meters per square second. \blacksquare

<h3>What is the astronaut's linear speed?</h3>

By definition of <em>uniform circular</em> motion, we have the following formula for the <em>linear</em> velocity of the astronaut (<em>v</em>):

v = \sqrt{a\cdot r}   (1)

Where <em>r</em> is the radius of the human centrifuge, in meters.

If we know that a = 30.990\,\frac{m}{s^{2}} and r = 12\,m, then linear velocity of the astronaut is:

v = \sqrt{\left(30.990\,\frac{m}{s^{2}} \right)\cdot (12\,m)}

<em>v ≈ 19.284 m/s</em>

The astronaut has a <em>linear</em> speed of approximately 19.284 meters per second. \blacksquare

<h3>What is the astronaut's angular speed? </h3>

The <em>angular</em> speed of the astronaut (ω), in radians per second, is found by the following <em>kinematic</em> relationship:

\omega = \frac{v}{R}   (1)

If we know that <em>v ≈ 19.284 m/s</em> and <em>R = 12 m</em>, then the angular speed is:

\omega = \frac{19.284\,\frac{m}{s} }{12\,m}

<em>ω = 1.607 rad/s (15.346 rev/m)</em>

The <em>angular</em> speed of the astronaut is 1.607 radians per second (15.346 revolutions per minute). \blacksquare

To learn more on centripetal forces, we kindly invite to check this verified question: brainly.com/question/11324711

6 0
2 years ago
What type of light can go through aluminum foil
EleoNora [17]

Answer:

DmxmxmdmdExplanation:sejwjsjskdkdkdekskekememd

8 0
3 years ago
Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the fre
vekshin1

This question is incomplete; here is the complete question:

Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the frequency as 3 hertz, which statement about the wave is accurate?

A. The wave has traveled 32.4 cm in 3 seconds.

B. The wave has traveled 32.4 cm in 9 seconds.

C. The wave has traveled 97.2 cm in 3 seconds.

D. The wave has traveled 97.2 cm in 1 second.

The answer to this question is D. The wave has traveled 97.2 cm in 1 second.

Explanation:

The frequency of a wave, which is in this case 3 hertz, represents the number of waves that go through a point during 1 second. According to this, if the frequency of the wave is 3 hertz this means in 1 second there were 3 waves. Moreover, if you multiply the wavelength (32.4cm) by the frequency (3) you will know the distance the wave traveled in 1 second: 32.4 x 3 =  97.2 cm. This makes option D the correct one as the distance in 1 second was 97.2 cm.

8 0
3 years ago
Consider a positive charge Q and a point B twice as far away from Q as point A. What is the ratio of the electric field strength
Vikentia [17]

Answer:

\frac{E_{A}}{E_{B}}=4

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

E=\frac{F}{q}.

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

F=\frac{kQq}{r^{2}}.

By substitution we get that

E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}

Now, letting E_{A} be the electric field at point A, letting E_{B} be the electric field at point B, and letting R be the distance from the charge to A:

E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}.

The ration of the electric fields is

\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4

This means that at half the distance, the electric field is four times stronger.

4 0
3 years ago
A lava lamp is a decorative item found in some homes. The lamp portion contains a waxy substance in a liquid medium. When the la
dangina [55]

Answer:

Conduction, convection, and radiation

Explanation:

Conduction is the process in which energy is transferred from a hot body to a cooler body. Convection is a process by which heat is transferred as a result of the movement of heated fluid in the form of air or water. Radiation refers to the transmission of energy through a material medium.

Based on the information provided, a lava lamp could be used to provide an example of conduction, convection, and radiation.

7 0
3 years ago
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