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BartSMP [9]
4 years ago
11

The work w done when lifting an object varies jointly with the object's mass m and the height h that the object is lifted. the w

ork done when a 120-kilogram object is lifted 1.8 meters is 2116.8 joules. how much work is done when lifting a 100-kilogram object 1.4 meters
Physics
1 answer:
Galina-37 [17]4 years ago
7 0
Work done can be computed using the formula:

W=Fd

Where:

W = work (J)
F = Force (N)
d = Distance (d)

Looking at the given, you know that you do not have a value for force, so you will have to solve for it. 

F = ma

Where:

F = Force
m = mass
a = acceleration

Because the object is being lifted, the acceleration will rely on gravity. Acceleration due to gravity is a constant 9.8 m/s^2. Let's list our given first:

F = ?
m = 100kg
a = 9.8m/s^2

Put that into our equation and solve:
F=ma
F=(100kg)(9.8m/s^{2})
F=980kg.m/s^{2}

Our force is then 980 N. 

Now that we have force we can solve for Work. The given for work is as follows:
F= 980N
d = 1.4m

Put that into our formula and solve:
W = Fd
W = (980N)(1.4m)
W = 1,372J

The work done is 1,372J.
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A certain CD has a playing time of 74.0 minutes. When the music starts, the CD is rotating at an angular speed of 480 revolution
Rashid [163]

Answer: 0.00636\ rad/s^2

Explanation:

Given

CD has a playing time of t=74\ min\ or\  74\times 60\ s

Initial angular speed of CD is 480\ rpm

Final angular speed of DC is 210\ rpm

Angular speed, when rpm is given

\omega =\dfrac{2\pi N}{60}

\omega_i=\dfrac{2\pi \times 480}{60}\\\\\Rightarrow \omega_i=16\pi \ rad/s

Final speed

\Rightarrow \omega_f=\dfrac{2\pi \times 210}{60}\\\\\Rightarrow \omega_f=7\pi \ rad/s

Using equation of angular motion

\Rightarrow \omega_f=\omega_i+\alpha t

Insert the values

\Rightarrow 7\pi =16\pi +\alpha \times 74\times 60\\\Rightarrow -9\pi =\alpha \cdot (4440)\\\\\Rightarrow \alpha=-\dfrac{9\pi}{4440}\\\\\Rightarrow \alpha=-0.00636\ rad/s^2

Magnitude of angular acceleration is 0.00636\ rad/s^2

3 0
3 years ago
Scientists are investigating how well different microphones capture and record sounds. They use tools that show how loud the hig
inna [77]

Answer:

Which microphone captures a singer's low sounds the loudest?

Explanation:

Scientific questions are the basis of any scientific experiment. An observation is followed by a scientific question. In this question, scientists are investigating how well different microphones capture and record sounds by using tools that show how loud the high and low sounds are and a tool that shows how much noise is recorded along with the actual sounds.

Hence, a scientific question should be that which will cover the whole aspect of the investigation i.e. the question should contain a general scope of the problem. Hence, a scientific question that could be included in this investigation is:

Which microphone captures a singer's low sounds the loudest?

This is because this question contains every part being investigated i.e. How different microphones capture sounds and records them (whether low or high).

7 0
3 years ago
What type of energy is formed when chemical bonds are broken?
torisob [31]
A. Chemical energy is formed when chemical bonds are broken. 
8 0
3 years ago
Charge q1 = +2.00 μC is at -0.500 m along the x axis. Charge q2 = -2.00 μC is at 0.500 m along the x axis. Charge q3 = 2.00 μC i
Kobotan [32]

The magnitude of <em>electrical</em> force on charge q_{3} due to the others is 0.102 newtons.

<h3>How to calculate the electrical force experimented on a particle</h3>

The vector <em>position</em> of each particle respect to origin are described below:

\vec r_{1} = (-0.500, 0)\,[m]

\vec r_{2} = (+0.500, 0)\,[m]

\vec r_{3} = (0, +0.500)\,[m]

Then, distances of the former two particles particles respect to the latter one are found now:

\vec r_{13} = (+0.500, +0.500)\,[m]

r_{13} = \sqrt{\vec r_{13}\,\bullet\,\vec r_{13}} = \sqrt{(0.500\,m)^{2}+(0.500\,m)^{2}}

r_{13} =\frac{\sqrt{2}}{2}\,m

\vec r_{23} = (-0.500, +0.500)\,[m]

r_{23} = \sqrt{\vec r_{23}\,\bullet \,\vec r_{23}} = \sqrt{(-0.500\,m)^{2}+(0.500\,m)^{2}}

r_{23} =\frac{\sqrt{2}}{2}\,m

The resultant force is found by Coulomb's law and principle of superposition:

\vec R = \vec F_{13}+\vec F_{23} (1)

Please notice that particles with charges of <em>same</em> sign attract each other and particles with charges of <em>opposite</em> sign repeal each other.

\vec R = \frac{k\cdot q_{1}\cdot q_{3}}{r_{13}^{2}}\cdot \vec u_{13}  +\frac{k\cdot q_{2}\cdot q_{3}}{r_{23}^{2}}\cdot \vec u_{23} (2)

Where:

  • k - Electrostatic constant, in newton-square meters per square Coulomb.
  • q_{1}, q_{2}, q_{3} - Electric charges, in Coulombs.
  • r_{13}, r_{23} - Distances between particles, in meters.
  • \vec u_{13}, \vec u_{23} - Unit vectors, no unit.

If we know that k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}, q_{1} = 2\times 10^{-6}\,C, q_{2} = 2\times 10^{-6}\,C, q_{3} = 2\times 10^{-6}\,C, r_{13} =\frac{\sqrt{2}}{2}\,m, r_{23} =\frac{\sqrt{2}}{2}\,m, \vec u_{13} = \left(-\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}  \right) and \vec u_{23} = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right), then the vector force on charge q_{3} is:

\vec R = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right) + \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right)

\vec R = 0.072\cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right) + 0.072\cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right)\,[N]

\vec R = 0.072\cdot \left(0, -\sqrt{2}\right)\,[N]

And the magnitude of the <em>electrical</em> force on charge q_{3} (R), in newtons, due to the others is found by Pythagorean theorem:

R = 0.102\,N

The magnitude of <em>electrical</em> force on charge q_{3} due to the others is 0.102 newtons. \blacksquare

To learn more on Coulomb's law, we kindly invite to check this verified question: brainly.com/question/506926

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2 years ago
The function of the eardrum in the middle ear is to
lozanna [386]
B. vibrate with the frequency of the received sound<span />
7 0
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