answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Gekata [30.6K]
1 year ago
14

(Double points) A machine receives electricity that enables it to deliver a total of 8,542 N of force for the completion of its

task. The machine's useful output is 7,023 N. What is the machine's efficiency?
Physics
1 answer:
storchak [24]1 year ago
4 0

Answer: machine's efficiency = 82.2%

Explanation:

Efficiency of a machine is the capability of a machine to convert input to output without waste.

It can be expressed as

Efficiency = output/ input × 100%

Output = 7,023N

Imput = 8,542N

Efficiency = 7,023N/8,542N × 100%

Efficiency = 82.2%

You might be interested in
Which of the following forces exists between objects even in the absence of direct physical contact
den301095 [7]

Answer: TRUST ME I GOT IT WRONG the answer is B

Explanation:

3 0
2 years ago
Read 2 more answers
"In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression
zavuch27 [327]

Answer:

The value of R is 1.72\times10^{11}\ m.

(B) is correct option.

Explanation:

Given that,

In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression

R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}

We need to calculate this for value of R

R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}

R=1.89\times10^{11}\ m

So, The nearest option of the value of R is 1.72\times10^{11}\ m

Hence, The value of R is 1.72\times10^{11}\ m.

6 0
2 years ago
Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i
Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

V = 4\frac{kQ}{r}

here

r = distance of corner of the square from it center

r = \frac{a}{\sqrt2}

r = \frac{10nm}{\sqrt2} = 7.07 nm

Q = e = -1.6 * 10^{-19} C

now the net potential is given as

V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}

V = 0.815 V

now potential energy of alpha particle at this position

U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J

Now at the mid point of one of the side

Electrostatic potential is given as

V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}

here we know that

r_1 = \frac{a}{2} = 5 nm

r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}

r_2 = 11.2 nm

now potential is given as

V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}

V = -0.576 - 0.257 = -0.833 V

now final potential energy is given as

U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J

Now work done in this process is given as

W = U_f - U_i

W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}

W = -7 * 10^{-22} J

8 0
1 year ago
Suppose that a sound source is emitting waves uniformly in all directions. If you move to a point twice as far away from the sou
Helen [10]

Answer:

<em>d. unchanged.</em>

Explanation:

The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.

In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from

v = fλ

that the frequency is tied to the wave, and does not change throughout the waveform.

where v is the speed of the sound wave

f is the frequency

λ is the wavelength of the sound wave.

4 0
1 year ago
A spaceship of mass 8600 kg is returning to Earth with its engine turned off. Consider only the gravitational field of Earth. Le
Katyanochek1 [597]

Answer:

\Delta KE = 4.20\times 10^{13}\ J

Explanation:

given,

mass of spaceship(m) = 8600 Kg

Mass of earth = 5.972 x 10²⁴ Kg

position of movement of space ship

R₁ = 7300 Km

R₂ = 6700 Km

the kinetic energy of the spaceship increases by = ?

Increase in Kinetic energy = decrease in potential energy

    \Delta KE = GMm (\dfrac{1}{R_2}-\dfrac{1}{R_1})

    \Delta KE = GMm (\dfrac{R_1-R_2}{R_2R_1})

    \Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{7300 - 6700}{7300 \times 6700})

    \Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{600}{48910000})

    \Delta KE = 4.20\times 10^{13}\ J

5 0
1 year ago
Other questions:
  • A spring driven dart gun propels a 10g dart. It is cocked by exerting a force of 20N over a distance of 5cm. With what speed wil
    6·2 answers
  • A star is located at a distance of about 100 million light years from Earth. An astronomer plans to measure the distance of the
    11·1 answer
  • A biker travels at an average speed of 18 km/hr along a 0.30 km straight segment of a bike path. How much time (in hours) does t
    12·1 answer
  • An object is 6.0 cm in front of a converging lens with a focal length of 10 cm.Use ray tracing to determine the location of the
    9·1 answer
  • In Physics lab, Ellen has been given the task of constructing a simple motor. She must find common household items at home and b
    11·2 answers
  • A magnetic field is directed perpendicular to the plane of a 0.15-m × 0.30-m rectangular coil consisting of 240 loops of wire. T
    7·1 answer
  • Which of the following statements is/are true? Check all that apply. Check all that apply. The work done by a nonconservative fo
    13·1 answer
  • On an amusement park ride, passengers are seated in a horizontal circle of radius 7.5 m. The seats begin from rest and are unifo
    13·1 answer
  • The velocity of a car increases from 2.0 m/s to 16.0 m/s in a time period of 3.5 s. What was the average acceleration?
    13·2 answers
  • You skip north for 12 minutes to your best friend's house that is 1.5 kilometers away. What is your average velocity?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!