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gulaghasi [49]
3 years ago
10

Marie lifts a 5 kg mass from the floor and puts in on a table one metre high. What is the work done?

Physics
1 answer:
sergey [27]3 years ago
8 0

Answer:

The work done was 49.05 N*m.

Explanation:

The work done by a a force is given by:

W = F*d

We first need to calculate the force necessary to lift the box from the ground, the force needed for is the weigh of the box, which is shown below:

F = m*g\\F = 5*9.81\\F = 49.05 \text{ N}

Therefore the work done by lifting the box and placing it on the table is:

W = 49.05*1 \\W = 49.05 \text{ N*m}

The work done was 49.05 N*m.

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Why is it that the two possible faults with inductors are short circuit and open circuit?
lukranit [14]
These are the most common type of faults not just inductors but also with other elements too like resistors,transformers, generators etc.
open circuit fault means the flow of current is disrupted some how in the circuit and the circuit stops operating. and for short circuit fault the current in the system will be pretty high and this short circuit current or fault current will always run back to the fault location, if the inductor got short circuited somehow then the fault current will only run through it because it will then provide a very low impedence path
5 0
3 years ago
A major contribution of Johannes Kepler to the development of modern astronomy was:________.
fgiga [73]

Answer:

<h2>The answer is  planetary motion</h2>

Explanation:

According to Johannes Kepler, the laws governing planetary motion

states that:

1. The orbit of a planet is an ellipse with the Sun at one of the two foci.

2. A line segment joining a planet and the Sun sweeps out equal areas          

   during equal intervals of time.

3. The square of a planet's orbital period is proportional to the cube of the semi-major of its orbit.

Johannes Kepler was a German astronomer, mathematician, and astrologer

Born: 27 December 1571, Weil der Stadt, Germany

Died: 15 November 1630

8 0
3 years ago
"When fire stopping material is used where more than ____________________ nonmetallic sheathed cables pass through wood framing
GenaCL600 [577]

Answer: When fire stopping material is used where more than 2 non-metallic sheathed cables pass through wood framing members, their ampacities must be adjusted, according to 310.15"

Answer is 2

8 0
3 years ago
The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a
mario62 [17]

Answer:

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a)Given I = 55A−(0.65A/s2)t²

I = dQ/dt

dQ = I×dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

Q = [55t – 0.65/3×t³]

Q between t=0 and t= 7.5s

Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.

3 0
3 years ago
Read 2 more answers
A device for training astronauts and jet fighter pilots is designed to rotate the trainee in a horizontal circle of radius 11.0
kvv77 [185]

The velocity of the trainee is 29 m/s or 0.42 rev/s

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration (m / s²)v = final velocity (m / s)</em>

<em>u = initial velocity (m / s)</em>

<em>t = time taken (s)</em>

<em>d = distance (m)</em>

Centripetal Acceleration of circular motion could be calculated using following formula:

\large {\boxed {a_s = v^2 / R} }

<em>a = centripetal acceleration ( m/s² )</em>

<em>v = velocity ( m/s )</em>

<em>R = radius of circle ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Radius of horizontal circle = R = 11.0 m

Force Felt by the Trainee = F = 7.80w

<u>Unknown:</u>

Velocity of Rotation = v = ?

<u>Solution:</u>

F = ma

F = m\frac{v^2}{R}

7.80w = m\frac{v^2}{R}

7.80mg = m\frac{v^2}{R}

7.80g = \frac{v^2}{R}

7.80 \times 9.8 = \frac{v^2}{11.0}

v^2 = 840.84

v \approx 29 ~m/s

\omega = \frac{v}{R}  → in rad/s

\omega = \frac{v}{2 \pi R}  → in rev/s

\omega = \frac{29}{2 \pi \times 11.0}

\omega \approx 0.42 ~ rev/s

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Uniform Circular Motion : brainly.com/question/2562955
  • Trajectory Motion : brainly.com/question/8656387

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Circular , Ball , Centripetal

6 0
3 years ago
Read 2 more answers
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