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lesya692 [45]
2 years ago
7

The measured value of mass M in an experiment is M = 0.743 ± 0.005kg. The error in 2M is

Physics
1 answer:
omeli [17]2 years ago
8 0

Given the measured value of mass M in the experiment the error in 2M is  ; dM

<u>Given data :</u>

Measured value of mass M = ( 0.743 ± 0.005 ) kg

<h3>The error in 2M in the measured value </h3>

The error in 2M in the measured value M = 0.743 ± 0.005 kg is dM while the error in M² will be 2dM

Hence we can conclude that the measured value of mass M in the experiment the error in 2M is  ; dM.

Learn more about error in 2M : brainly.com/question/6650225

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What are the unit for acceleration
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<h3>Answer</h3>

m/s^2 (meter per sec square)

Explanation:

acc = change in velocity/time

= distance/time

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time

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7 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Sonbull [250]

Question 1  

In order to do work, the force vector must be

in the same direction as the displacement vector and the motion.

Question 2  

In which of the following cases is work being done on an object?

Pulling a trailer up a hill

Question 3  

Which situation is an example of NOT doing work?

carrying a box

Question 4  

Work is measured in

Joules

Question 5  

To find the work done, the force exerted and distance moved are multiplied. A couch is moved twice before you are happy with its placement. The same force was used to move the couch both times. If more work is done the first time it is moved, what do you know about the distance it was moved?

When more work was done, the couch was moved further.

Question 6  

A weight lifter raises a 1600 N barbell to a height of 2.0 meters. How much work was done? W = Fd

3200 Joules

Question 7  

You and a friend (Alex) are at a a tree-top adventure park .. . . part of the course requires you to climb up a rope. You both climb the same rope in the same amount of time. However, the tension in the rope is greater

when Alex climbs. Who did the most work?

Alex did - more tension means more force - more force means more work was done  

Question 8  

Doing work at a faster rate creates power.

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Question 9  

In one challenge on the Titan Games, competitors have to lift 200 pounds up a long ramp. Angel is able to move the weight in 42 seconds. Anthony gets it done in only 38 seconds. Which statement is true?

Anthony has more power than Angel.  

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Your family is moving to a new apartment. While lifting a box 83 Joules of work is done to put the box on a truck, you exert an upward force of 75 N for 3 s. How much power is required to do this? (Hint: You only need two of the 3 numbers given!) Power = Work / time

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<em>*100% CORRECT ANSWERS</em>

4 0
3 years ago
Pls help asap. will mark as brainliest if correct
Deffense [45]

Answer:

Hey mate, here is your answer answer. Hope it helps you

Explanation:

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2. C- Electrical energy.The term electrical activity means that the food itself has the power to generate electric energy that persists some period of time. This work presents a purely renewable energy as energy comes. the zinc reacts with food tissue.

8 0
3 years ago
Read 2 more answers
An electromagnet is a device in which moving electric charges (current) in a coil of wire create a magnet. What’s one advantage
docker41 [41]
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Hope this helps!
5 0
2 years ago
1)Determine, in terms of unit vectors, the resultant of the five forces illustrated in the figure, Consider F1=20 N, F2= 12 N, F
LiRa [457]

Explanation:

1) F₁ lies in a plane perpendicular to the xy plane, 60° from the x axis.  The angle between F₁ and the +z axis is 30°.  Therefore, the vector is:

<F₁> = 20 (sin 30° cos 60° i + sin 30° sin 60° j + cos 30° k)

<F₁> = 20 (¼ i + ¼√3 j + ½√3 k)

<F₁> = 5 i + 5√3 j + 10√3 k

F₂ is in the xy plane.  Its slope is -24/7.  The vector is:

<F₂> = 12 (-⁷/₂₅ i + ²⁴/₂₅ j + 0 k)

<F₂> = -3.36 i + 11.52 j

F₃ is parallel to the +x axis.  The vector is:

<F₃> = 17 (i + 0 j + 0 k)

<F₃> = 17 i

F₄ is parallel to the -z axis.  The vector is:

<F₄> = 15 (0 i + 0 j − k)

<F₄> = -15 k

F₅ is in the xy plane.  It forms a 15° angle with the -y axis.  The vector is:

<F₅> = 9 (-sin 15° i − cos 15° j + 0 k)

<F₅> = -9 sin 15° i − 9 cos 15° j

The resultant vector is therefore:

<F> = (5 − 3.36 + 17 − 9 sin 15°) i + (5√3 + 11.52 − 9 cos 15°) j + (10√3 − 15) k

<F> = 16.31 i + 11.49 j + 2.32 k

2) Sum of forces at point B in the x direction:

∑F = ma

Tbc cos 40° − ¹⁵/₁₇ Tab = 0

Tbc cos 40° = ¹⁵/₁₇ Tab

Tbc = 1.15 Tab

Sum of forces at point B in the y direction:

∑F = ma

Tbc sin 40° + ⁸/₁₇ Tab − mAg = 0

Tbc sin 40° + ⁸/₁₇ Tab = (2 kg) (10 m/s²)

(1.15 Tab) sin 40° + ⁸/₁₇ Tab = 20 N

1.21 Tab = 20 N

Tab = 16.52 N

Tbc = 19.02 N

Sum of forces at point C in the x direction:

∑F = ma

Tcd sin 25° − Tbc cos 40° = 0

Tcd sin 25° = Tbc cos 40°

Tcd = 1.81 Tbc

Tcd = 34.48 N

3(a) When the crane is on the verge of tipping, the center of gravity is directly over point F.  Relative to point A:

3.7 m = [ (390 kg) (0.9 m) + (90 kg) (9 m cos θ + 1.7 m) + (80 kg) (9 m cos θ + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + 810 kgm cos θ + 153 kgm + 720 kgm cos θ + 232 kgm

1336 kgm = 1530 kgm cos θ

θ = 29.17°

3(b) 3.7 m = [ (390 kg) (0.9 m) + (90 kg) (x + 1.7 m) + (80 kg) (x + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + (90 kg) x + 153 kgm + (80 kg) x + 232 kgm

1336 kgm = (170 kg) x

x = 7.86 m

4) Find the lengths of the cables.

Lab = √((2 m)² + (3 m)² + (5 m)²)

Lab = √38 m

Lac = √((2 m)² + (3 m)² + (5 m)²)

Lac = √38 m

Lde = √((2 m)² + (3 m)²)

Lde = √13 m

Sum of forces in the x direction:

∑F = ma

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Sum of forces in the y direction:

∑F = ma

2/√38 Fab − 2/√38 Fac = 0

Fab = Fac

Sum of forces in the z direction:

∑F = ma

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

Sum of moments about the y-axis:

∑τ = Iα

(3/√38 Fab) (5 m) + (3/√38 Fac) (5 m) + (3/√13 Fde) (2 m) − (mg) (2 m) = 0

Substitute Fab = Fac and simplify:

6/√38 Fab + 3/√13 Fde − mg = 0

30/√38 Fab + 6/√13 Fde − 2mg = 0

Double first equation:

12/√38 Fab + 6/√13 Fde − 2mg = 0

Subtract from the second equation:

28/√38 Fab = 0

Fab = 0

Fac = 0

Solve for Fde:

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

3/√13 Fde = mg

3/√13 Fde = (1.7 kg) (10 m/s²)

Fde = 20.43 N

Solve for Rx:

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Rx = 2/√13 Fde

Rx = 11.33 N

8 0
3 years ago
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