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vitfil [10]
3 years ago
11

A 3.8 kg object is lifted 12 meters. approximately how much work is performed during the lifting.

Physics
2 answers:
Dmitry [639]3 years ago
7 0

work done=446.9 J . so option (c) is correct.

Explanation:

the formula for work done is given by

W= F d

F= force= mg where m= mass and g= acceleration due to gravity

F= 3.8 (9.8)=37.24 J

so W=37.24 (12)

W=446.9 J

zimovet [89]3 years ago
6 0

Answer:

Explanation:

446.9 j

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What force is required to accelerate a body with a mass of 15 kilograms at a rate of 8 m/s²?
Dominik [7]
Force is defined as Mass multiplied by Acceleration, or F = MA.
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3 years ago
Three charges, qA is +6.0 μC, qB is –5.0 μC, and qc is +6.0 μC, are located at the corners of a square with each side length at
lesya [120]

The net electric field at point D is determined as 3.95 x 10⁷ N/C.

<h3>Electric field at D due to charge A</h3>

E = kq/r²

where;

  • r is the distance between A and D
  • q is charge A

E(AD) = (9 x 10⁹ x 6 x 10⁻⁶)/(0.05²)

E(AD) = 2.16 x 10⁷ i N/C

<h3>Electric field at D due to charge B</h3>

E = kq/r²

where;

  • r is the distance between A and B
  • q is charge B

r² = 5²  + 5²

r² = 50

r = √50

r = 7.07 cm

E(BD) = (9 x 10⁹ x 5 x 10⁻⁶)/(0.0707²)

E(BD) = 9 x 10⁶ N/C

in x - direction = 9 x 10⁶ N/C x cos(45) = 6.36 x 10⁶ i  N/C

in y - direction = 9 x 10⁶ N/C x sin(45) = 6.36 x 10⁶j  N/C

<h3>Electric field at D due to charge C</h3>

E = kq/r²

where;

  • r is the distance between C and D
  • q is charge C

E(CD) = (9 x 10⁹ x 6 x 10⁻⁶)/(0.05²)

E(CD) = 2.16 x 10⁷ j N/C

<h3>Net electric field in x direction</h3>

Ei = 2.16 x 10⁷ i N/C  + 6.36 x 10⁶ i  N/C

Ei = 2.796 x 10⁷ i N/C

<h3>Net electric field in y direction</h3>

Ej = 2.16 x 10⁷ j N/C  + 6.36 x 10⁶j  N/C

Ej = 2.796 x 10⁷ j N/C

<h3>Resultant electric field at D</h3>

E = √Ei² + Ej²

E = √[(2.796 x 10⁷)² + (2.796 x 10⁷)²]

E = 3.95 x 10⁷ N/C

Thus, the net electric field at point D is determined as 3.95 x 10⁷ N/C.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

7 0
2 years ago
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