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Strike441 [17]
4 years ago
13

A box is pulled with a horizontal force of 500N and moves 5m. What is the work done?

Physics
2 answers:
shepuryov [24]4 years ago
8 0
The formula for work is N x m, which will equal in joules.
500 x 5 = 2,500 joules
Murljashka [212]4 years ago
8 0

Answer: Just multiply both numbers to get your answer

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If you could observe atoms and molecules with the naked eye, what would you look for as conclusive evidence of a chemical reacti
Blizzard [7]
<span>If you could observe atoms and molecules with the naked eye, an evidence of a chemical reaction would be a formation of a new substance where atoms and molecules bond with each other forming new compounds. Hope this helps. Have a nice day.</span>
3 0
3 years ago
The lamp cord is 85cm long and comprises cupper wire. Calculate the wire‘s resistance?
DaniilM [7]

Answer:

R = 0.0015Ω

Explanation:

The formula for calculating the resistivity of a material is expressed as;

ρ = RA/l

R is the resistance

ρ is the resistivity

A is the area of the wire

l is the length of the wire

Given

l = 85cm = 0.85m

A = πr²

A = 3.14*0.0018²

A = 0.0000101736m²

ρ = 1.75 × 10-8Ωm.

Substitute into the formula

1.75 × 10-8 = 0.0000101736R/0.85

1.4875× 10-8 = 0.0000101736R

R = 1.4875× 10-8/0.0000101736

R = 0.0015Ω

3 0
3 years ago
a proton travelling along the x-axis is slowed by a uniform electric field E. at x = 20 cm, the proton has a speed of 3.5x10^6 m
AveGali [126]

Answer:

The magnitude of the electric field is 3.8 × 10⁵⁸ N/C and it is in the negative x- direction.

Explanation:

From work-kinetic energy principles, the kinetic energy change of the proton equals the work done by the electric field.

So, ΔK = W

1/2m(v₂² - v₁²) = qEd where m = mass of proton = 1.673 × 10⁻²⁷ kg, v₁ = initial speed of proton = 3.5 × 10⁶ m/s, v₂ = final speed of proton = 0 m/s, q = proton charge = + e = 1.602 × 10⁻¹⁹ C, E = electric field and d = distance moved by proton = x₂ - x₁ , x₁ = 20 cm and x₂ = 80 cm. So, d = 80 cm - 20 cm = 60 cm = 0.6 m

1/2m(v₂² - v₁²) = qEd

E = (v₂² - v₁²)/2mqd

substituting the values of the variables into E, we have

E = ((0 m/s)² - (3.5 × 10⁶ m/s)²)/(2 × 1.673 × 10⁻²⁷ kg × 1.602 × 10⁻¹⁹ C × 0.6 m)

E = - 12.25 × 10¹² m²/s² ÷ 3.22 × 10⁻⁴⁶

E = -3.8 × 10⁵⁸ N/C

So, the magnitude of the electric field is 3.8 × 10⁵⁸ N/C and it is in the negative x- direction.

6 0
3 years ago
PLEASE ANSWER THIS I AM SO STUCK The increasing trend of energy requirements in the U.S.: a) is directly proportional to populat
stiks02 [169]

Answer:

It should be A

Explanation:

the amount of energy a population has should be based on how many people there are. if more people come into the population, the more energy it should have. if people leave the population, the less energy it should have.

7 0
3 years ago
a 2000kg car initially traveling at a speed of 15 m/s is accelerated by a constant force of 10000 n for 3 seconds. the new speed
Juliette [100K]
  • Mass of the car (m) = 2000 Kg
  • Initial velocity (u) = 15 m/s
  • Force (F) = 10000 N
  • Time (t) = 3 s
  • Let the acceleration be a.
  • By using the formula, F = ma, we get,
  • 10000 N = 2000 Kg × a
  • or, a = 10000 N ÷ 2000 Kg
  • or, a = 5 m/s^2
  • Let the final velocity be v.
  • By using the formula, v = u + at, we get,
  • v = 15 m/s + 5 m/s^2 × 3 s
  • or, v = 15 m/s + 15 m/s
  • or, v = 30 m/s

<u>Answer</u><u>:</u>

<em><u>The </u></em><em><u>new </u></em><em><u>sp</u></em><em><u>e</u></em><em><u>ed </u></em><em><u>of </u></em><em><u>the </u></em><em><u>car </u></em><em><u>is </u></em><em><u>3</u></em><em><u>0</u></em><em><u> </u></em><em><u>m/</u></em><em><u>s.</u></em>

Hope you could get an idea from here.

Doubt clarification - use comment section.

8 0
3 years ago
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