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BlackZzzverrR [31]
3 years ago
10

If you used 650 W of power lifting a 300 N weight in 2 seconds how high did you lift?

Physics
1 answer:
dimulka [17.4K]3 years ago
7 0

Answer:

4.33m

Explanation:

Power = work done/ time

work done = power × time =650 × 2 = 1300J

work done = force × distance

distance = work done/force

distance = 1300/300 = 4.33m

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What is the magnitude of the electric field strength between them, if the potential 7.95 cm from the zero volt plate (and 2.05 c
Dmitriy789 [7]

Answer:

ΔVab = Ed

ΔVab = Va-Vb = Va-V0 = Va

E = Va/ d

   = 413V / 0.0795 m

    = 5194.97 V/M

Explanation:

the potential difference between two uniform plates is calculated by the formula of electric field.

6 0
3 years ago
What is water potential???​
melisa1 [442]

Answer:

Water potential is the potential energy of water per unit volume relative to pure water in reference conditions. Water potential quantifies the tendency of water to move from one area to another due to osmosis, gravity, mechanical pressure and matrix effects such as capillary action.

3 0
3 years ago
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A rectangular copper strip 1.5cm wide and 0.10cn thick carries a current of 5.0A. Find the Hall voltage for a 1.2T magnetic fiel
Scrat [10]

Answer:

4.4345× 10^-7V

Explanation:

The computation of the half voltage for a 1.2T magnetic field applied is shown below

The volume of one mole of copper is

v = m ÷p

= 63.5 ÷ 8.92

= 7.12cm

Now the density of free electrons in copper is

n = Na ÷ V

= 6.02 × 10^23 ÷ 7.12

= 8.456× 10^28/m^3

Now the half voltage is

= IB ÷ nqt

= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)

= 4.4345× 10^-7V

7 0
3 years ago
3. According to the article, why did Europeans so quickly accept that the sun did move and change?
Sphinxa [80]

Answer:A

Explanation:because it is good

5 0
2 years ago
Read 2 more answers
A cyclist rides 6.2 km east, then 9.28 km in a direction 27.27 degrees west of north, then 7.99 km west. A. What is the magnitud
Pani-rosa [81]

Answer:

Explanation:

given,

cyclist ride  6.2 km east and then 9.28 km in the direction of 27.27° west of north and then 7.99 km west.

vertical component = 9.28 cos∅

                                = 9.28 cos 27.27°

                                = 8.24 km

horizontal axis component = 9.28 sin ∅

                                             = 9.28 sin 27.27°

                                             = 4.5 km

distance of the final point from the origin

                            = 7.99 -(6.2-4.5)

                            = 6.29 km

displacement

d = \sqrt{6.29^2+8.24^2}

d = 10.37 km

b) tan \theta = \dfrac{6.29}{8.24}

θ = 37.36°

4 0
3 years ago
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