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faust18 [17]
3 years ago
9

Stored energy or energy of position that can be released or harnessed to do work

Physics
1 answer:
Alekssandra [29.7K]3 years ago
3 0

Answer:

A. potential energy

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An electric current is flowing through a long cylindrical conductor with radius a. The current density J is uniform in the cylin
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Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

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URGENT!!!! WILL GIVE BRAINLIEST!!!!!! 15 POINTS
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It's the angle made by the incident ray when it's perpendicular to the surface. (Perpendicular lines are the lines that form a graph or like a 90-degree angle)
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Read 2 more answers
Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot
Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² =  2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ =  m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁  = 4.427 - 2v₂  ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂  = 8.854

v₂ = 8.854 / 3

v₂  = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁  = - 1.477 m/s

v₁  = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

mgh_{max} = \frac{1}{2}mv_{max}^2

(a) The maximum height achieved by the first ball (v₁  = 1.477 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max}  = 0.11 \ m

(b) The maximum height achieved by the second ball (v₂  = 2.95 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max}  = 0.44 \ m

6 0
3 years ago
You launch a ball at an angle of 35 degrees above the horizontal with an initial velocity of 38 m/s. What is the time the ball w
gayaneshka [121]

Vf=Vi+at

0=38+(-9.8)(?)

?=38-0+(-9.8)

?=28.2 s

5 0
2 years ago
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