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frozen [14]
3 years ago
7

The electric field strength E 0 is measured at a perpendicular distance R from an infinitely large, thin sheet that contains a u

niform positive charge density σ . In terms of the stated properties and the permittivity of free space ε 0 , write an expression for the electric field strength E at a perpendicular distance 2 R from the sheet.
Physics
1 answer:
Brilliant_brown [7]3 years ago
7 0

Answer:

Electric Field Strength E₀ = E₀ = Constant

Explanation:

The Electric Filed Strength E₀ to an Infinite uniformly charge large sheet is constant how far is it i.e. it is independent of the distance away from the uniformly charge sheet.

Formula: E₀ = σ / 2 ε₀  

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With what speed does the can move immediately after the collision? Answer in units of m/s.
Ratling [72]

Answer:

1.74 m/s

Explanation:

From the question, we are given that the mass of the an object, m1= 2.7 kilogram(kg) and the mass of the can,m(can) is 0.72 Kilogram (kg). The velocity of the mass of an object(m1) , V1 is 1.1 metre per seconds(m/s) and the velocity of the mass of can[m(can)], V(can) is unknown- this is what we are to find.

Therefore, using the formula below, we can calculate the speed of the can, V(can);

===> Mass of object,m1 × velocity of object, V1 = mass of the can[m(can)] × velocity is of the can[V(can)].----------------------------------------------------(1).

Since the question says the collision was elastic, we use the formula below

Slotting in the given values into the equation (1) above, we have;

1/2×M1×V^2(initial velocity of the first object) + 1/2 ×M(can)×V^2(final velocy of the first object)= 1/2 × M1 × V^2 m( initial velocity of the first object).

Therefore, final velocity of the can= 2M1V1/M1+M2.

==> 2×2.7×1.1/ 2.7 + 0.72.

The velocity of the can after collision = 1.74 m/s

7 0
3 years ago
A hanging wire made of an alloy of titanium with diameter 0.05 cm is initially 2.7 m long. When a 15 kg mass is hung from it, th
Svet_ta [14]

Answer:

Explanation:

Young's modulus of elasticity Y = stress / strain

stress = force / cross sectional area

= weight of 15 kg / π r²

= 15 x 9.8 / 3.14 x ( .025 x 10⁻² )²

stress = 74.9 x 10⁷ N / m²

strain = Δ L / L , Δ L is change in length and L is original length

Putting the values

strain = .0168 / 2.7 =.006222

Young's modulus of elasticity Y  = 74.9 x 10⁷ / .006222

= 120.88 x 10⁹ N / m² .

8 0
3 years ago
Which notation is better to use? (Choose between 4,000,000,000,000,000 m and 4.0 × 1015 m)
ipn [44]

Answer:

4 x 10¹⁵

Explanation:

5 0
3 years ago
A jack is used to______ the force.​
noname [10]

Answer:

A jack is used to <u>lift</u> the force.

I think that's the answer. I don't really understand the question.

4 0
3 years ago
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A stone is dropped off a cliff and it falls for 4.8 m/s
valina [46]

Answer:

you cant caculate this because there is no question

Explanation:

7 0
3 years ago
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