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Law Incorporation [45]
4 years ago
5

Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th

e right hand and has magnetic field of 4.00 N/C .If an ion of charge +2e is released at rest at the left hand plate .What its K.E when reaches the right hand plate?tell the correct answer with explanation..(a)4ev (B) 32ev (C)64ev(D)16ev
Physics
1 answer:
yuradex [85]4 years ago
3 0

Answer:

C

Explanation:

Formula E=F/C also E=V/d

In this case use the second formula; E=V/d

Data given; E=4N/C d=8m

So v=E X d

     V=4x8=32V

k.e=eV= 2X32=64eV

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When would you use a compound microscope?
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 a compound microscope is used for viewing samples at high magnification<span> 40 - 1000x, which is achieved by the combined effect of two sets of lenses: the ocular lens in the eyepiece and the objective lenses close to the sample.</span>
8 0
3 years ago
4) The components of Vector A are given as follows Ax = -2.4 Ay = + 3.8 What is the magnitude of the A, and what is the angle th
Monica [59]

Answer:

A = 4.49

α = 57.72°

Explanation:

Knowing the magnitude of x & y of a vector we can determine the total magnitude of a vector.

A=\sqrt{A_{x}^{2} +A_{y}^{2} } \\A=\sqrt{(2.4)^{2} +(3.8)^{2} }\\A=4.49

The angle tangent can be used to determine the angle.

tan(\alpha )=\frac{3.8}{2.4}\\tan(\alpha ) =1.5833\\\alpha =tan^{-1}(1.5833) \\\alpha = 57.72 (deg)

5 0
3 years ago
A solid conducting sphere with radius R = 0.390 m carries a net charge of +0.650 nC.
Tom [10]

Answer:

38.5 N/C

Explanation:

The electric field generated by a charged sphere at a point outside the sphere is equivalent to the electric field generated by a single point charge, and it is given by

E=k\frac{Q}{r^2}

where

k=9\cdot 10^9 Nm^2C^{-2} is the Coulomb's constant

Q is the net charge

r is the distance from the centre of the sphere

In this problem, we have

Q=+0.650 nC=0.65\cdot 10^{-9}C

r=0.390 m

Substituting into the equation, we find

E=(9\cdot 10^9)\frac{(0.65\cdot 10^{-9}C)}{(0.390m)^2}=38.5 N/C

6 0
3 years ago
Three blocks are arranged in a stack on a frictionless horizontal surface. The bottom block has a mass of 37.0 kg. A block of ma
Alex777 [14]

Answer:

N₂ = 503.8 N

Explanation:

given,

mass of bottom block = 37 Kg

mass of middle block = 18 Kg

mass of the top block = 16 Kg

force acting on the top block = 170 N

force on the block at top

N₁ be the normal force from block at middle

now,

N₁ = 170 + m g

N₁ = 170 + 16 x 9.8

now, force on block at middle

N₂ be the normal force exerted by the bottom block

N₂ = N₁ + m₂ g

N₂ = 326.8 + 18 x 9.8

N₂ = 503.8 N

hence, normal force by bottom block is equal to N₂ = 503.8 N

7 0
4 years ago
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Hope it help!
4 0
4 years ago
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