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skad [1K]
3 years ago
10

A test charge is placed at a distance of 2.5 × 10-2 meters from a charge of 6.4 × 10-5 coulombs. What is the electric field at t

he test charge? (k = 9.0 × 109 newton·meter2/coulomb2)
Physics
2 answers:
Novosadov [1.4K]3 years ago
8 0
Using the formula: E = kQ / d² where E is the electric field, Q is the test charge in coulomb, and d is the distance. 

E = kQ / d²

k = 9 x 10^9 N-m²/C²
Q = 6.4 x 10^-5 C
d = 2.5 x 10^-2 m

Substituting the given values to the equation, we have:
E = (9 x 10^9)(6.4 x 10^-5) / (2.5 x 10^-2) ²

Electric field at the test charge is 921600000 N/C
Sveta_85 [38]3 years ago
8 0

Answer:

E.  9.2 × 108 newtons/coulomb.

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Where D = density
m = mass
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now we divide D on both sides giving us
V = m/D 

We know our mass which is 600g and our density is 3.00 g/cm^3
so
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3 years ago
a heavy jar sits on top of a 3.4 m shelf with a gravitational potential energy of 180 j. What is the mass of the jar?
sp2606 [1]

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Explanation:

g.p.e= mgh

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6 0
3 years ago
Select the correct answer.
VladimirAG [237]

Answer:

It's effective temperature.

Explanation:

8 0
3 years ago
What kind of system does not allow matter or energy to enter or exit?
Vilka [71]

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Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0
Alisiya [41]

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

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The modulus vector can be found using the Pythagorean theorem

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