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adelina 88 [10]
2 years ago
11

A force of 20 N is exerted by an electric field on a test charge of 8.0 x 10² C at a point, P. What is the electric field streng

th
at point P?
0.004 N/C
Ob
1.6 N/C
Oc
160 N/C
Od
250 N/C
Physics
1 answer:
kenny6666 [7]2 years ago
5 0

Answer:

the answer is equal to 1.6N/C

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Who made advances to our understanding of the conscious and unconscious mind?
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3 years ago
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A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po
aleksandrvk [35]

Answer:

Explanation:

Let the equilibrium position of third charge be x distance from q₁.

Force on third charge due to q₁

= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²

Force on third charge due to q₂

= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

Both the force will act in opposite direction and for balancing , they should be equal.

9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

5  / x² = 2 / ( .4 - x )²

Taking square root on both sides

2.236 / x = 1.414 / .4 - x

2.236 ( .4 - x ) = 1.414 x

.8944 - 2.236 x = 1.414 x

.8944 = 3.65 x

x = .245 m

24.5 cm

So the third charge should be at a distance of 24.5 cm from q₁ .

4 0
2 years ago
How much energy is needed to melt 5 g of ice? The specific latent heat of melting for water is 334000 J/kg.
Katen [24]

Answer:

The needed energy to melt of ice is 1670 J.

Explanation:

Given that,

Mass of ice = 5 g

Specific latent heat = 334000 J/kg

We need to calculate the energy

Using formula of energy

Q=mL

Where, m = mass

L = latent heat

Put the value into the formula

Q=5\times10^{-3}\times334000

Q=1670\ J

Hence, The needed energy to melt of ice is 1670 J.

5 0
3 years ago
If the air pressure is doubled, the speed of sound
Tom [10]

it remains unchanged

6 0
3 years ago
If 90 grams of gold has a volume of 5 cm3, calculate the density (include the units and show your work or you get no credit
ehidna [41]

Answer:

The density of gold is of 18 grams per cm3.

Explanation:

The mass density of a homogeneous material expresses how much mass of that material is present in a given volume. Since the density of an object is obtained by dividing its mass by its volume, to obtain the density of gold, its 90 grams of mass must be divided by its 5 cm3 volume, performing the following calculation:

90/5 = X

18 = X

Thus, the density of gold is 18 grams per cm3.

6 0
3 years ago
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