What is the commercial unit of energy.

A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

b)

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

c)

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

$q_{2} = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC (3)$

6 0

3 years ago

A sample of metallic frewium weighs 185N on a spring scale in air. When immersed in pure water, the frewium pulls on the scale w

balu736 [363]

Wow ! This one could have some twists and turns in it.
Fasten your seat belt. It's going to be a boompy ride.

-- The buoyant force is precisely the missing 30N .

-- In order to calculate the density of the frewium sample, we need to know
its mass and its volume. Then, density = mass/volume .

-- From the weight of the sample in air, we can closely calculate its mass.

 Weight = (mass) x (gravity)
 185N = (mass) x (9.81 m/s²)
 Mass = (185N) / (9.81 m/s²) = 18.858 kilograms of frewium

-- For its volume, we need to calculate the volume of the displaced water.

The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³. So the volume of the
displaced water (in cm³) is the same as the number of grams in it.

The weight of the displaced water is 30N, and weight = (mass) (gravity).

 30N = (mass of the displaced water) x (9.81 m/s²)

 Mass = (30N) / (9.81 m/s²) = 3.058 kilograms

 Volume of displaced water = 3,058 cm³

Finally, density of the frewium sample = (mass)/(volume)

 Density = (18,858 grams) / (3,058 cm³) = 6.167 gm/cm³ (rounded)

================================================

I'm thinking that this must be the hard way to do it,
because I noticed that

 (weight in air) / (buoyant force) = 185N / 30N = 6.1666...

So apparently . . .

 (density of a sample) / (density of water) =

 (weight of the sample in air) / (buoyant force in water) .

I never knew that, but it's a good factoid to keep in my tool-box.

3 0

3 years ago

Help asap please thank you

Tom [10]

Answer:

the answer its time

v = d/t

8 0

3 years ago

Read 2 more answers

In boxing, the use of 16-ounce gloves rather than 12-ounce gloves reduces the chance of injury because the force is distributed

mr Goodwill [35]

Answer:

true

Explanation:

Here we have assumed that increasing the mass of a glove will increase the surface area.

Injury is caused by the application of pressure at a point on the body. The application of pressure takes place via the area of the gloves. Pressure is given by

$P=\dfrac{F}{A}$