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3 years ago

9

Find the force between 2C and -1C separated by a distance 1m. Indicate if it is an attractive force (negative) or a repulsive fo

rce (positivie).

1 answer:

telo118 [61]3 years ago

6 0

Answer:

Explanation:

Its definitely an Attractive force since the two charges are Unlike.

From Coulombs Law

F=kq1q2/R²

Given

K=9x10^9

R=1m

q1=2C

q2=-1C

F=(9x10^9 x 2 x -1)/1²

F= - 1.8x10^10N. (Attractive).

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Two concentric circular loops of wire lie on a tabletop, one inside the other. The inner wire has a diameter of 20.0 cm and carr

Harrizon [31]

Answer:

Explanation:

Magnetic field due to circular wire at the center = μ₀ I / 2 r

I is current and r is radius . μ₀ = 4π x 10⁻⁷.

field B₁ due to inner loop

B₁ = 4π x 10⁻⁷ x 12 / 2 x .20

= 376.8 x 10⁻⁷

Field due to outer loop

B₂ = 4π x 10⁻⁷ x I / 2 x .30

For equilibrium

B₁ = B₂

376.8 x 10⁻⁷ = 4π x 10⁻⁷ x I / 2 x .30

I = 18 A.

The direction should be opposite to that in the inner wire . It should be anti-clockwise.

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3 years ago

Weather stays the same all the time true or false

Vlad1618 [11]

This question is false because it is not hot or cold every single day

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3 years ago

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At a distance r1 from a point charge, the magnitude of the electric field created by the charge is 395 N/C. At a distance r2 fro

Ksenya-84 [330]

Answer:

r₂/r₁ = 1.82

Explanation:

The electric field due to a point charge, has the following expression:

$E =\frac{k*q}{r^{2}}$

For a distance r₁, the magnitude of the electric field is 395 N/C, so we can solve for r₁², as follows:

r₁² = $\frac{k*q}{395 N/C} }$ (1)

For a distance r2, the magnitude of the electric field is 119 N/C, so we can solve for r₂², as follows:

r₂² = $\frac{k*q}{119 N/C} }$

We can find the quotient r₂/r₁, from (1) and (2):

r₂/r₁ = $\sqrt{395/119}$ = 1.82

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4 years ago

wide tube that extends down from the bag of solution, which hangs from a pole so that the fluid level is 90.0 cm above the needl

Bingel [31]

Answer:

The average gauge pressure inside the vein is 110270.58 Pa

Explanation:

This question can be solved using the Bernoulli's Equation. First, in order to determine the outlet pressure of the needle, we need to find the total pressure exerted by the atmosphere and the fluid.

$P_f: fluid's\ pressure\\P_f= \rho g h=1025\frac{kg}{m^3} \times 9.8 \frac{m}{s^2} \times 0.9 m=9040.5 Pa \\P_T: total\ pressure\\P_T=P_{atm}+P_f\\P_T=101325 Pa + 9040.5 Pa=110275.5 Pa\\$

Then, we have to find the fluid's outlet velocity with the transversal area of the needle, as follows:

$S: transversal\ area \\S= \pi r^2=\pi (0.200 \times 10^{-3})^2=5.65 \times 10^{-7} m^2\\v=\frac{F}{S}=\frac{5.55 \times 10^{-8} \frac{m^3}{s}}{5.65 \times 10^{-7} m^2}=0.98\times 10^{-1} \frac{m}{s}$

As we have all the information, we can complete the Bernoulli's expression and solve to find the outlet pressure as follows:

$P_T-P_{out}=\frac{1}{2} \rho v^2\\P_{out}=P_T-\frac{1}{2} \rho v^2=110275.5 Pa-\frac{1}{2} 1025\frac{kg}{m^3} (0.98\times 10^{-1} \frac{m}{s})^2=110275.5 Pa-4.92 Pa =110270.58 Pa$

6 0

3 years ago

Which has greater kinetic energy, a car traveling at 30 km/HR or a car of half the mass traveling at 60 km/HR? a. The 60 km/HR c

artcher [175]

Answer:

a.The 60 km/HR car

Explanation:

Kinetic Energy: This can be defined as the energy of a body due to motion. The S.I unit of kinetic energy is Joules (J).

It can be expressed mathematically as

Ek = 1/2mv²......................... Equation 1

Where Ek = kinetic energy, m = mass, v = velocity.

(i) A car travelling at 30 km/hr, with a mass of m,

Ek = 1/2(m)(30)²

Ek = 450m J.

(ii) A car travelling at 60 km/hr, with a mass of m/2

Ek = 1/2(m/2)(60)²

Ek = 900m J.

Thus , the car travelling at 60 km/hr at half mass has a greater kinetic energy to the car traveling at 30 km/hr at full mass.

The right option is a.The 60 km/HR car

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4 years ago