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On the graph of voltage versus current, which line represents a 2.0 Ω resistor?

Vikki [24]

Answer:

Explanation:

According to ohm's law V = IR where;

V i sthe supply voltage (in volts)

I = supply current (in amperes)

R = resistance (in ohms)

In order to calculate the line that is equal to 2ohms, we need to calculate the slope of each line using the formula.

For line B, R = ΔV/ΔI

R = V₂-V₁/I₂-I₁

R = 14.0-4.0/7.0-2.0

R = 10.0/5.0

R = 2.0ohms

Since the slope of line B is equal to 2 ohms, this shows that the line B is the one that represents the 2ohms resistor.

3 0

4 years ago

Which statement describes a chemical property related to the particles that make up the material

Stells [14]

Answer: The correct answer would be C. Salt crystals are hard and brittle because they are made up of atoms that combine by ionic bonding.

Explanation: All the other ones are chemical changes because a new substance is formed.

7 0

3 years ago

What effect does a tripling of the net force have upon the acceleration of the object?

dsp73

The formula of net Force is:F = mawhere m is the mass of the objecta is the acceleration of the object
thus, if we triple the net force applied to the object:
3F = maa = 3F / m
The acceleration is also tripled since the force is directly proportional to the acceleration.

4 0

4 years ago

An auto repair technician who specializes in the installation, troubleshooting, and repair of heating and air conditioning syste

Arisa [49]

C is the correct answer

7 0

3 years ago

Two charges, each 9 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field on the x axis at

bearhunter [10]

a) Electric field at x = -2 m: 21,060 N/C to the left

b) Electric field at x = 2 m: 18,000 N/C to the right

c) Electric field at x = 6 m: 18,000 N/C to the left

d) Electric field at x = 10 m: 21,060 N/C to the right

e) Electric field is zero at x = 4 m

Explanation:

a)

The electric field produced by a single-point charge is given by

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

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

Here we have two charges of

$q=9\mu C = 9\cdot 10^{-6} C$

each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.

We call the charge at x = 0 as $q_0$ , and the charge at x = 8 m as $q_8$ .

For a point located at x = -2 m, both the fields $E_0$ and $E_8$ produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:

$E=-\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=-kq(\frac{1}{(-2)^2}+\frac{1}{(8-(-2))^2})=-21060 N/C$

b)

In this case, we are analyzing a point located at

x = 2 m

The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(2)^2}-\frac{1}{(8-2)^2})=18000 N/C$

And since the sign is positive, the direction is to the right.

c)

In this case, we are considering a point located at

x = 6 m

The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(6)^2}-\frac{1}{(8-6)^2})=-18000 N/C$

And the negative sign indicates that the electric field in this case is towards the left.

d)

In this case, we are considering a point located at

x = 10 m

This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:

$E=\frac{kq_0}{(0-x)^2}+\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(10)^2}+\frac{1}{(8-(10))^2})=21060 N/C$

And the positive sign means the field is to the right.

e)

We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(-x)^2}-\frac{1}{(8-x)^2})=0$