List five uses of electromagnets identified in the article.

Redox reactions: Redox reactions: rarely involve enzymes. do not occur under physiological conditions. are characterized by one

Gnoma [55]

Answer:

are characterized by one substance gaining an electron while another substance loses an electron

Explanation:

Redox reaction in chemical reaction in which one substance gaining an electron while another substance loses an electron. This means that one element is oxidize by losing an electron while the other is reduced by gaining an electron. The one oxidized is called reducing agent while the one reduced is called oxidizing agent.

8 0

4 years ago

What is the melting point of substance A?

Misha Larkins [42]

Answer:

Solids are easily recognized by their ability to retain a fixed shape and definite volume. Particles making

up a solid are held together in a rigid form. They are not free to move about or slide past one another and

the solid does not have the ability to flow. (Although the particles of a solid do not move position to position, they do have motion in that they are constantly vibrating.

To change the temperature of a solid, heat energy must be added. The amount of heat energy that changes

the temperature of 1.0 g of a solid by 1.0°C is called its specific heat (c). Each substance has its own

specific heat. The specific heat of ice is 2.1 Joules/g°C. In other words we must supply 1.0 gram of ice

with 2.1 Joules of heat energy to raise its temperature by 1.0 °C.

The general equation for calculating heat energy to change the temperature of a solid is:

Heat = Mass x Specific Heat (solid) x Temperature Change

Q = m c DT

10 g 10 g 10 g 10 g 10 g 10 g

Calculate the heat necessary to change 10 g of ice(s) at -20 °C to 10 g of ice(s) at 0°C. (A-B)

Q = mc∆T = (10 g) (2.1 J/g°C) (20°C) = 420 J

If you continue to add heat energy once the temperature of the ice reaches 0°C , the heat absorbed is called

the heat of fusion (Lf). This heat is used to cause a change of phase (from a solid to a liquid). This heat is

increasing the potential energy of the molecules of the solid. No temperature change takes place. Each

substance has its own heat of fusion. The heat of fusion for ice is 340 Joules/g. Exactly the same amount

of heat is given up when 1.0 g of water is changed to ice. This heat is called the heat of crystallization.

The general equation for calculating heat energy to change a solid to a liquid is:

Heat = Mass x Heat of Fusion

Q = m Lf

Calculate the heat necessary to change 10 g of ice(s) at 0°C to 10 g of water(l) at 0°C.(B-C)

Explanation:

Q = mLf = (10 g)( 340 J/g) = 3400 J

3 0

3 years ago

What is the kinetic energy of a 2000 kg car moving at 72 km/h?

gtnhenbr [62]

kinetic energy is given by 1 /2

$\times mass \: \times \: velocity {?}^{2}$ then you have to replace the figures

7 0

4 years ago

You and your friend Peter are putting new shingles on a roof pitched at 21∘. You're sitting on the very top of the roof when Pet

Marat540 [252]

Answer:

3.69 m/s

Explanation:

Forces :

mgsin Θ - mumgcosΘ = ma

g x sinΘ - mu x g x cosΘ = a

9.8 x sin 21 - 0.53 x 9.8 x cos 21 = a

a = -1.337 m/s²

so you have final velocity = 0 m/s

initial velocity = ? m/s

Given d = 5.1 m

By kinematics

vf² = vo² + 2ad

0 = vo² + 2 x -1.337*5.1

vo = 3.69 m/s

8 0

4 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

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