275mm=cm conversions

A 10cm long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire

soldi70 [24.7K]

Answer:

a) $v=4.0m/s$

b) $B=2.958T$

Explanation:

From the question we are told that:

Wire Length $l=10cm=>0.10m$

Resistance $R=0.35$

Force $F=1.0N$

Power $P=4.0W$

a)

Generally the equation for Power is mathematically given by

$P=Fv$

Therefore

$v=\frac{P}{F}$

$v=\frac{4.0}{1.0}$

$v=4.0m/s$

b)

Generally the equation for Magnetic Field is mathematically given by

$B=\frac{\sqrt{PR}}{vl}$

$B=\frac{\sqrt{4*0.35}}{4*0.10}$

$B=2.958T$

5 0

3 years ago

The north end of a compass points towards

3241004551 [841]

Opposite attract , this is the magnetic south pole

option (b) ✔

8 0

2 years ago

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How does radiation transfer thermal energy from the Sun to Earth?

Svetradugi [14.3K]

Answer:

The thermal energy is carried by electromagnetic waves

Explanation:

There are three types of transfer of heat (thermal energy):

- Conduction: conduction occurs when two objects/two substances are in contact with each other. The heat is transferred from the hotter object to the colder object by the collisions between the molecules of the two mediums.

- Convection: convection occurs when a fluid is heated by an external source of heat. The part of the fluid closer to the heat source gets warmer, therefore it becomes less dense and it rises, and it is replaced by the colder part of the fluid, which is colder. Then, this part of fluid is heated as well, so it gets warmer, it rises, etc.. in a cycle.

- Radiation: radiation occurs when thermal energy is carried by electromagnetic waves. Since electromagnetic waves do not need a medium to propagate, this is the only method of heat transfer that can occur through a vacuum (so, in space as well).

Indeed, the Sun emits a lot of electromagnetic radiation, which travels through space and eventually reaches the Earth, heating it.

4 0

3 years ago

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Guys I'm in kind of a PICKLE!!!!!! I know people say it a lot but I will give Brainiest to the best explained answer. Determine

Flauer [41]

Answer:

E≅1.2×10^7 N/C

Explanation:

First off I'd like to say that I'm taking "net electric field" to mean that they don't want this answer to be put into vector component form and instead want magnitudes. Sometimes the wording of these questions throws me off, so sorry ahead of time if that's what they want from you!

Edit: I ended up adding it anyways ;P

Since we are observing the net electric field acting at q1, we need to use the formula: $E=k\frac{q}{r^{2} }$

And since we are observing the effects of multiple charges at once...

E=ΣE, which just means wee need to add all the observed electric fields together:

ΣE= $k\frac{q2}{r^{2} } +k\frac{q3}{r^{2} }$

Since we are observing [static] electric fields here, we don't actually need q1's charge. (Though if you wanted to find the net force you would.) Now, before we start plugging values in, let's acknowledge what we know. We know that:

q2=q3
they are the same distance from q1

These are actually really nice to have, because now we can simplify our expression to:

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

Now let's plug in our values and get an answer out.

E= 2(8.99×10^9)(4×10^-5)/(0.24)

Plugging all that in, I get:

E≅1.2×10^7 N/C

If you end up needing the net force, F=(q1)(E). That is, you just multiply the electric field by the value of q1. And again, if your teacher wants the answer in vector component form, then the answer will look different.

Let me know what doesn't make sense, or if I got something wrong. Good luck with AP Phy.!

Edit: I put the component form for my answer in the attachment. I also noticed a small calculator related error in my original answer. I updated that to match the new one.

6 0

3 years ago

Use the following half-life graph to answer the following question:

Digiron [165]

Answer:

B. 3

Explanation:

The half-life of a radioisotope is the time it needs for the mass of the radioisotope to halve with respect to its original value.

In this problem, the initial mass of the radioisotope at t=0 is

m0 = 50.0 mg

We see that after t = 1 min, the mass of the isotope is

m(1 min) = 25.0 mg

so, exactly half the initial mass: this means that 1 minute is exactly the half-life of this radioisotope.

So, the amount of mass left after each minute is the following:

m (1 min ) = 25.0 mg (1 half-life)

m (2 min) = 12.5 mg (2 half-lives)

m (3 min) = 6.25 mg (3 half-lives)

so, when we are left with 6.25 mg of isotope, 3 minutes have passed, which means that 3 half-lives have passed.

7 0

3 years ago