Help me slove this problem 115 divided by 2

What time is it in Orlando, FL when it is 10:00 a.m. in Los Angeles, CA?

vlabodo [156]

Answer:

1:00 pm

Explanation:

CA is pacific time zone

FL is Eastern

pacific time is 3 hours less

5 0

3 years ago

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An electron is moving at a speed of 4.8 × 10 6 m/s at an angle of 30.0° with respect to a uniform magnetic field of 9.2 × 10 –4

omeli [17]

Answer:

r = 5.94 10⁻² m

Explanation:

The magnetic force is given by the ratio

F =q v x B

The bold indicate vectors and the vector product determines that the force is perpendicular to the other two vectors, so the modulus of the velocity does not change, but its direction, therefore the acceleration in Newton's second law is centripetal

F = m a

The magnitude of the force is

F = q v B sin θ

The centripetal acceleration is

a = v² / r

Let's replace

q v B sin θ = m v² / r

r = m v / (q B sin θ)

Let's calculate

r = 9.11 10⁻³¹ 4.8 10⁶ / (1.60 10⁻¹⁹ 9.2 10⁻⁴ sin 30)

r = 5.94 10⁻² m

6 0

3 years ago

Do animals fart and do all animals blink and do animals have the same trait of humans

pantera1 [17]

Uh yes they do fart and blink. Like very living person, and animal. They do have the same traits, as that to humans, but their biology is set up differently. They are stronger and quicker.
I'm not going to state that humans are animals, because I don't agree. Humans are NOT animals!!! But that's my belief, oppose as you like.
I think the best way to answer this question, is to imagine being an animal. Possibly a wild animal. What do you do? How do you find food? Where do you sleep? See, usually humans know how to be disgusted about living in unsanitary and unsafe environments. I mean looking at the traits, they're biology is just set up differently from humans.

I hope this helps.<span />

8 0

4 years ago

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A motorcycle is capable of accelerating at 5.1 m/s starting feom rest how far can it travel im 1.5 seconds

saul85 [17]

5.1 m/ s

~Multiply 1.5 on both sides since you’re trying to finding out how far it can travel in 1.5 seconds

5.1 x 1.5= 7.65

s x 1.5= 1.5

So a motorcycle can travel 7.65 m in 1.5 seconds.

6 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

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