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3 years ago

What forces moves electrons through an electrical outlet

Physics

1 answer:

aliya0001 [1]3 years ago

4 0

Answer:

electromotive force

Explanation:

This force is called electromotive force, or EMF. Sometimes it is convenient to think of EMF as electrical pressure. In other words, it is the force that makes electrons move in a certain direction within a conductor

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At a height of ten meters above the surface of a freshwater lake, a sound pulse is generated. The echo from the bottom of the la

konstantin123 [22]

Answer:

Explanation:

Velocity of sound in air at 20 degree = 343 m/s

Velocity of sound in water at 20 degree = 1470 m/s

Time taken in to and fro movement in air

=( 2 x 10) / 343 = 0.0583 s

Rest of the time is

.171 - .0583 = .1127 s

This time is taken to cover distance in water. If d be the depth of lake

2d / velocity = time taken

2 d / 1470 = .1127

d = 82.83 m

5 0

3 years ago

The best type of exercise for weight loss is

lbvjy [14]

Answer:

finding a workout.

Explanation:

6 0

3 years ago

A flat sheet of ice has a thickness of 1.4 cm. It is on top of a flat sheet of crown glass that has a thickness of 3.0 cm. Light

MAXImum [283]

Answer:

t = 2.13 10-10 s , d = 6.39 cm

Explanation:

For this exercise we use the definition of refractive index

n = c / v

Where n is the refraction index, c the speed of light and v the speed in the material medium.

The refractive indices of ice and crown glass are 1.13 and 1.52, respectively, therefore the speed of the beam in the material medium is

v = c / n

As the beam strikes perpendicularly, the beam path is equal to the distance of the leaves, there is no refraction, so we can use the uniform motion relationships

v = d / t

t = d / v

t = d n / c

Let's look for the times on each sheet

Ice

t₁ = 1.4 10⁻² 1.31 / 3 10⁸

t₁ = 0.6113 10⁻¹⁰ s

Crown glass (BK7)

t₂ = 3.0 10⁻² 1.52 / 3.0 10⁸

t₂ = 1.52 10⁻¹⁰ s

Time is a scalar therefore it is additive

t = t₁ + t₂

t = (0.6113 + 1.52) 10⁻¹⁰

t = 2.13 10-10 s

The distance traveled by this time in a vacuum would be

d = c t

d = 3 10⁸ 2.13 10⁻¹⁰

d = 6.39 10⁻² m

d = 6.39 cm

3 0

3 years ago

Choose all facts that increase the orbital velocity of a vessel around planet B. Bigger mass of planet B smaller mass of planet

telo118 [61]

Answer:

- Bigger mass of planet B

- orbiting closer to planet B

Explanation:

The orbital velocity of the vessel around the planet can be found by equalizing the force of gravity between the vessel and the planet and the centripetal force:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

m is the mass of the vessel

M is the mass of the planet

r is the distance between the vessel and the centre of the planet

v is the orbital velocity of the vessel

Re-arranging the formula, we find an expression for v:

$v=\sqrt{\frac{GM}{r}}$

We see that:

- the bigger the mass of the planet, M, the bigger the velocity

- the bigger the distance between the vessel and the planet, r, the smaller the velocity

So, the correct choices that increase the orbital velocity are:

- Bigger mass of planet B

- orbiting closer to planet B

6 0

4 years ago

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If t

coldgirl [10]

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

Y axis

N- W = 0

N = W = mg

X axis

-Fr = m a

-μ N = m a

-μ mg = ma

a = μ g

a = - 0.32 9.8

a = - 3.14 m/s²

We calculate the distance using the kinematics equations

Vf² = Vo² + 2 a x

x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

x = ( 0 - 15²) / 2 (-3.14)

x= 35.8 m

The shortest braking distance is 35.8 m

7 0

3 years ago