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3 years ago

Please help asap. thank you !!

Physics

1 answer:

kogti [31]3 years ago

3 0

Answer:

Explanation:

insulators do not let their valance (out side elections) move about easily.

series circuits provide only 1 path for the electrons to travel.

Conductors allow their outside electrons to travel easily.

Imagine yourself in a wresting match. Your opponent only allows you to move if he lets you. That what resistance is.

Answers

1. insulators

2. Series circuits

3. conductors

4. resistance

In that order.

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The train traveled 500 kilometers north to Odessa in 2 hours . What’s the train’s speed ? What is the velocity ?

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Explanation:

Speed is distance over time.

500 km / 2 hr = 250 km/hr

Velocity is speed and direction.

250 km/hr north

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3 years ago

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3 years ago

A 10 kg box slides down a lane inclined at an angle θ = 30 . The plane has a friction of coefficient 0.1. The box starts from th

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Answer:

The distance the box traveled down the plane is 19.28 m

Explanation:

The angle of repose, α, is given by the relation;

tan⁻¹(μ) = α

tan⁻¹(0.1) = 5.7°

Therefore, we have;

M·g·sin(θ) - μ·N = M·a

Where:

M = Mass of the box = 10 kg

g = Acceleration due to gravity = 9.81 m/s²

θ = Angle of inclination of the plane = 30°°

μ = Coefficient of friction = 0.1

a = Acceleration of the box along the incline plane

N = Normal force due to the weight of the box = M·g·cos(θ)

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48.2 = 10 ×a

a = 48.2/10 = 4.82 m/s²

The distance, s, traveled by the box is given by the relation;

s = u·t + 1/2×a·t²

Where:

u = Initial velocity = 0 m/s

t = Time of motion = 2.0 s

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3 years ago

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Answer:

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Explanation:

is this for that crossword puzzle lol

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4 years ago

Light of wavelength 550 nm falls on a

Brums [2.3K]

Answer:

The first diffraction maximum fringe will be at approximately 2.7 meters from the central maximum.

Explanation:

We can describe single slit diffraction phenomenon with the equation:

$a\sin\theta=m\lambda$ (1)

with θ the angular position of the minimum of order m respect the central maximum, a the slit width and λ the wavelength of the incident light. Because the distances between the first minima and the central maximum ( $y_{m}$ ) are small compared to the distance between the screen and the slit (x), we can approximate $\sin\theta\approx\tan\theta=\frac{y}{x}$ , using this on (1):

$a\frac{y_{m}}{x}=m\lambda$

solving for y

$y_{m}= \frac{mx\lambda}{a}$

Note that $y_{m}$ is the distance between a minimum and the central maximum but we need the position of a maximum not a minimum, here we can use the fact that a maximum is approximately between two minima, so the first diffraction maximum fringe is between the minima of order 1 and 2, so we should find $y_{1}$ , $y_{2}$ add them and divide by two: