Waves that make up the visible part of the electromagnetic spectrum have

A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is

lbvjy [14]

Answer:

$a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}$

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

a)

$a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\$

b)

$a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}$

8 0

3 years ago

Which sentences describe the cyclone device?

Yakvenalex [24]

Answer:

✓ A cyclone device accumulates fine particulates from the air by making a dirty air stream flow in a spiral path inside a cylindrical chamber.

✘ It consists of several long and narrow fabric filter bags suspended upside-down in a large enclosure.

✓ When dirty air enters the chamber, the larger particulates strike the chamber wall and fall into a conical dust hopper at the bottom.

✘ Fans blow dirt-filled air upward from the bottom of the enclosure, trapping dirt particles inside the filter bags and releasing clean air from the top.

✓ The top of the chamber has an outlet that lets out cleaned air.

Basically, any of these choices that have the word "filter" are wrong. The point of the cyclone device is to separate the particles without the use of filters. You can tell the right answers based on the picture attached below.

3 0

3 years ago

Read 2 more answers

A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

yaroslaw [1]

The electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.

To determine σ:

σ = Q/A

Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:

σ = Q/d²

Make this substitution in the equation for E:

E = Q/(2ε₀d²)

We see that E is inversely proportional to the square of d:

E ∝ 1/d²

The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:

$E_{new} = E/4$