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2 years ago

An advertiser who calls a basement apartment a “garden apartment” is using the rhetorical device called

1 answer:

3 0

I'd say B or D, depending on where the place is. If it's in South Kensington, London, then maybe D. If it's in a deprived area then I'd say B.

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Tori experiments with pulleys in physics class. She applies 70 newtons of force to a single pulley to lift a bowling ball. By ad

e-lub [12.9K]

The minimum input force she'll need to lift the ball is 35 N.

Explanation:

Mechanical advantage of a single pulley is 1. As, she applies 70 N of force to lift the bowling ball, so the output force(weight of the ball) is also 70 N.

Now, adding another pulley gives a mechanical advantage of 2. We have,

M.A = (Output Force)/(Input Force)

Substituting the values we get,

$2=\frac{70}{F_{i}}$

$F_{i}=\frac{70}{2}$ = 35 N

Input force equals to 35 N needs to be applied.

6 0

2 years ago

Read 2 more answers

A visitor to the observation deck of a skyscraper manages to drop a penny over the edge. As the penny falls faster, the force du

pentagon [3]

If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.

8 0

2 years ago

Read 2 more answers

The resistivity of a silver wire with a radius of 5.04 × 10–4 m is 1.59 × 10–8 ω · m. if the length of the wire is 3.00 m, what

Alexxx [7]

<span>5.98 x 10^-2 ohms. Resistance is defined as: R = rl/A where R = resistance in ohms r = resistivity (given as 1.59x10^-8) l = length of wire. A = Cross sectional area of wire. So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives: R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2) R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7) R = (4.77 x 10^-8) / (7.98015 x 10^-7) R = 5.98 x 10^-2 ohms So that wire has a resistance of 5.98 x 10^-2 ohms.</span>

4 0

2 years ago

An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start

FrozenT [24]

Answer:

The total distance traveled is 28 inches.
The displacement is 2 inches to the east.

Explanation:

Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector $\hat{i}$ pointing in the west direction, the ant start at position

$\vec{r}_0 = 16 \ inch \ \hat{i}$

Then, moves to

$\vec{r}_1 = 29 \ inch \ \hat{i}$

so, the distance traveled here is

$d_1 = |\vec{r}_1 - \vec{r}_0 | = | 29 \ inch \ \hat{i} - 16 \ inch \ \hat{i} |$

$d_1 = | 13 \ inch \ \hat{i} |$

$d_1 = 13 \ inch$

after this, the ant travels to

$\vec{r}_2 = 14 \ inch \ \hat{i}$

so, the distance traveled here is

$d_2 = |\vec{r}_2 - \vec{r}_1 | = | 14 \ inch \ \hat{i} - 29 \ inch \ \hat{i} |$

$d_2 = | - 15 \ inch \ \hat{i} |$

$d_2 = 15 \ inch$

The total distance traveled will be:

$d_1 + d_2 = 13 \ inch + 15 \ inch = 28 \ inch$

The displacement is the final position vector minus the initial position vector:

$\vec{D}=\vec{r}_2 - \vec{r}_1$

$\vec{D}= 14 \ inch \ \hat{i} - 16 \ inch \ \hat{i}$

$\vec{D}= - 2 \ inch \ \hat{i}$

This is 2 inches to the east.

6 0

2 years ago

Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. H

kirill [66]

Answer:

U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

dU = - F. dr

the esxresion for strength is

F = B / r³

let's replace

∫ dU = - ∫ B / r³ dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

U = B / 2r²

we substitute the value of B = 2

U = 1 / r²

5 0

2 years ago