All categories

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.

You might be interested in

An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

$V_{f1}=0.8511m/s$

B) $7.9Kg\rightarrow 11.2m/s$

$V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}$

$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

7 0

3 years ago

A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of

sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height $\Delta h$ depends on the square of the time:

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

where

$\Delta h$ is the change in the ball's height.
$g$ is the acceleration due to gravity.
$t$ is the time for which the ball is in the air.
$v_0$ is the initial vertical velocity of the ball.

The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, $v_0 = 0$ .

Solve for $t$ .

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ;

$\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2}$ ;

$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

$t \approx 0.440315\;\text{s}$ .

What's the initial horizontal velocity of the ball?

Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0

3 years ago

The current theory of the structure of the

Elza [17]

Answer: 91.4 J

Explanation:

Kinetic energy is the energy possessed by a body due to virtue of its motion.

K.E. = 0.5 m v²

Mass of the continent is given, m = 1.819 × 10²¹ kg

Side of the block of continent, s = 4150 km = 4150000 m

Depth of the block of continent, d = 38 km = 38000 m

(Mass = density × volume

m = 2780 kg/m³× (4150 × 10³ m)²× 38 × 10³ m = 1.819 × 10²¹ kg)

The continent is moving at the rate of, v = 1 cm /year = 0.01 m / 31556926 s = 3.17 × 10⁻¹⁰ m/s

⇒ K.E. = 0.5 × 1.819 × 10²¹ kg × (3.17 × 10⁻¹⁰ m/s)²= 91.4 J

Hence, mass of the continent has 91.4 J of kinetic energy.

5 0

3 years ago

The velocity of blood flow is the lowest in capillaries because

Varvara68 [4.7K]

Answer:

the total cross-sectional area of the capillaries is greater than the total cross-sectional area of the arteries or any other part of the circulatory system.

Explanation:

Blood velocity is not the same in all areas. In the capillaries it is where there is less speed, while in arteries and veins it is quite similar. This is due to the cross-sectional area of each of the vessels. It is a mistake to think of a vein, artery or capillary individually. We have to put them all together to see that the total area of the capillaries is 100 times larger than that of the arteries or veins. Blood flowing through arteries or veins is going faster because there is less area.

Blood velocity is inversely proportional to each of the areas of its territories. The greater the area, the lower the speed.

6 0

2 years ago

The cement bag shown in the figure has a mass of 50 kg and the small pulley has a mass of 2 kg. If the cement bag is pulled upwa

GuDViN [60]

For the cement bag we can say as per its force diagram we will have

$2F - mg = ma$