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2 years ago

9

What . Slides are placed upside down in a slide projector to correct for the _______ image produced. A. magnified B. inverted C.

small size of the D. virtual

2 answers:

Amanda [17]2 years ago

6 0

Penn Foster STudents: inverted

Yuri [45]2 years ago

5 0

<span>The correct answer is B. Inverted image. This is because of all the lenses and light refractions and what not. The same things happens with our eyes except our brains fix the inverted image automatically. Since there are no brains in a projector, you have to fix it on your own by putting it in reverse.</span>

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The force of gas particles against the walls of a container is called ________

Liono4ka [1.6K]

Answer:

the answer is pressure

8 0

2 years ago

The 5.5 million vinyl long-playing (LP) records sold in the United States per year pales in comparison with the 1.26 billion dig

miv72 [106K]

Answer:

decline

Explanation:

Based on the scenario being described within the question it can be said that these types of firms are in the decline stage of the product life cycle. This stage refers to when a product has already passed it's peak potential and sales begin to decline until production is ultimately halted and the product dies off. Which is exactly what is happening to the LP's since everyone has moved on to digital downloads.

4 0

2 years ago

You leave the doctor's office after your annual checkup and recall that you weighed 688 N in her office. You then get into an el

Lapatulllka [165]

Answer:

$a=0.5418\ m.s^{-2}$ upwards

$a=1.283\ m.s^{-2}$ downwards

Explanation:

Given:

weight of the person, $w=688\ N$

So, the mass of the person:

$m=\frac{w}{g}$

$m=\frac{688}{9.81}$

$m=70.132\ kg$

Now if the apparent weight in the elevator, $w_a= 726\ N$

<u>Then the difference between the two weights is :</u>

$\Delta w=w_a-w$

$\Delta w=726-688$

$\Delta w=38\ N$ is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

$a=\frac{\Delta w}{m}$

$a=\frac{38}{70.132}$

$a=0.5418\ m.s^{-2}$ upwards, because the normal reaction that due to the weight of the body is increased here.

Now if the apparent weight in the elevator, $w_a= 598\ N$

<u>Then the difference between the two weights is :</u>

$\Delta w=w-w_a$

$\Delta w=688-598$

$\Delta w=90\ N$ is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

$a=\frac{\Delta w}{m}$

$a=\frac{90}{70.132}$

$a=1.283\ m.s^{-2}$ downwards, because the normal reaction that due to the weight of the body is decreased here.

7 0

3 years ago

Read 2 more answers

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

0

0

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0

2 years ago

The Jensens dccided to spend their family vaeation white water rafting. During one segment of their trip down the river, the raf

torisob [31]

Answer:

24,187.04 J ≈ 24,200 J

Explanation:

mass (m) = 544 kg

initial speed (u) = 6.75 m/s

final speed (v) = 15.2 m/s

change in height (Δh) = -14 m (negative sign is because there is a decrease in height )

acceleration due to gravity (g) = 9.8 m/s^{2}

How much work was done on the raft by non conservative forces?

work done = change in energy of the system = change in kinetic energy + change in potential energy

work done = ( $0.5m(v^{2} -u^{2} )$ ) + (mgΔh)

work done = ( $0.5x544x(15.2^{2} -6.75^{2} )$ ) + (544 x 9.8 x (-14))

work done = 50449.76 - 74,636.8

work done = 24,187.04 J ≈ 24,200 J

7 0

3 years ago