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2 years ago

11

Which best describes cost-effectiveness

1 answer:

Rufina [12.5K]2 years ago

3 0

Answer:

Maintaining the lowest expenses for making a product.

Explanation:

Cost efficiency is the ability to use less resources (cost) to achieve greater output while cost effectiveness is the ability to use resources (cost) to achieve the objectives (outcomes).

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Why pressure above the surface is greater than in the air

nexus9112 [7]

Answer:

At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels. ... Since most of the atmosphere's molecules are held close to the earth's surface by the force of gravity, air pressure decreases rapidly at first, then more slowly at higher levels.

Explanation:

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3 years ago

NEED ASAP PLEASE !!

Svetllana [295]

The balloon was 30.65 meters above ground.

ANSWER: A

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3 years ago

Read 2 more answers

A 27.0 g marble sliding to the right at 56.8 cm/s overtakes and collides elastically with a 13.5 g marble moving in the same dir

Nataly [62]

Answer: $v_1=28.4 cm/s$

Explanation:

Given

mass of marble $m_1=27 gm$

velocity of marble $u_1=56.8 cm/s \approx 0.568 m/s$

mass of second marble $m_2=13.5 gm$

Velocity of second marble $u_2=14.2 cm/s \approx 0.142 m/s$

After collision 13.5 gm marble moves to the right at i.e. $v_2=71 cm/s$

Conserving momentum

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$27\times 56.8+13.5\times 14.2=27\times v_1+13.5\times 71$

$1533.6+191.7=27\cdot v_1+958.5$

$27\cdot v_1=766.8$

$v_1=\frac{766.8}{27}=28.4 cm/s$

7 0

3 years ago

This object is located 13.0 cm to the left of the lens and the image forms at 20.8 cm to the right of the lens.

Alina [70]

Answer:

0.125 cm

Explanation:

1/f = 1/d¡ + 1/d。

Find the focal point

(13.0^-1 + 20.8^-1) = 0.125 m

Focal point = 0.125 m

6 0

2 years ago

Read 2 more answers

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

3 years ago