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3 years ago

7

Which is the result of using a machine?

1 answer:

Eddi Din [679]3 years ago

3 0

Correct Answers is A.

The machines gives us some mechanical advantage. This means the mechanical average makes the work output greater than the work input
Simple most example is a lever. The force applied is smaller and the output work is larger as compared to input.

Option B cannot be true, as there must be a force to get some work done.
Option C and D are inverse of what a machine is designed for. A small force can be exerted through a large distance to have a large force exerted through a small distance. Common Example of this principle is a screw opener.

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Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of

mario62 [17]

I don't know if you need to complete this question or do it otherwise, however, I managed to find on the Internet on several places this completion of your sentence:
<span>Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of q = 2 x 10</span>⁶ W/m³.
I'm not sure whether that is the answer you were looking for, but that's what I found.

5 0

2 years ago

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

bogdanovich [222]

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 height

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb's 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 final 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 only 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

The 0.5kg soccer ball moves toward the net with a force of 4N. What is its acceleration?

Eduardwww [97]

Answer:

8 m/s²

Explanation:

Given,

Force ( F ) = 4 N

Mass ( m ) = 0.5 kg

To find : -

Acceleration ( a ) = ?

Formula : -

F = ma

a = F / m

= 4 / 0.5

= 40 / 5

a = 8 m/s²

It's acceleration is 8 m/s².

6 0

2 years ago

If a car is traveling on the highway at a constant velocity, the force that pushes the car forward must be A. equal to the weigh

Gala2k [10]

the correct answer is c

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3 years ago

Change in momentum is equal to force times the amount of time the force is applied? True or False?

harina [27]

The answer to this question is false

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2 years ago