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3 years ago

What is the difference between thrust and applied force?

Physics

1 answer:

Tamiku [17]3 years ago

7 0

Answer:

Thrust acts on the accelerated object in the direction opposite to the applied force hence it accelerates the object in the direction opposite to the applied force. ... Its magnitude is equal to that of applied force. It always increases the velocity of the object.

Explanation:

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Compare green and orange light from the visible spectrum. You are currently in a labeling module. Turn off browse mode or quick

77julia77 [94]

Green: nm 495–570. Yellow: nm 570–590. 590–620 nm for orange. Red: 620-750 nm (400–484 THz frequency)

Solids' molecules are strongly attracted to one another. As a result, the molecules are barely moving and tightly packed. Because of this, shape and volume are fixed.

The forces of attraction and repulsion in liquids are comparable. Compared to the solid state, they move a little bit more. They then assume the shape of the container while still having a fixed capacity.

The attraction forces between the molecules in gases are quite weak. They move quite freely and grow in an effort to fill as much space as they can. Consequently, their volume and shape vary (adopt the shape of the container).

You can learn more about states of the matter here:

brainly.com/question/18538345

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6 0

2 years ago

Estimate the monthly cost of using a 700 W refrigerator that runs for 10 hours a day if the cost per kilowatt hours is cents $.2

Mrac [35]

Answer:

$4.2

Explanation:

Given data

Power= 700W

time= 10 hours

Cost per kilowatt hours is cents $0.20

Let us find the number of hours in a month

=24*30

=720 hours

Energ= power*time

Energy= 700/1000*30

Energy= 7*3

Enery= 21 kwh

1 kwh= 0.2

21kwh= x

cross multiply

x=21*0.2

x= $4.2

4 0

2 years ago

You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee

Archy [21]

Answer:

V = 4.81 V

Explanation:

As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_{rint} = I* r_{int}$

The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_{b} - V_{rint} = 5.03 V = 6.0 V - 0.383 A* r_{int}$

We can solve for rint, as follows:

$r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega$

When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_{b} - V_{rint} = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V$

As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

5 0

3 years ago

Which formulas have been correctly rearranged to solve for radius? Check all that apply. r = GM central/v^2 r =fcm/v^2 r =ac/v^2

jek_recluse [69]

The orbital radius is: $r=\frac{GM}{v^2}$

Explanation:

The problem is asking to find the radius of the orbit of a satellite around a planet, given the orbital speed of the satellite.

For a satellite in orbit around a planet, the gravitational force provides the required centripetal force to keep it in circular motion, therefore we can write:

$\frac{GMm}{r^2}=m\frac{v^2}{r}$