Match the following terms or processors with the correct definition?

Physics

1 answer:

lidiya [134]4 years ago

5 0

Answers:
Power - Measure of energy delivered by the circuit per unit time

Transistor - Can be used as a switch or amplifier

Voltage - A measure of the electrical potential difference across or between two points

Ohm’s Law - For a given voltage, current and resistance are inversely proportional

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A 3.00 kg ball is dropped from the roof of a building 176.4 m high. While the ball is falling to Earth, a horizontal wind exerts

Katen [24]

Answer:

a.)How far from the building does the ball hit the ground?

72 m

b.)How long does it take to hit the ground?

6 s

c.)What is its speed when it hits the ground?

63.48 m/s

Explanation:

When an object is in free fall with no air resistance present or wind acting on it it only is under the influence of gravity. Gravity acts straight down and accelerates the object all the way to the ground. If there is air resistance or a wind force on the object it will have an acceleration that isn't 9.8 m/s^2

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4 years ago

A 2kg mass is initially at rest on a frictionless surface. A 45N force acts at an angle of 300

raketka [301]

Answer:

a) The work done by the force is 136.400 joules.

b) The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

Explanation:

The correct statement is shown below:

A 2-kg mass is initially at reat on a frictionless surface. A 45 N force acts at an angle of 30º to the horizontal for a distance of 3.5 meters:

a) Find the work done by the force.

b) Find the speed of the mass at the end of the 3.5 meters.

a) Given that a external constant force is acting on the mass on a frictionless surface, the work done by the force ( $W_{F}$ ), measured in joules, is:

$W_{F} = F\cdot \Delta s \cdot \cos \theta$ (1)

Where:

$F$ - External constant force exerted on the mass, measured in newtons.

$\Delta s$ - Horizontal travelled distance, measured in meters.

$\theta$ - Direction of the external force regarding the horizontal, measured in sexagesimal degrees.

If we know that $F = 45\,N$ , $\Delta s = 3.5\,m$ and $\theta = 30^{\circ}$ , then the work done by the force is:

$W_{F} = (45\,N)\cdot (3.5\,m)\cdot \cos 30^{\circ}$

$W_{F} = 136.400\,J$

The work done by the force is 136.400 joules.

b) The final speed of the mass at the end ot the 3.5 meter is calculated by means of the Work-Energy Theorem, which means that the work done on the mass is transformed into a translational kinetic energy. That is:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$ (2)