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3 years ago

Which are examples of a medium?

Physics

2 answers:

guajiro [1.7K]3 years ago

6 0

Answer:

An example? I mean if we're going for an example, I'd say middle, Or balanced.

Ne4ueva [31]3 years ago

4 0

Answer:

One example of a medium is air.

Explanation:

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Acceleration is often measured in what unit? A. newtons B. kilograms C. meters per second D. meters per second squared

spayn [35]

D. meters per second squared

3 0

4 years ago

Read 2 more answers

When AC Electric is producing 120 volts with 20 amps, What controls the the 120 volts at 60cycles a second?

lubasha [3.4K]

Answer:

Cycles per second is dependent on the construction of the alternator and the 120 volts is dependent upon the current and resistance in the circuit according to the ohms law.

Explanation:

We are given with AC of 120 volts, 20 amperes and 60 hertz frequency.

<u>According to the Ohm's law, we find its resistance:</u>

$R=\frac{V}{I}$

$R=\frac{120}{20}$

$R=6\ \Omega$

So, this 6 ohm resistance controls the current controls the magnitude of the AC current, while the frequency of the current remains constant and depends upon the construction and rotational speed of the armature of the alternator producing the current.

Here the value of frequency is the number of times the current changes its direction or the polarity in one second.

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4 years ago

Suppose the acceleration of the particle moving gin a circle of radius "r" with the uniform speed "v" is proportional to some po

garik1379 [7]

So i say the power or the radius cortex in my equinox when i look at this V is the determination of V

5 0

3 years ago

How long would it take an object to reach the ground from the top of a building that is 470 feet tall? Round to the nearest tent

Zinaida [17]

Answer:

It would take the object 5.4 s to reach the ground.

Explanation:

Hi there!

The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:

h = h0 + v0 · t + 1/2 · g · t²

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).

t = time

Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:

h = h0 + 1/2 · g · t²

We have to find the time at which h = 0:

0 = 470 ft - 1/2 · 32.2 ft/s² · t²

Solving for t:

-470 ft = -16.1 ft/s² · t²

-470 ft / -16.1 ft/s² = t²

t = 5.4 s

3 0

3 years ago

An object moves in a straight path. Its position x as a function of time t is presented by the equation x(t) = at – bt2+c, where

ankoles [38]

Answer:

The answer is below

Explanation:

Given that:

x(t) = at – bt2+c

a) x(t) = at – bt2+c

Substituting a = 1.4 m/s, b = 0.06 m/s2 and c =50 m gives:

x(t) = 1.4t - 0.06t² + 50

At t = 5, x(5) = 1.4(5) - 0.06(5)² + 50 = 55.5 m

At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

$v=\frac{x(5)-x(0)}{5-0}\\ \\v=\frac{55.5-50}{5-0}=1.1\\ \\v=1.1\ m/s$

b) x(t) = 1.4t - 0.06t² + 50

At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m

At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

$v=\frac{x(10)-x(0)}{10-0}\\ \\v=\frac{58-50}{10-0}=0.8\\ \\v=0.8\ m/s$

c) x(t) = 1.4t - 0.06t² + 50

At t = 15, x(5) = 1.4(15) - 0.06(15)² + 50 = 57.5 m

At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m

The average velocity (v) is given as:

$v=\frac{x(15)-x(10)}{15-10}\\ \\v=\frac{57.5-58}{15-10}=0.1\\ \\v=0.1\ m/s$

6 0

3 years ago