A pendulum takes 10 seconds to swing through 2 complete cycles.a. How long does it take to complete one cycle?
b. What is its period?
c. What is its frequency?

Answers

Answer 1
Answer:

Answer:

a. 5 seconds

b. 5 seconds

c. 0.2 Hz

Explanation:

a) For 2 complete cycles, time taken is 10 s.

Thus, for 1 complete cycle, the pendulum would take = 10 s÷ 2 = 5 s

b) The time period of a pendulum can be defined as time to complete 1 cycle/oscillation. This is calculated in part a. Thus, time period of the pendulum is 5 s.

c) The frequency of a pendulum can be define number of cycles/oscillations in 1 s. It takes 5 s to complete 1 cycle.

In 1 s, number of cycles = (1 ÷ 5) s⁻¹ = 0.2 Hz.


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The sentence Walking the wire, my dog waited for the squirrel to fall is an example of _____.split infinitive
parallel structure
dangling participle
none of these

Answers

The answer is dangling participle. The dog wasn't walking the wire, the squirrel was. It is modifying the wrong noun.

Answer:

dangling participle

Explanation:

A constant net torque is applied to a rotating object. Which of the following best describes the object's motion?a. The object will rotate with increasing angular acceleration.
b. The object will have a decreasing moment of inertia.
c. The object will rotate with constant angular velocity.
d. The object will have an increasing moment of inertia.
e. The object will rotate with decreasing angular acceleration.
f. The object will rotate with constant angular acceleration.

Answers

Answer:

f. The object will rotate with constant angular acceleration.

Explanation:

We know that

τ = I α

τ=Torque

I=Moment of inertia of the object

α =Angular acceleration

The moment of inertia is constant for an object and it does not depends on the applied torque.

If applied torque is constant then the angular acceleration of the object will be constant because moment of inertia is constant.

If angular speed is constant then the angular acceleration will,be zero.But in this case object is having angular acceleration that is why it have varying angular speed.

Therefore option "f" is correct.

Question #4 Multiple Choice Which of the following is a similarity between an industrial metrologist and a legal metrologist? Both are employed by private companies Both ensure confidence in the quality of measurements Both ensure adequate functioning of measurement instruments Both develop measurement standards based on research​

Answers

Answer:

choice B

Explanation:i took the test

Answer:

ITS D

Explanation:

if its wrong i will comeback and re do it

Venus's average distance from the sun is 0.72 AU and Saturn's is 9.54 AU. Calculate the orbital velocity of Venus and Saturn around the sun. (The mass of the sun is 1.99 X 10^30 KG. an AU is 1.50 X 10^11 M.

Answers

The orbital velocity of Venus and Saturn around the sun is 8.854 x 10^8.

What is orbital velocity?

Orbital velocity is the speed required to achieve orbit around a celestial body, such as a planet or a star.

So you can compute the force of gravity on the planet

GMm/r^2

or

(6.674 * 10^(-11))(1.99 * 10^(30))m / d^2(1.5 * 10^(11))^2

Where d is the average distance of the planet from the sun and m is the mass of the planet (I will keep these as symbols so doing Saturn and Venus will be simple substitutions into one formula in the end).

Now realize that if they are in a stable circular orbit, then this force must provide the necessary centripetal force mv^2/r or mv^2 / d(1.5 * 10^(11))

So we get:

(6.674 * 10^(-11))(1.99 * 10^(30))m / d^2(1.5 * 10^(11))^2 = mv^2 / d(1.5 * 10^(11))

The m's cancel out as does one

1/d(1.5 * 10^(11))(6.674 * 10^(-11))(1.99 * 10^(30)) / d(1.5 * 10^(11)) = v^2

Evaluating and square rooting yeilds:

v = √(((8.854 * 10^8)/d))

Now plug in 0.72 for venus and 9.54 for saturn

Venus = 35067.39 m/s

Saturn = 9633.75 m/s

Some precision was lost rounding to 8.854 x 10⁸, so if more precision is required type the whole thing from before into a calculator. Anyways the process is sound.

Learn more about orbital velocity here:

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So you can compute the force of gravity on the planet GMm/r^2
or (6.674 X 10^−11)(1.99 X 10^30)m / d^2(1.5 X 10^11)^2
Where d is the average distance of the planet form the sun and m is the mass of the planet (I will keep these as symbols so doing Saturn and Venus will be simple substitution into one formula in the end).

Now realize that if they are in stable circular orbit, then this force must provide the necessary centripetal force mv^2/r or mv^2 / d(1.5 X 10^11)

So we get:

(6.674 X 10^−11)(1.99 X 10^30)m / d^2(1.5 X 10^11)^2 = mv^2 / d(1.5 X 10^11)
The m's cancel out as does one 1/d(1.5 X 10^11)
(6.674 X 10^−11)(1.99 X 10^30) / d(1.5 X 10^11) = v^2
Evaluating and square rooting yeilds:
v = sqrt((8.854 X 10^8)/d)

Now plug in 0.72 for venus and 9.54 for saturn

Venus = 35067.39 m/s
Saturn = 9633.75 m/s

Some precision was lost rounding to 8.854 X 10^8, so if more presion is required just type the whole thing from before into a calculator.  Anyways the process is sound

Two people pull as hard as they can on horizontal ropes attached to a boat that has a mass of 200 kg. If they pull in the same direction, the boat has an acceleration of 1.42 m/s2 to the right. If they pull in opposite directions, the boat has an acceleration of 0.509 m/s2 to the left. What is the magnitude of the force each person exerts on the boat

Answers

Answer: 192.9N and 91.1N

Explanation:

Let the force exerted by one person = X

Let the force exerted by the other person = y

Then, we can create equations

When both persons pull in the same direction, then

X + Y = 200 * 1.42

When both persons pull I'm opposite directions, then

X - Y = 200 * 0.509

Now we have

X + Y = 284

X - Y = 101.8

When we add the 2 equations together, we get

2X = 385.8

X = 385.8/2

X = 192.9

If x = 192.9, then, y = 284 - x

Y = 91.1

Therefore, one person exerts 192.9N, and the other exerts 91.1N

What are air pressure humidity clouds and temperature called

Answers

Answer:

Elements of weather

Explanation:

Weather can be defined as the atmospheric conditions of a particular area over a short period of time.

Air pressure, humidity, clouds and temperature are all called elements of weather. The other elements of weather are;

- Wind

- Precipitation.

Air pressure, humidity, clouds, and temperature are all part of weather and climate studies. They represent the weight of air above a point on the earth's surface, the amount of water vapor in the atmosphere, collections of airborne water droplets or ice crystals, and the heat energy in the atmosphere, respectively.

Air pressure, humidity, clouds, and temperature are all components of weather and climate studies, key elements of geography and meteorology. Air pressure is the force exerted by the weight of air above any given point on Earth's surface. Humidity refers to the amount of water vapor in the atmosphere. Clouds are collections of water droplets or ice crystals that form in the Earth's atmosphere when water vapor condenses. Temperature is a measure of the amount of heat energy in the atmosphere, usually expressed in degrees Celsius or Fahrenheit. The combination of all these factors form different weather patterns and influence the Earth's overall climate.

Learn more about Weather and Climate Components here:

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