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lilavasa [31]
3 years ago
10

A 0.5 kg object is whirled on the end of a string that is 1.2 m long at a speed of 7.5 m/s. Calculate the angular momentum of th

e
object.
(Round your response to the nearest tenth, if necessary, and include the base unit abbreviation with your answer.)
Physics
1 answer:
WINSTONCH [101]3 years ago
3 0

Answer:

4.5kg m^2 s−1

Explanation:

Step one:

the expression for angular momentum is given as

L = mvr

m = mass  = 0.5kg

v = velocity  = 7.5m/s

r = radius= 1.2m

Required

L = angular momentum

Step two:

substitute into the expression

L=0.5*7.5*1.2

L=4.5kg m^2 s−1

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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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Three examples that prove newton's law of motion​
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If you have a skateboard and you skate into a tree on accident the same amount of force you put onto that tree when you was on the skateboard will come back at you when you bounce back
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3 years ago
On a hot day, the temperature of a 65,000-L swimming pool increases by 1.20°C. What is the net heat transfer during this heating
vichka [17]

Answer:

326149.2 KJ

Explanation:

The heat transfer toward and object that suffered an increase in temperature can be calculated using the expression:

Q = m*cv*ΔT

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A 65000 L swimming pool will have a mass of:

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The specific heat capacity at constant volume of water is equal to 4.1814 KJ/KgC.

We replace the data and get:

Q = m*cv*ΔT = 65000 kg * 4.1814 KJ/KgC * 1.2°C = 326149.2 KJ

3 0
3 years ago
A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
Rasek [7]

Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

4 0
3 years ago
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cluponka [151]

Answer:

Momentum is always conserved, and kinetic energy may be conserved.

Explanation:

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If the body separates and move with a common velocity after collision, the collision is elastic but if they sticks together after collision, the collision is inelastic.

Either ways the momentum of the bodies are always conserved since they will always move with a common velocity after collision but their kinetic energy may or may not be conserved after collision, it all depends whether they separates or stick together after collision and since we are not told in question whether or not they separate, we can conclude that their kinetic energy "may" be conserved.

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