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ivanzaharov [21]
3 years ago
15

When a pendulum is at the position all the way to the left when it is swinging (at the top of the arc), what is true of the kine

tic and potential energy? A) The potential energy is zero, and the kinetic energy is maximized. B) The kinetic energy is zero, and the potential energy is maximized. C) Both the kinetic and potential energy are zero. D) Both the kinetic and potential energy are at their maximum. D) There is no energy at this position.
Physics
1 answer:
JulsSmile [24]3 years ago
7 0
Choice - B is the correct one.

At the top of the arc, at one end of the swing:
-- it's not going to get any higher, so the potential energy is maximum
-- it stops moving for an instant, so the kinetic energy is zero

At the bottom of the arc, in the center of the swing:
-- it's not going to get any lower, so the potential energy is minimum
-- it's not going to move any faster, so the kinetic energy is maximum
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Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and
vova2212 [387]

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}

Here

P = Pressure

T = Temperature

\gamma =The ratio of specific heats. For air normally is 1.4.

Our values are given as,

P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K

Therefore replacing we have,

(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}

(\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}

Solving for P_2,

P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}

P_2 = 44.15Lbf/in^2

Therefore the maximum theoretical pressure at the exit is 44.15Lbf/in^2

5 0
4 years ago
What is the speed of a car that travels 48 km in 2 hours?
ankoles [38]
Speed=distance/time

distance =48km
time=2 hours

speed=48/2
=24km/h


speed=24km/h

FORMULAS

speed = distance/time
time = distance/speed
distance = speed×time

3 0
4 years ago
A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x, t) = (5
barxatty [35]

Answer:

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t]y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t], where the origin is at the left end of the string, the x-axis is along the string, and the y-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation y(x, t) for this string if it were vibrating in its eighth harmonic?

4 0
3 years ago
A object with a mass of 1.5 kg is lifted from the ground to a height of 0.22 m what is the objects potential energy
svet-max [94.6K]

Answer:

<h2>3.3 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 1.5 × 10 × 0.22

We have the final answer as

<h3>3.3 J</h3>

Hope this helps you

5 0
3 years ago
During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel? Use the formula =1/2 *
marta [7]
Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s

To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s

Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m

This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.

But anyways, good luck!



4 0
3 years ago
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