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noname [10]
2 years ago
9

Betty weighs 420 n and she is sitting on a playground swing seat that hangs 0.40 m above the ground. tom pulls the swing back an

d releases it from rest when the seat is 1.0 m above the ground. how fast is betty moving when the swing passes through its lowest position
Physics
1 answer:
Svet_ta [14]2 years ago
8 0
Weight = mass * gravity
420 = mass * 9.8
mass of Betty = 42.857 kg

Difference in height = 1 - 0.45 = 0.55 meters

Total energy = Kinetic energy + potential energy

At the highest point, the kinetic energy is zero while the potential energy is maximum, therefore, we can get the total energy as follows:
Total energy = 0 + mgh
Total energy = 42.857*9.8*0.55 = 231 Joules

At the lowest point, the potential energy is zero while the kinetic energy is maximum. Therefore:
Total energy = 0.5 * m * (v)^2 + 0
231 = 0.5 * (42.857) * (velocity)^2
(velocity)^2 = 10.78
velocity = 3.28 meters/sec
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A 15.0-gram lead ball at 25.0°C was heated with 40.5 joules of heat. Given the specific heat of lead is 0.128 J/g∙°C, what is th
mr Goodwill [35]

Answer:

T=4985.5^{\circ}K

Explanation:

The equation that relates heat Q with the temperature change T-T_0 of a substance of mass <em>m </em>and specific heat <em>c </em>is Q=mc(T-T_0).

We want to calculate the final temperature <em>T, </em>so we have:

T=\frac{Q}{mc}+T_0

Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add 273^{\circ} to our temperature in ^{\circ}C to have it in ^{\circ}K as it must be):

T=\frac{Q}{mc}+T_0=\frac{40.5J}{(15g)(0.128J/g^{\circ}C)}+(298^{\circ}K)=4985.5^{\circ}K

3 0
1 year ago
where again p is the phonon momentum, E is the photon energy and c is the speed of light. When you divide the photon energy foun
creativ13 [48]

Answer:

Yes

Explanation:

p = momentum of photon

E = energy of photon

c = velocity of light

Units of p = kg m /s

Units of E = kg m^2 / s^2

Units of E / p = {kg m^2 / s^2} / {kg m /s} = m/s

It is the unit of speed, so by the division of energy to the momentum, we get the speed. yes it is correct.

8 0
2 years ago
Juan and Kuri are on a carousel. Juan is closer to the center of the carousel than Kuri. Which statement describes their tangent
Licemer1 [7]

Answer:

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

Explanation:

Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution

As we know that the distance moved in one revolution is given as

d = 2\pi r

also the time period of revolution for both will remain same as they move with the time period of carousel

Now we can say that the speed is given as

v = \frac{2\pi r}{T}

so Juan will have less tangential speed. so correct answer will be

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

6 0
2 years ago
Read 2 more answers
American Football Field Uses A Field That Is 100.0 Yd Long, Whereas A Soccer Field Is 100.0m Long. Which Field Is Longer And By
postnew [5]
Note that
1 yd = 0.9144 m

Therefore,
The length of an American Football field is
(100 yds)*(09144 m/yd) = 91.44 m

Because the soccer field is 110 m long, its length exceeds the American Football Field by
100 - 91.44 = 8.56 m
or
(8.56/.9144) =  9.36 yd
This difference is equivalent to (8.56/91.44)*100 = 9.4%

Answer:
The Soccer Field is longer by
8.56 m, or
9.36 yd, or
9.4%
4 0
2 years ago
A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of th
icang [17]

To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

\theta= \theta_0 +\omega t

From our given data we know that,

\omega = 76\frac{rev}{min}

\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})

\omega = 7.958rad/s

Moreover we know that

\theta_0 = 0.47 rad

Therefore for time t=8.1s we have,

\theta= \theta_0+ \omega t

\theta= 0.47+(7.958)(8.1)

\theta = 64.9298rad

That number in revolution is:

\theta = 64.9298rad(\frac{1rev}{2\pi})

\theta = 15.108 Revolutions

Here, we see that there are 15 complete revolutions

And 0.108 revolutions i not complete, so the tunable rotation is

\theta_{net} = 0.108*2\pi=0.216\pi

Therefore the angle of the speck at a time 8.1s is 0.216\pi

4 0
2 years ago
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