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babunello [35]
3 years ago
15

The bob (weight) at the end of a pendulum has a mass of 0.3 kilograms. The bob is pulled to position B and allowed to swing. It

goes all the way to position C and swings back.
The potential energy of the bob at position B is >>1.47 joules. If the maximum height of the bob is 0.45 meters when it swings back, ___ joules of energy was transformed to thermal energy.
>>>>>Use g = 9.8 m/s2 and PE = m × g × h.<<<<<

It's NOT 1.323!!!

Physics
1 answer:
Ivahew [28]3 years ago
3 0

Answer:

0.147 J

Explanation:

The total energy that has been transformed into thermal energy is equal to the loss of gravitational potential energy between the initial situation (bob at h=0.5 m above the ground) and the final situation (bob back but at h=0.45 m above the ground).

Therefore, we have

E_{thermal}=\Delta U=mgh_1 - mgh_2 = mg(h_1 -h_2)

where

m = 0.3 kg is the mass of the bob

g = 9.8 m/s^2

h1 = 0.5 m is the initial height

h2 = 0.45 m is the final height

Substituting, we find the thermal energy

E_{thermal}=(0.3 kg)(9.8 m/s^2)(0.5 m-0.45 m)=0.147 J

Therefore, the energy transformed into thermal energy is 0.147 J.

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A proton and an electron are released from rest, with only the electrostatic force acting. Which of the following statements mus
lbvjy [14]

Answer:

their electrical potential energy decreases. True, because of the negative sign as the distance decreases, it becomes more negative.

The kinetic energy increases. True, the force is attractive as the distance decreases the force increases, so acceleration and therefore the speed

Explanation:

The only force between the proton and the electron is electric

      Fe = k q1q2 / r2

      Fe = - k e2 / r2

We can see that it is an attractive force (negative sign)

The electric power energy is

       U = k q1 q2 / r

       U = -k e2 / r

 

The kinetic energy is

       K = ½ m v2

With the expressions for each term we can analyze the sentences :

Their electric potential energy increases. False,

their electrical potential energy decreases. True, because of the negative sign as the distance decreases, it becomes more negative.

The kinetic energy increases. True, the force is attractive as the distance decreases the force increases, so acceleration and therefore the speed

Kinetic energy decreases False

The acceleration decreases.  False, as the force increases so does the acceleration

7 0
3 years ago
a 150 N force is used to pull a wooden box across a wooden surface at a constant velocity. what is the mass of the box?
IRINA_888 [86]

Answer:

The mass of the box:

m =  60 kg

Explanation:

Given:

F = 150 N

g = 10 m/s²

_________

m - ?

Coefficient of friction wood on wood:

μ = 0.25

Friction force:

F₁ = μ*m*g

Newton's Third Law:

F = F₁

F = μ*m*g

The mass of the box:

m = F / ( μ*g) = 150 / (0.25*10) =  60 kg

7 0
1 year ago
The major contributions of Maury included:
scoray [572]
The best and most correct answer among the choices provided by your question is the second choice.

<span>The major contributions of Maury included mapping the ocean bottom.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
6 0
3 years ago
The effect of itching power what’s the Independent variable and dependent and constants?
kenny6666 [7]
The independent variable is what type of powder is used, the dependent variable is how long itching lasts and the constant is the amount of powder used (50 grams)
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3 years ago
A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 rad/s2 until its angular speed is 1820 rev/min. (a) What
kati45 [8]

Explanation:

The given data is as follows.

           radius (r) = 3.25 cm,    \alpha = 11.6 rad/s^{2}

Now, we will calculate the tangential acceleration as follows.

          a_{tangential} = \alpha \times r

Putting the given values into the above formula as follows.

         a_{tangential} = \alpha \times r

                      = 11.6 rad/s^{2} \times 3.25 cm

                      = 37.7 rad cm/s^{2}

Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 rad cm/s^{2}.

6 0
3 years ago
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