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solmaris [256]
3 years ago
13

When a pendulum is at the midpoint of its oscillation, hanging straight down, which statement is true?

Physics
2 answers:
Svetach [21]3 years ago
5 0
When a pendulum is at the midpoint of its oscillation, hanging straight down ...

-- that's the fastest it's going to swing, so its kinetic energy is maximum;
and
-- that's the lowest it's going to get, so its potential energy is minimum.

'c' is your choice.
Misha Larkins [42]3 years ago
3 0

c. Kinetic energy is maximum and potential energy is minimum.

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A box of mass 12 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.42. What is th
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The maximum static force that can be applied is equal to the normal force*the frictional force. the normal force on the box is equal to mg since the floor is flat using 9.81m/s^2 for gravity 12kg*9.81m/s^2 = 118N multiplying the normal force by the frictional force you get a 118*.42= 49.6N so overcome the force of static friction on the box a minimum of 49.6N would need to be applied.
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2 years ago
5. The rate at which an organism uses energy, measured in humans at complete<br> rest.
yuradex [85]

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n. Abbr. BMR. The rate at which energy is used by an organism at complete rest, measured in humans by the heat given off per unit time, and expressed as the calories released per kilogram of body weight or per square meter of body surface per hour.

Explanation:

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2 years ago
A baseball of radius r = 5.2 cm is at room temperature T = 20.8 C. The baseball has emissivity of ε = 0.86 and the Stefan-Boltzm
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Answer:

P = 12.37 \frac{J}{s} = 12.37 Watts

Explanation:

Previous concepts

The Thermal radiation is one of "3 mechanisms who allows to bodies exchange energy".

The thermal radiation formula is given by:

\frac{P}{A} = \epsilon \sigma T^4

Where \sigma = 5.67 x10^{-8} \frac{J}{sm^2 K^4}

If we solve for P we got:

P = A \epsilon \sigma T^4

Since we have a baseball ball considered as a sphere the superficial area is given by:

A = 4\pi r^2

Solution to the problem

And if we replace this into our equation of P we got:

P = (4\pi r^2) \epsilon \sigma T^4

And we can reorder this like that:

P = 4 \epsilon \pi \sigma r^2 T^4

We can convert the radius to meters and we got:

r= 5.2 cm*\frac{1m}{100 cm}=0.052 m

Now we can convert the temperature to Kelvin and we got:

T = 20.8 +273.15 = 293.95 K

\epsilon = 0.86 the emissivity given

And now we can replace into the formula for P and we got:

P = 4*0.86*\pi *(5.67x10^{-8} \frac{J}{s m^2 K^4}) (0.052m)^2 (293.95 K)^4

P = 12.37 \frac{J}{s} = 12.37 Watts

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