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OleMash [197]
3 years ago
6

An example of ________ would be the horizontal distance a soccer ball moves between being kicked and touching the ground again.

Physics
1 answer:
Marina CMI [18]3 years ago
4 0

Answer:

Range.

Explanation:

Range in projectile motion can be defined as the horizontal distance traveled by an object.

An example of range would be the horizontal distance a soccer ball moves between being kicked and touching the ground again.

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Twice as fast, so multiply the kinetic energy of the slower plane by two.
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Which does a reference point provide? Check all that apply.
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A reference point is a point in geometry or general which is used to determine or refer to another point. In reference to this point, measurements can be done. Like, a reference point in space is used to measure distance or displacement to another point. A reference point in time is used to measure the time interval of an event. A reference in space-time can be used to measure the speed or detect object's motion. Hence, the following points apply:

→A position from which to measure future distance  

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A reference point provides a way/method to infer information about the object.


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How would seasons on earth be different if the earth was not tilted on its axis
MakcuM [25]

If the Earth's axis were 'straight' ... pefectly perpendicular to the ecliptic
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Please Help <br><br> Conservation of energy
Sonja [21]

Answer:

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Explanation:

8 0
3 years ago
A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn
EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, C=5.00 \mu F, the charged stored on the capacitor at the end of the process is

Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

V=\frac{Q}{C}

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

C=\frac{\epsilon_0 \epsilon_r A}{d}

where:

\epsilon_0 is the permittivity of free space

\epsilon_r is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: C'=\frac{C}{2}. The new voltage across the plate is given by

V'=\frac{Q}{C'}

and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V

So, the new voltage is

V'=2 (12.0 V)=24.0 V

(c) (ii) 3.0 V

The area of each plate of the capacitor is given by:

A=\pi r^2

where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be

A'=\pi (2r)^2 = 4 \pi r^2 = 4A

While the separation between the plate was unchanged (d); so, the new capacitance will be

C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C

So, the capacitance has increased by a factor 4; therefore, the new voltage is

V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}

which means

V'=\frac{12.0 V}{4}=3.0 V

3 0
3 years ago
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