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Jobisdone [24]
3 years ago
13

If this plastic cup is heated to its melting

Physics
1 answer:
Oduvanchick [21]3 years ago
5 0

Answer:

A. The particles will begin to move enough

that they slide past each other.

Explanation:

When the plastic cup is heated, the Kinetic energy of its particles starts increasing. As the temperature rises, the kinetic energy keeps increasing. With the increase of K.E, the particles start moving faster and faster. When the temperature finally reaches the melting point, the K.E of the molecules is enough to break the bonds and slide past each other.

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On a balanced seesaw, a boy three times as heavy as his partner sits
slega [8]

Answer:

1/3 the distance from the fulcrum

Explanation:

On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:

W_1 d_1 = W_2 d_2

where

W1 is the weight of the boy

d1 is its distance from the fulcrum

W2 is the weight of his partner

d2 is the distance of the partner from the fulcrum

In this problem, we know that the boy is three times as heavy as his partner, so

W_1 = 3 W_2

If we substitute this into the equation, we find:

(3 W_2) d_1 = W_2 d_2

and by simplifying:

3 d_1 = d_2\\d_1 = \frac{1}{3}d_2

which means that the boy sits at 1/3 the distance from the fulcrum.

8 0
3 years ago
A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation betw
TEA [102]

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. Q, the amount of charge stored in this capacitor, will stay the same.

The formula \displaystyle Q = C\, V relates the electric potential across a capacitor to:

  • Q, the charge stored in the capacitor, and
  • C, the capacitance of this capacitor.

While Q stays the same, moving the two plates apart could affect the potential V by changing the capacitance C of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

\displaystyle C = \frac{\epsilon\, A}{d},

where

  • \epsilon is the permittivity of the material between the two plates.
  • A is the area of each of the two plates.
  • d is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of \epsilon. Neither will that change the area of the two plates.

However, as d (the distance between the two plates) increases, the value of \displaystyle C = \frac{\epsilon\, A}{d} will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula \displaystyle Q = C\, V can be rewritten as:

V = \displaystyle \frac{Q}{C}.

The value of Q (charge stored in this capacitor) stays the same. As the value of C becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.  

3 0
3 years ago
Whats the speed of darkness?
lianna [129]
The same speed of light, the fastest possible

darkness is just the absence of light therefore once light leaves darkness returns. making darkness have the same speed of light
3 0
3 years ago
Using Rayleigh's criterion, what is the smallest separation between two pointlike objects that a person could clearly resolve at
atroni [7]

 Answer:

y = 52.44 10⁻⁶  m

Explanation:

It is Rayleigh's principle that two points are resolved if the maximum of the diffraction pattern of one matches the minimum the diffraction pattern of the other

Based on this principle we must find the angle of the first minimum of the diffraction expression

         a sin θ= m λ

The first minimum occurs for m = 1

       sin θ =  λ  / a

Now let's use trigonometry the object is a distance L = 0.205 m

        tan θ = y / L

Since the angles are very small, let's approximate

        tan θ = sin θ/cos θ = sin  θ

        sin θ = y / L

We substitute in the diffraction equation

         y / L =  λ  / a

         y =  λ  L / a

Let's calculate

        y = 550 10⁻⁹ 0.205 / 2.15 10⁻³

        y = 52.44 10⁻⁶ m

7 0
3 years ago
In a cup game if the teams have the same score at the end of the match, 30 minutes of ------- are played.
Alina [70]

Answer:

second lag

Explanation:

If in a cup game, a specified time limit is assigned to both teams to score high. If both teams are unable to score or if score of both the teams is equal then there is another second lag played where each team tries to score high. Even if in second lag both teams fail to score higher than other the last third lag is played or else game is declared draw.

5 0
2 years ago
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