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Elden [556K]
3 years ago
13

What is the answer ?

Physics
1 answer:
Fofino [41]3 years ago
7 0
P= mv
p=(.5)(20)
p=10 kg/ms
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Along some shorelines, incoming waves cause the water to simply rise and fall rather that form a surf zone. What does this tell
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Answer:

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Explanation:

Along some shorelines, incoming waves cause the water to simply rise and fall and not form surfs because of the steepness of the shorelines. Long period waves wont form surfs at steep shores because of the breaking and unstability of waves. Wave breaks at the shallow waters. The breaking of the waves depends upon H/L ratio.

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A(n) ______ is matter that is composed of one type of atom.<br><br><br><br><br>answerd::: An Element
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A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2
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Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S  

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1) 22.4 \mu C

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

C=\frac{Q_0}{V_0}

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

C=4.4\mu F = 4.4\cdot 10^{-6}F

V_0 = 5.1 V

Solving for Q_0,

Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) 20.0\mu C

To find the charge on the capacitor at any other time t, we use the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

where

Q_0 = 22.4 \mu C

t is the time

R=2.0 \Omega is the resistance

C=4.4\mu F is the capacitance

Therefore, at time t=1 \mu s, we have:

Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C

3) 0.08 \mu C

As before, we use again the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

However, here the time to consider is

t=50 \mu C

Substituting into the formula,

Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C

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