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Alex777 [14]
3 years ago
5

41. What effects the efficiency of a pulley?

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
6 0

Answer:

e. only a and b

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The four best metallic conductors are: A. gold B. cadmium C. alloys D. copper E. silver F. sodium G. aluminum H. iron I have to
RoseWind [281]

Answer:

Explanation:

I can give you four for certain.

Copper

Gold

Silver

Aluminum

7 0
3 years ago
Just as in a bar magnet, the magnetic field of a wire with a current forms a cylinder shape around the wire.
Ne4ueva [31]
Yes, the magnetic field of a wire carrying a current forms a cylinder shape around the wire, but that's not similar to the field of a bar magnet.
7 0
3 years ago
A group of organisms that is able to interbreed is called a
uranmaximum [27]

Answer:

D species

Explanation:

According to the most widely used species definition, the biological species concept, a species is a group of organisms that can potentially interbreed, or mate, with one another to produce viable, fertile offspring. In this definition, members of the same species must have the potential to interbreed.

8 0
3 years ago
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
Grace [21]

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S  

2) t = 1 mu S

3) t = 50 mu S

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

4 0
3 years ago
Does anyone know delta and star math ?
Goshia [24]
Google it and click on the first website. 
7 0
4 years ago
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