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3 years ago

Do transistors amplfy a.c or d.c

Physics

2 answers:

Papessa [141]3 years ago

7 0

Answer:

Yes,by using direct coupling(transistors) amplifier we will amplify low frequency (DC) signals. A transistor is certainly capable of taking a small-current input signal and controlling a high-current output at the same voltage, thereby amplifying the power of the input signal whether it's AC or DC.

Can you give me brainliest?please?

Gnoma [55]3 years ago

5 0

Explanation:

thet amplify DC, because of the voltage ( small current input signal)

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What do you think will happen to Charlie now that he is smart? Explain.

postnew [5]

.... I don’t know but, he will be able to make smarter choices, he will be able to think before he does something, honestly don’t know

7 0

2 years ago

PLEASE HELP!!!!!!! MT TIME FOR MY TEST IS ALMOST OVER!!!

pav-90 [236]

Answer:

50 N

Explanation:

Let the natural length of the spring = L

100 = k(40 - L) (1)

200 = k(60 - L) (2)

(2)/(1): 2 = (60 - L)/(40 - L)

60 - L = 2(40 - L)

60 - L = 80 - 2L

2L - L = 80 - 60

L = 20

Sub it into (1):

100 = k(40 - 20) = 20k

k = 100/20 = 5 N/in

Now

X = k(30 - L) = 5(30 - 20) = 50 N

3 0

2 years ago

A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular t

Natasha_Volkova [10]

Answer:

v=12.5 i + 12.5 j m/s

Explanation:

Given that

m₁=m₂ = m

m₃ = 2 m

Given that speed of the two pieces

u₁=- 25 j m/s

u₂ =- 25 i m/s

Lets take the speed of the third mass = v m/s

From linear momentum conservation

Pi= Pf

0 = m₁u₁+m₂u₂ + m₃ v

0 = -25 j m - 25 i m + 2 m v

2 v=25 j + 25 i m/s

v=12.5 i + 12.5 j m/s

Therefore the speed of the third mass will be v=12.5 i + 12.5 j m/s

4 0

3 years ago

A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$