Ask question

Ask question

All categories

3 years ago

15

Technician A says the small base circuit of a transistor controls current flow. Technician B says the small emitter circuit cont

rols current flow. Who is right?

1 answer:

Amiraneli [1.4K]3 years ago

6 0

Answer:

A

Explanation:

You might be interested in

Technician A ay that acid rain doe the mot harm when it firt fall on a finih. Technician B ay that hard water potting i uually j

Tamiku [17]

Technician B is right say that hard water potting i usually jut a Surface problem that can be wahed off.

What do you mean by Hard water?

The amount of dissolved calcium and magnesium in the water determines its hardness. Calcium and magnesium are the main dissolved minerals in hard water. The last time you washed your hands, you might have actually felt the effects of hard water.

What do you mean by acid rain?

Any type of precipitation that contains acidic elements, such as sulfuric or nitric acid, that falls to the ground from the atmosphere in wet or dry forms is referred to as acid rain, also known as acid deposition. Rain, snow, fog, hail, and even acidic dust can fall under this category.

Some plants are sensitive to excessive moisture around their root zone, so it may be necessary to increase drainage when growing plants in pots. Additionally, standing water at the bottom of the pot can cause root rot.

Many university agriculture extension agencies have thoroughly debunked the old garden myth that adding rocks to the bottom of a pot will increase drainage.

Learn more about hard water click here:

brainly.com/question/28178305

#SPJ4

6 0

1 year ago

a) A total charge Q = 23.6 μC is deposited uniformly on the surface of a hollow sphere with radius R = 26.1 cm. Use ε0 = 8.85419

dusya [7]

Answer:

(a) E = 0 N/C

(b) E = 0 N/C

(c) E = 7.78 x10^5 N/C

Explanation:

We are given a hollow sphere with following parameters:

Q = total charge on its surface = 23.6 μC = 23.6 x 10^-6 C

R = radius of sphere = 26.1 cm = 0.261 m

Permittivity of free space = ε0 = 8.85419 X 10−12 C²/Nm²

The formula for the electric field intensity is:

E = (1/4πεo)(Q/r²)

where, r = the distance from center of sphere where the intensity is to be found.

(a)

At the center of the sphere r = 0. Also, there is no charge inside the sphere to produce an electric field. Thus the electric field at center is zero.

<u>E = 0 N/C</u>

(b)

Since, the distance R/2 from center lies inside the sphere. Therefore, the intensity at that point will be zero, due to absence of charge inside the sphere (q = 0 C).

<u>E = 0 N/C</u>

(c)

Since, the distance of 52.2 cm is outside the circle. So, now we use the formula to calculate the Electric Field:

E = (1/4πεo)[(23.6 x 10^-6 C)/(0.522m)²]

<u>E = 7.78 x10^5 N/C</u>

4 0

3 years ago

How is engine power expressed?

mars1129 [50]

Answer: Engine power is the power that an engine can put out. It can be expressed in power units, most commonly kilowatt, pferdestärke (metric horsepower), or horsepower.

Explanation: (I hope this helped!! ^^)

5 0

2 years ago

Read 2 more answers

Initially, a pump pressure of __________ pounds per square inch should be used to maintain a sprinkler or standpipe system.

jasenka [17]

<u>Answer:</u>

<u>of 150 pounds per square inch</u>

Explanation:

Note that the unit for measuring water pressure is called <u> pounds per square inch (psi)</u>

In the case of sprinklers and standpipe systems, a pressure <u>of 150 pounds per square inch</u> was used initially.

6 0

3 years ago

Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and

Pavlova-9 [17]

Answer:

Part A:

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

Explanation:

FP Instructions=50*106=5300

INT Instructions=110*106=11660

L/S Instructions=80*106=8480

Branch Instructions=16*106=1696

Calculating Execution Time:

Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

Execution Time= $\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}$

Execution Time= $2.7136*10^{-5}\ s$

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

$New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4$

New Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

8 0

3 years ago