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3 years ago

Which of the following is a metric unit of mass? Question 2 options: kilogram pound liter ampere

Physics

2 answers:

denpristay [2]3 years ago

6 0

kilogram is the correct answer

coldgirl [10]3 years ago

6 0

Kilogram is a metric unit of mass.

Answer: Option A

<u>Explanation:</u>

Mass is used to denote the weight of the object hence it is represented in Kilograms. Liters are used to represent the quantity of liquids in weight, pounds are another unit for weight.

Ampere is the SI Unit for current passing through a conductor. Hence Option A is valid and B, C, D are irrelevant. Metric system is used as a common platform for engineers and scientists.

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A bug crawls 2.25 m along the base of a wall. Upon reaching a corner, the bugs direction of travel changes from south to west. T

denis-greek [22]

Answer:

The magnitude of the bugs displacement is 3.87 m

Explanation:

An illustrative diagram for the scenario is given in the attachment below.

In the diagram, the bug's displacement is given by x. The diagram shows a right angle triangle with x as the hypotenuse. We can determine x from the Pythagorean theorem which states that " the square of the hypotenuse equals sum of squares of the other two sides". That is

x² = 2.25² + 3.15²

x² = 5.0625 + 9.9225

x² = 14.985

x = √14.985

x = 3.87 m

Hence, the magnitude of the bugs displacement is 3.87 m.

8 0

3 years ago

Imagine that an uncharged pith ball is brought into the electrostatic field of a charged rod. The side of the pith ball closest

nataly862011 [7]

As per the question the uncharged pith ball is taken towards the electrostatic field of a charged rod.

It is told that the side of the pith ball is closest to the negatively charged end of the rod .

The pith ball was initially neutral.The electrons in the atom were oriented in their respective orbits.

Whenever it is taken towards the negative end of the rod,the electrons present in the ball will not stick to their orbit .Basically the closest electrons will suffer electrostatic repulsive force and will come out their orbits.They will move towards the other end of the pith ball.Hence the closest side of the pith ball will acquire positive charges and other side will be negative.

The bound charges nearest to the charged rod is called induced positive charges and bound negative charges present at the other end are called free bound charges.The induced charges and free charges are equal in amount.Hence the pith ball is neutral.

Hence from above we see that only option B is right.

5 0

3 years ago

Read 2 more answers

Which spacecraft carried the first Spacelab? -Columbia -Challenger -Discovery -Atlantis

Nat2105 [25]

The Atlantis spacecraft carried the first space lab.

Answer: Option 4

<u>Explanation: </u>

The first space lab is named as ATLAS 1 which is the abbreviation of Atmospheric Laboratory for Applications and Science. It is a short space lab set up in space to observe the atmospheric changes and other scientific experiments in the outer atmosphere of Earth from space.

It contains hi-tech instruments and facilities. It was a part of Phase I of NASA’s mission to planet Earth. This helped in better understanding of Earth’s outer and inner atmosphere. So, the spacecraft used to carry the ATLAS 1 is named as Atlantis.

4 0

3 years ago

Find the voltage gain vO/vS and current gain iO/iX for the circuit for g = 0.0025 S. Then, for vS = 4 V, find the power supplied

Lelechka [254]

Answer:

Incomplete question, no circuit diagram.

Check attachment for further explanation and circuit diagram

Explanation:

We want to find Vo/Vs and Io/Ix and also power delivered to the 2 kΩ resistor

Given that,

g= 0.0025 S (i.e conductance )

Vs= 4 V

R1 = 1 kΩ = 1000 Ω

R2 = 3 kΩ = 3000 Ω

R3 = 10 kΩ = 10000 Ω

R4 = 500Ω = 0.5 kΩ

R5 = 2 kΩ = 2000 Ω

a. At Loop 1: let use voltage divider rule to get Vx

Then, Vx = R2/(R1+R2) • Vs

Vx=3/(1+3) •Vs

Vx=¾ Vs.

The small signal current is given as

Is=g•Vx, since Vx=¾Vs

Then, Is= 0.025×¾Vs

Is=3/1600 Vs. Equation 1

Note: Ressistor R4 and R5 are in series, then the equivalent resistance of R4 and R5 is given as

Req= R4+R5

Req=2+0.5=2.5 kΩ

So, using current divider rule between R3 and the equivalent resistance of R4 and R5.

Therefore, Io= R3/(R3+Req) • Is

Io= R3/(R3+Req) • Is equation 2

Note : using ohms law on resistor R5,

V=iR. , R5=2 kΩ

Vo=IoR5

Vo=2Io

Io=Vo/2. Equation 3

Substitute equation 1 and 3 into 2

Io= R3/(R3+Req) • Is

½Vo = 10/(10+2.5) • 3/1600 Vs

½Vo = 10/12.5 • 3/1600 Vs

½Vo = 3/2000 Vs

Vo/Vs = 3/2000 × 2

Vo/Vs = 1 / 1000

The voltage output gain is

Vo/Vs = 1 / 1000

b. From equation 2

Io= R3/(R3+Req) • Is

Also, applying ohms law to resistor R2,

Vx = Ix• R2, R2=3kΩ

Vx = 3•Ix

Given that, Is= g•Vx

Is=0.0025(3•Ix)

Is= 3/400 Ix

Then, Io= R3/(R3+Req) • Is

Io= 10/(10+2.5) • 3/400 Ix

Io= 10/12.5 • 3/400 Ix

Io= 3/500 Ix

Io/Ix= 3/500

The current gain is

Io/Ix= 3/500

c. Output power

Power is given as

P=I²R

Then, output power at Resistor 5 is

Po = Io²•R5

R5=2000 Ω

From loop 1: using KVL, sum of voltage in a loop is zero

-Vs+1000Ix+3000Ix=0

4000Ix=Vs

Since Vs=4

Then, 4000Ix=4

Ix =4/4000

Ix = 1/1000 A

Since, Io/Ix = 3/500

Then, Io = 3/500 • Ix

Io=3/500 × 1/1000

Io= 6×10^-6 A

Therefore,

Po=Io²•R5. ,R5=2000

Po= (6×10^-6l² × 2000

Po=7.2×10^-8 W

Po=72×10^-9 W

Po=72 nW

The output power at resistor R5 is

72 nW

6 0

3 years ago

The half-life of element X is 20 years. If there are 48 g initially a) How much is there after 80 years

Gre4nikov [31]

Answer:

After 80 years there will be 3 g of element X remaining

Explanation:

Given;

the half life of element X = 20 year

initial mass of element X = 48 g

a) How much is there after 80 years

0 year --------------------------> 48 g

20 years -----------------------> (48g / 2) = 24 g

40 years ------------------------> 12 g

60 years ------------------------> 6 g

80 years --------------------------> 3 g

Therefore, after 80 years there will be 3 g of element X remaining.

3 0

4 years ago