20 examples of scalar quantity​

Answers

Answer 1
Answer:

Answer:

Length

Time

Mass

Temperature

Energy

Direct Current (DC)

Frequency

Volume

Speed

Amount of substance

Luminous Intensity

Density

Concentration

Refractive Index

Work

Pressure

Power

Charge

Electric Potential

Entropy


Related Questions

Part A A microphone is located on the line connecting two speakers hat are 0 932 m apart and oscillating in phase. The microphone is 2 83 m from the midpoint of the two speakers What are the lowest two trequencies that produce an interflerence maximum at the microphone's location? Enter your answers numerically separated by a comma
Please give the answer
Which two processes allow water to enter the atmosphere
At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room temperature is 20° C.)
If R = 12 cm, M = 430 g, and m = 60 g , find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.)

A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to move it 4.00 m to the side.(a) What is the magnitude of F when the crate is in this final position? During the crate's displacement, what are
(b) the total work done on it,
(c) the work done by the gravitational force on the crate, and
(d) the work done by the pull on the crate from the rope?
(e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force F does on the crate.
(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?

Answers

Answer:

(a) magnitude of F = 797 N

(b)the total work done  W = 0

(c)work done by the gravitational force =  -1.55 kJ

(d)the work done by the pull  = 0

(e) work your force F does on the crate = 1.55 kJ

Explanation:

Given:

Mass of the crate, m =  220 kg

Length of the rope, L = 14.0m

Distance, d =  4.00m

(a) What is the magnitude of F when the crate is in this final position

Let us first determine vertical angle as follows

=>Sin \theta = (d )/(L)

=> \theta = Sin^(-1) (d)/(L) =

Now substituting thje values

=> \theta = Sin^(-1) (4)/(12) =

=> \theta = Sin^(-1) (1)/(3)

=> \theta = Sin^(-1)(0.333)

=> \theta = 19.5^(\circ)

Now the tension in the string resolve into components

The vertical component supports the weight

=>Tcos\theta = mg

=>T = (mg)/(cos\theta)

=>T = (230 * 9.8 )/(cos(19.5))

=>T = (2254 )/(cos(19.5))

=>T = (2254 )/(0.9426)

=>T =2391N

Therefore the horizontal force

F = TSin(19.5)

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

c) The work done by the gravitational force on the crate

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values                                                            

= -230 * 9.8* 12 ( 1 - cos(19.5) )

= -230 * 9.8* 12 ( 1 - 0.9426) )

= -230 * 9.8* 12 (0.0574)

= -230 * 9.8* 0.6888

=  -230 * 6.750

= -1552.55 J

The work done by gravity = -1.55 kJ

d) the work done by the pull on the crate from the rope

Since the pull  is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg  

WF = -(-1.55)

WF = 1.55 kJ

(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)

Here the work done by force is not equal to F*d  

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)      

A novice scuba diver practicing in a swimming pool takes enough air from his tank to fully expand his lungs before abandoning the tank at depth L and swimming to the surface. He ignores instructions and fails to exhale during his ascent. When he reaches the surface, the difference between the external pressure on him and the air pressure in his lungs is 9.3 kPa. From what depth does he start?

Answers

Answer:

11.625 m.

Explanation:

Difference of pressure will be due to hydro-static pressure due to column of water of height L.

Pressure of water column  = L d g , where L is depth ,

d is density of water = 1000kg /m³

g = 9.8 ms²

Pressure difference = 9.3 kPa = 9300 Pa

So Ldg = 9300

L X 1000 X 0.8 =9300

800 L = 9300

L = 11.625 m.

What force causes oppositely charged particles to attract each other? A. Magnetic force B. Compression
C. Electrical Force D. Gravity

Answers

Answer:

It is electrical force

Explanation:

i got it wrong on A P E X  with magnetic hope this helps!

Answer:

The electromagnetic force.

Part of your electrical load is a 60-W light that is on continuously. By what percentage can your energy consumption be reduced by turning this light off

Answers

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete question in the attached file.

The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.

\to (\$60)/((30 \ days)/(24\ hours)) = \$0.08 / kwh.

Thus, (\$0.08)/(\$0.12) = 0.694 \ kW *  0.694 \ kW  * 1000 = 694 \ W.

The electricity used is continuously 694W over 30 days.

If just resistor loads (no reagents) were assumed,

\to I = (P)/(V)= (694\ W)/(120\ V) = 5.78\ A

Energy usage reduction percentage = ((60\ W)/(694\ W) * 100\%)

This bulb accounts for 8.64\% of the energy used, hence it saves when you switch it off.

Which of the following is not a risk associated with using legal drugs without medical supervision

Answers

Answer:

paying too much on the black market instead of getting a prescription

Explanation:

i just took the quiz

Answer:

Paying too much on the black market instead of getting a prescription

Explanation:

The rest of the options are risks associated with using legal drugs without medical supervision.

Solenoid A has total number of turns N length L and diameter D. Solenoid B has total number of turns 2N, length 2L and diameter 2D. Inductance of solenoid A is 8 times inductance of solenoid B
1/4 of inductance of solenoid B
same as inductance of solenoid B
1/8 of inductance of solenoid B
four times of inductance of solenoid B

Answers

Answer:

∴Inductance of solenoid A is \frac18 of inductance of solenoid B.

Explanation:

Inductance of a solenoid is

L=N\frac\phi I

 =N(B.A)/(I)

 =N(\mu_0NI)/(l.I)A

 =(\mu_0N^2A)/(l)

 =(\mu_0N^2)/(l).\pi(\frac d2)^2

 =\mu_0\pi(N^2d^2)/(4l)

N= number of turns

l = length of the solenoid

d= diameter of the solenoid

A=cross section area

B=magnetic induction

\phi = magnetic flux

I= Current

Given that, Solenoid A has total number of turns N, length L and diameter D

The inductance of solenoid A is

=\mu_0\pi(N^2D^2)/(4L)

Solenoid B has total number of turns 2N, length  2L and diameter 2D

The inductance of solenoid B is

=\mu_0\pi((2N)^2(2D)^2)/(4.2L)

=\mu_0\pi(16 N^2D^2)/(4.2L)

Therefore,

\frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=(\mu_0\pi(N^2D^2)/(4L))/(\mu_0\pi(16 N^2D^2)/(4.2L))

\Rightarrow \frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac18

\Rightarrow  {\textrm{Inductance of A}}=\frac18* {\textrm{Inductance of B}}

∴Inductance of solenoid A is \frac18 of inductance of solenoid B.

Hi there!

We can begin by calculating the inductance of a solenoid.

Recall:
L = (\Phi _B)/(i)

L = Inductance (H)
φ = Magnetic Flux (Wb)

i = Current (A)

We can solve for the inductance of a solenoid. We know that its magnetic field is equivalent to:
B = \mu _0 (N)/(L)i

And that the magnetic flux is equivalent to:
\Phi _B = \int B \cdot dA =  B \cdot A

Thus, the magnetic flux is equivalent to:
\Phi _B = \mu _0 (N)/(L)iA

The area for the solenoid is the # of loops multiplied by the cross-section area, so:
A_(total)= N * A

\Phi _B = \mu _0 (N^2)/(L)iA

Using this equation, we can find how it would change if the given parameters are altered:
\Phi_B ' = \mu_0 ((2N)^2)/(2L) i * 4A

**The area will quadruple since a circle's area is 2-D, and you are doubling its diameter.

\Phi'_B = (4)/(2) * 4(\mu_0 (N)/(L)iA) = 8\mu_0 (N)/(L)iA

Thus, Solenoid B is 8 times as large as Solenoid A.

Solenoid A is 1/8 of the inductance of solenoid B.