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2 years ago

Select the volume units that are greater than one liter.

Physics

2 answers:

AysviL [449]2 years ago

8 0

Answer: The correct answers are option (a)and(c).

Explanation:

1) In ,1 kilo Liter = 1,000 Liter

1 liter = 0.001 kilo Liter

2) In ,1 Mega Liter = 1,000,000

1 Liter = $10^{-6}$ Mega Liter

Where as milliliter, centiliter, deciliter and nano liters are smaller units that 1 liter

1 Liter = 1000 milliliter

1 Liter = 100 centiliter

1 Liter = 10 deciliter

1 Liter = $10^9$ nano liter

Hence, the correct answers are option (a)and(c).

Andreas93 [3]2 years ago

5 0

A.) kiloliter. 1 kiloliter = 1,000 liters
c.) megaliter. 1 megaliter = 1,000,000 liters

hope this helps

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Murrr4er [49]

Answer:C

Explanation:

It is given that hand strap moves from the vertical in the backward direction.

The direction of strap depends upon the acceleration of bus i.e. if bus is accelerating in forward direction then strap will move in backward direction and vice-versa.

The reason for moving backwards is due to the psuedo acting on strap which bends the strap in backward direction

angle of inclination is given by $\tan \theta =\frac{a}{g}$

where a=acceleration of bus

$\theta$ =inclination of strap from vertical

so we cannot conclude anything about the direction of the velocity of the bus

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2 years ago

A plane wall with constant properties is initially at a uniform temperature To. Suddenly, the surface at x = L is exposed to a c

Rzqust [24]

Answer:

The distribution is as depicted in the attached figure.

Explanation:

From the given data

The plane wall is initially with constant properties is initially at a uniform temperature, To.
Suddenly the surface x=L is exposed to convection process such that T∞>To.
The other surface x=0 is maintained at To

Uniform volumetric heating q' such that the steady state temperature exceeds T∞.

Assumptions which are valid are

There is only conduction in 1-D.
The system bears constant properties.
The volumetric heat generation is uniform

From the given data, the condition are as follows

Initial Condition

At t≤0

$T(x,0)=T_o$

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.

Boundary Conditions

At x=0

$T(0,t)=T_o$

This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.

At x=L

$-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]$

This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.

The temperature distribution along with the schematics are given in the attached figure.

Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.

It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.

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2 years ago

A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C

mario62 [17]

Answer:

3.62 m and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}$

$\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}$

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}$

$\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}$

x = - 1.4 m

6 0

2 years ago

charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric fiel

IRISSAK [1]

Answer:

Explanation:

Electric field due to charge at origin

= k Q / r²

k is a constant , Q is charge and r is distance

= 9 x 10⁹ x 5 x 10⁻⁶ / .5²

= 180 x 10³ N /C

In vector form

E₁ = 180 x 10³ j

Electric field due to q₂ charge

= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

= 30.33 x 10³ N / C

It will have negative slope θ with x axis

Tan θ = .5 / √.5² + .8²

= .5 / .94

θ = 28°

E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j

= 26.78 x 10³ i - 14.24 x 10³ j

Total electric field

E = E₁ + E₂

= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j

= 26.78 x 10³ i + 165.76 X 10³ j

magnitude

= √(26.78² + 165.76² ) x 10³ N /C

= 167.8 x 10³ N / C .

3 0

2 years ago

An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t

Lynna [10]

Answer:

2.5 m

Explanation:

Weight of billboard worker = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension in rope = 550 N

The sum of the torques will be

$-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m$

The position of the person will be 2.5 m

7 0

2 years ago