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murzikaleks [220]
3 years ago
11

If the position of a particle on the x-axis at time t is −5t2, then the average velocity of the particle for 0 ≤ t ≤ 3 is

Physics
1 answer:
Drupady [299]3 years ago
3 0

Answer:

v = 15 m / s

Explanation:

In this exercise we are given the position function

          x = 5 t²

and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval

          v= \frac{v_{f} - v_{o} }{t_{f} - t_{o} }

let's look for the displacements

        t = 0     x₀ = 0 m

        t = 3     x_{f} = 5 3 2

                     x_{f} = 45 m

 

we substitute

           v = \frac{45 -0}{3 - 0}

           v = 15 m / s

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Two separate but nearby coils are mounted along the same axis. A power supply controls the flow of current in the first coil, an
Anit [1.1K]

Answer:

b.only when the current in the first coil changes.

Explanation:

An induced current flow in the second coil only when there is a change in current in the first cool. A steady current will produce no change in flux (due to magnetic effect of a current) by the first coil, and according to Faraday, induced current is only produced when there is a change in flux linkage.

8 0
3 years ago
Two pipes move the same amount of ideal fluid in the same amount of time. One pipe has a 2 in. diameter; the other has a 3 in. d
KATRIN_1 [288]

Answer:

a) 3-in. pipe

Explanation:

Given that

Fluid flow is in same amount in the same time it means that volume flow rate is same for the pipes

Volume flow rate

Q = A V

A=Area ,V=Velocity

A=\dfrac{\pi}{4}d^2

If diameter d is more then the velocity will be less for same volume flow rate .We also Know that if pressure is more then the velocity will be less.

The second pipe 3 in diameter having more diameter then the velocity will be less but the pressure will be more.

That is why the 3 in diameter is having more pressure than 2 in diameter pipe.

Therefore the answer will be a.

a) 3-in diameter  pipe

6 0
3 years ago
Points A, B, and C form the vertices of a triangle in a nonuniform electrostatic field. The electrostatic work done on a particl
Trava [24]

Answer:

Explanation:

Let electric potential at A ,B and C be Va , Vb and Vc respectively.

Work done = charge x potential difference

Wab = q ( Va - Vb )

Wac =  q (  Va -  Vc )

Given

Wac = - Wab / 3

3Wac = - Wab

Now

Wbc = q ( Vb - Vc )

= q [ ( Va-Vc ) - ( Va - Vb )]  

= Wac - Wab

= Wac + 3Wac

= 4Wac

4 0
3 years ago
The membrane that surrounds a certain type of living cell has a surface area of 4.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
Nadusha1986 [10]

Answer:

The charge resides on the outer surface = 1.245 \times 10^{-12} C

Explanation:

Surface area of cell  (A) = 4.3\times 10^{-9}  m^{2}

Separation between two plate  (d) = 1.1 \times 10^{-8}  m  

Dielectric constant (k) = 4.2

Potential difference (\Delta V) = 85.7 \times 10^{-3} V

The capacitance of parallel plate capacitor in free space is given by,

           C = \frac{\epsilon_{o} A }{d}

Where \epsilon_{o}  = permittivity of free space = 8.85 \times 10^{-12}

The Capacitance of capacitor is increase by k times when it placed in dielectric medium.

C_{dielectric}  = \frac{k \epsilon_{o} A }{d}

And we know that, C = \frac{Q}{ \Delta V}

So charge on the outer surface is given by,

      Q = \frac{k \epsilon A \Delta V }{d}

      Q = \frac{4.2 \times 4.3 \times 10^{-9} \times 8.85 \times 10^{-12} \times 85.7\times 10^{-3}   }{1.1 \times 10^{-8} }

      Q = 1.245 \times 10^{-12}

3 0
3 years ago
Mercury has a density of 13.56 g/mL. How many kilograms of mercury would you expect to fit in a cylindrical glass cup with a bot
vovangra [49]

Answer:

263.152kg

Explanation:

<em>The density of a substance is related to its mass and volume as follows;</em>

density = mass / volume      

mass = density x volume       -------------(i)

The substance in question here is <em>mercury </em>which has;

density = 13.56g/mL = 13.56g/cm³

Since the mercury is going to be put in the cylindrical glass, the volume of the cylindrical glass is going to be equal to the volume of the mercury that will be put.

And we know that the;

volume of a cylinder = πr²h

<em>Where;</em>

π = 3.142

r = bottom radius of the cylinder = 5.75inches

h = height of the cylinder = 0.950ft

<em>For uniformity, let's convert the radius and height of the cylinder to their corresponding values in cm</em>

r  = 5.75 inches = 5.75 x 2.54 cm = 14.605cm

h = 0.950 ft = 0.950 x 30.48 cm = 28.956cm

<em>Therefore, the volume of the cylinder;</em>

v = 3.142 x (14.605cm)² x 28.956cm = 19406.5cm³

v = 19406.5cm³ [This is also the volume of the mercury necessary to fit the cylinder]

<em>Now the following value has been found;</em>

volume = 19406.5cm³

<em>Substitute the values of density and volume into equation (i)  as follows;</em>

mass = 19406.5cm³ x 13.56g/cm³

mass = 263152.14g

<em>Convert the result to kg by dividing by 1000</em>

mass = 263.152kg

Therefore, 263.152kg kilograms of mercury would fit in the cylindrical glass.

3 0
3 years ago
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