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yawa3891 [41]
2 years ago
12

Study the velocity vs. time graph shown.

Physics
1 answer:
laiz [17]2 years ago
6 0

For velocity vs Time graphs, the displacement of the object from 2 seconds to 6 seconds is 30 m.

<h3>What is displacement?</h3>

The displacement is the shortest distance travelled by the particle. It is the vector quantity which re[presents both the magnitude and direction.

In velocity time graphs, the displacement is the area under the curve of the graph on the x axis.

'

A line starts at (0, 2) and ends at (6, 8) in v-t graph

Displacement is equal to the area of a triangle and a rectangle formed under the line.

Area = 1/2 base x height + length x breadth

Area = 1/2 x 6x 6 + 6x2

Area = 18 +12

Area = 30 m

Thus, the displacement is 30 m.

Learn more about displacement.

brainly.com/question/11934397

#SPJ1

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As she was trying to study, Tanisha asked her roommate to lower the radio. Her roommate had turned the radio up originally from
STatiana [176]

Answer:

just-noticeable difference

Explanation:

Using principles of psychology and physics, a branch of experimental psychology called psychophysics has been created. This field is focused on the sensation, the sense and the perception of stimuli. Within this branch it has been called just-noticeable difference to the amount that must be changed of some stimulus so that this difference is noticeable, that is to say, the threshold at which the change is perceived.

4 0
3 years ago
From the frame of reference of car 2, what is the velocity of car
Alborosie
It’s is the last one d
5 0
3 years ago
Saved
Anon25 [30]

Total distance moved by bead is 1.952 cm.

Explanation:

Let first consider all data that are given in question.

1.    F = 8 N                       ...force acting on string

2.   f  = 2 Hz                     ...frequency of system

3.   β = 4 cm = 0.04 m    ...wavelength of wave formed due to vibration           4.   A =  1 cm  = 0.01 m     ...Amplitude of vibration

Under certain conditions, waves can bounce back and forth through a particular region, effectively becoming stationary. These are called standing waves.

Here,it is due to vibration induced in spring due to tension induced in string

Standing wave equation is given by

y = (x,t) = 2A * sinK x * cos (wt)                ...(1)

Let first find, value of K, x, w, t

k = 2 * pi / beta                                          ....(2)

Where β is wavelength in meters

                   K is wave number

k = 2 * pi / 0.01

k = 628.31 m^{-1}

now, let us find value of w

W = 2 x pi x f                               ....(3)

                where f is frequency in hertz

W = 2 x pi x 2

W = 4 x pi

W = 0.08 \frac{m}{s}

y = (x,t) = 2A * sinK x * cos (wt)

now, let us find value of v that is wave speed

Notice that some x-positions of the resultant wave are always zero no matter what the phase relationship is.  These positions are called nodes.

Finding the positions where the sine function equals zero provides the positions of the nodes.

In our case, and      

K * x = pi

x = 0.04 / 2

x = 0.02

y = (0.02,1) = 2(0.01) * sin pi  * cos (12.5664 * 1)

y = (0.02,1) = 2(0.01) * -1  * cos 0.9761

Y = 1.952 cm

Finally, when bead is at middle of the string, total distance after stretch covered  is 1.952 cm.

3 0
3 years ago
Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance
siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

       

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

          E_ {total} = E₁ + E₂

          E_{ total} = k \frac{Q}{x_1^2} + k  \frac{Q}{x_2^2}

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = k  \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )

            E_ {total} = k \frac{Q}{x_1^2}   ( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

             x₁² =  (x₁ + d)²

           

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

              E_ {total} = -k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}

this expression holds for the points located at

                  -r₁ <x₁ <r₁

3) Point between the two spheres

                E_ {total} = k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

                  E_ {total} = + k \frac{Q}{(d-x_1)^2}+ k Q / (d-x1) 2

point range

                  -r₂ <x₂ <r₂

             

5) Right side point

            E_ {total} = k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}

             E_ {total} = - k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )- k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0
3 years ago
How can electric energy be converted to heat energy?
Goshia [24]
To do that, you must pass electric current through a substance
that electrons have to spend energy to pass through. 
The substance will be one that gets warm and dissipates heat
when electric current flows through it.
We'll say that the substance has "resistance", which we can measure.
The amount of heat that appears when current flows through it
will be (current²)·(resistance).

A few examples of things used for that purpose:

-- resistors
-- burners on electric stoves
-- coils of resistor-wire in a toaster
-- aquarium heater
-- electric clothes iron
-- electric coffee pot
-- blow-dryer
-- electric hair-curling iron
-- skinny tungsten wire in a light-bulb .

5 0
4 years ago
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