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Elena L [17]
3 years ago
15

While riding TRAX at a constant speed of 16 m/s you walk toward the front of the car at 5 m/s.

Physics
1 answer:
AVprozaik [17]3 years ago
4 0

Speed of the TRAX is given as 16 m/s so let say it is given as

v_t = 16 m/s

now our speed towards the front end is given by 5 m/s so this is the relative speed of us with respect to TRAX

let say this speed is given as

v_1t = 5 m/s

now we need to find the speed with respect to someone standing outside the TRAX

so here we need to find the net speed in ground frame and hence we can use the formula of relative speed

v_{1t} = v_1 - v_t

v_1 = v_{1t} + v_t

v_1 = 5 + 16

v_1 = 21 m/s

so someone outside the TRAX will see our speed as 21 m/s

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A +5.00 pC charge is located on a sheet of paper.
emmainna [20.7K]

Answer:

a)    V = - x ( σ / 2ε₀)

c)  parallel to the flat sheet of paper

Explanation:

a) For this exercise we use the relationship between the electric field and the electric potential

          V = - ∫ E . dx        (1)

for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet

       Ф = ∫ E . dA = q_{int} /ε₀

the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product

          E A = q_{int} /ε₀

area let's use the concept of density

        σ = q_{int}/ A

       q_{int} = σ A

          E = σ /ε₀

as the leaf emits bonnet towards both sides, for only one side the field must be

          E = σ / 2ε₀

         we substitute in equation 1 and integrate

      V = - σ x / 2ε₀  

       V = - x ( σ / 2ε₀)

if the area of ​​the sheeta is 100 cm² = 10⁻² m²

      V = - x  (10⁻²/(2 8.85 10⁻¹²) = - x  ( 5.6 10⁻¹⁰)

       

      x = 1 cm     V = -1   V

      x = 2cm     V = -2   V

This value is relative to the loaded sheet if we combine our reference system the values ​​are inverted

       V ’= V (inf) - V

       x = 1 V = 5

       x = 2 V = 4

       x = 3 V = 3

   

These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.

 

In the attachment we can see a schematic representation of the equipotential surfaces

b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced

c) parallel to the flat sheet of paper

8 0
3 years ago
Healthy pregnant women should get at least 150 minutes of moderate-intensity aerobic activity, spread over a 7-day period, each
kirill115 [55]

Healthy pregnant women should get at least 150 minutes of moderate-intensity aerobic activity, spread over a 7-day period, each week - True

<u>Explanation:</u>

Women who are healthy but moderately active must be engaged with aerobic  activities moderately. they should be engaged with this activities at least 2 and a half hours in all 7 days of a week. They must be engaged with this activity during the pregnancy time and also in the period after their delivery.

During the time of pregnancy all women must be healthy and physically active. The ways for maintaining a very good health in the pregnancy time will be guided by the health care providers. Women should be highly active during and after pregnancy time. This helps in having a baby that is also very healthy.

3 0
3 years ago
A small smooth object slides from rest down a smooth inclined plane inclined at 30degrees horizontal.What is the acceleration do
asambeis [7]
The acceleration is given as:

a = g sin(30°) where g is the gravitational acceleration

For g = 10 m/s^2, we get

a = 10 sin(30°) = 10 * 1/2 = 5 m/s^2
7 0
3 years ago
If a force of 5 n to left and 9 n to right act on a object what is the net force
emmainna [20.7K]
4 n to the right is the net force
7 0
3 years ago
Godric and Savos are a few meters apart, at one end of the football field. Peter is at the other end of the field. Godric and Sa
Troyanec [42]

Answer:

From Savos point of view, Gadros appear not to be moving

Explanation:

The given parameters are;

The distance between Godric and Savos = A few meters

The location of Godric and Savos = One end of the field

The location pf Peter = The other end pf the field

The speed with which Godric and Savos are moving towards Peter = 4 m/s

Let the positive x-direction be the direction in which Godric and Savos are moving

Therefore;

The speed and direction with which Godric is moving = The speed and direction with which Savos is moving = 4·i m/s

From Savos point of view, the relative speed with which Godric is moving = (The speed and direction in which Godric is moving) - (The speed and direction in which Savos is moving) = 4·i m/s - 4·i m/s = 0 m/s

Therefore, from Savos point of view, the relative speed with which Godric is moving = 0 m/s or Gadros appears stationary or not to moving.

7 0
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