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solong [7]
4 years ago
12

if a bicyclist, with initial velocity of zero, steadily gain speed until reaching a final velocity of 26m/s, how far away did sh

e travel during the race (in the same amount of time?
Physics
1 answer:
Alja [10]4 years ago
5 0

Answer:

<h2><em>34.46m</em></h2>

Explanation:

Using one of the equation of motion to solve the question. According to the equation v² =u²+2as where;

v is the final velocity of the  bicyclist = 26m/s

u is the initial velocity of the  bicyclist = 0m/s

a is the acceleration due to gravity = 9.81m/s

s is the distance covered during travel

Substitute the given parameters into the formula above to get the distance traveled

26² = 0² + 2(9.81)s

676 = 19.62s

Divide both sides  by 19.62

676/19.62 = 19.62s/19.62

s = 34.46m

<em>The distance traveled by the  bicyclist during the race is 34.46m</em>

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Answer:

A

Explanation:

The process by which molten materials add new oceanic crust to the ocean floor.

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The impact front the fall, depending on the distance, has an effect on the impact of their bones crushing with the pressure from the fall.
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A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval Δt f
scoray [572]

Answer:

BA

0

\dfrac{NBA}{\Delta t}

Explanation:

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t = Time taken

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\phi_i=BAsin\theta\\\Rightarrow \phi_i=BAsin0\\\Rightarrow \phi_i=0

The magnetic flux is 0

Magnitude of emf is given by

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The magnitude of the average emf induced in the entire coil is \dfrac{NBA}{\Delta t}

4 0
3 years ago
Please help in finding the Total Resistence, Total Current, Req Current Center of Parallel and V40 olms
Blizzard [7]
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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
2 years ago
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