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viva [34]
3 years ago
13

A car starts from rest at a stop sign. It accelerates at 4.6 m/s^2 for 6.2 s , coasts for 2.1s , and then slows down at a rate o

f 3.3 m/s^2 for the next stop sign. How far apart the are the stop signs?
Physics
1 answer:
ki77a [65]3 years ago
5 0

Answer:

D = 271.54 m

Explanation:

given,

1. car accelerates at 4.6 m/s² for 6.2 s

2. constant speed for 2.1 s

3. slows down at 3.3 m/s²

distance travel for case 1

using equation of motion

 d_1 = u t +\dfrac{1}{2}at^2

 d_1 =\dfrac{1}{2}\times 4.6\times 6.2^2

      d₁ = 88.41 m

case 2

constant speed for 2.1 s now, we have to find velocity

v = u  + at

v = 0 + 4.6 x 6.2

v = 28.52 m/s

distance travel in case 2

d₂ = v x t

d₂ = 28.52 x 2.1 = 59.89 m

for case 3

distance travel by the car

v² = u² + 2 a s

final velocity if the car is zero

0² = 28.52² + 2 x (-3.3) x d₃

6.6 d₃ = 813.39

 d₃ = 123.24 m

total distance travel by the car

D = d₁ + d₂ + d₃

D = 88.41 + 59.89 + 123.24

D = 271.54 m

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Answer:

Extraneous

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Extraneous variables are any variables that you are not intentionally studying in your experiment or test

6 0
3 years ago
Two spherical shells have a common center. A -2.1 10-6 C charge is spread uniformly over the inner shell, which has a radius of
julsineya [31]

Answer:

a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing

b)  E_total = 1.89 10⁶ N / C, field is incoming radial

c) E_total = 0

Explanetion:

For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is

                Ф = E .dA = q_{int} /ε₀

inside the spherical shell there are no charges

The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.

We have two sphere shells with radii 0.050m and 0.15m respectively

a) point where you want to know the electric field d = 0.20 m

shell 1

the point is on the outside,d>ro,  therefore we can consider the charge to be concentrated in the center

            E₁ = k q₁ / d²

             

shell 2

the point is on the outside,d>ro

             E₂ = k q₂ / d²

the total camp is

              E_total = -E₁ + E₂

              E_total = k ( \frac{-q_1 + q_2}{d^2})

              E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²

              E_total = 6,525 10⁵ N /C

The field direction is radial and outgoing ti the shells

b) the calculation point is d = 0.10m

shell 1

point outside the shell d> ro

                 E₁ = k q₁ / d²

shell 2

the point is inside the shell d <ro

Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero

                E₂ = 0

               

                 E_total = E₁

                 E_total = k q₁ / d²

                 E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²

                 E_total = 1.89 10⁶ N / A

As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1

c) the point of interest d = 0.025 m

shell 1

point  is inside the shell d< ro

                 

as there are no charges inside

                     E₁ = 0

shell 2

point is inside the radius of the shell d <ro

                    E₂ = 0

the total field is

                    E_total = 0

3 0
2 years ago
What is a microwave?
bonufazy [111]

Answer:

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7 0
2 years ago
A block with m = 300 grams oscillates at the end of a linear spring with k = 6.5 N/m. (assume this is a top-down view and the bl
inna [77]

Answer:

a) T=1.35s

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first we need to get the angular frequency of the motion

so we have that

ω = √(k/m)

in this case motion is a simple harmonic so the period is defined by:

T= 2π / ω

T= 2π / √(k/m)

replacing the variables...

T= 2π / √(6.5/0.3)

T=1.35s   (period of the block's motion)

and...

α max = | ω²r max |

2 = (2π/1.35)² * r max

r max= 0.0923m

6 0
3 years ago
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Lostsunrise [7]

Two forces 3N and 4N act on a body in a direction due north From East, the equilibrant's angle is given by \theta=\tan ^{-1} \frac{3}{4}=36.8^{\circ}.

<h3>What are equilibrium and resultant force?</h3>

The equilibrium force is the balanced force when the net force acting is zero and is the exact opposite of the consequent force. The resultant force is one single force replaced by numerous forces.

<h3>Briefing:</h3>

3N and 4N are the two forces pulling on a body.

The forces work along the North and the East, which are perpendicular to one another.

The resultant of the forces, which is provided by the equilibrant force,

R  = √(3)²+(4)²

R = 5N

From East, the equilibrant's angle is given by

\theta=\tan ^{-1} \frac{3}{4}=36.8^{\circ}

To know more about equilibrium force visit:

brainly.com/question/12582625

#SPJ9

6 0
1 year ago
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