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Aliun [14]
2 years ago
13

A sports car can move 100.0 m in the first 4.5 s of constant acceleration.

Physics
2 answers:
Inessa [10]2 years ago
6 0
The car acceleration is 22.2 repeating meters per second squared

Troyanec [42]2 years ago
4 0

Answer:

Acceleration of the car, a=9.87\ m/s^2

Explanation:

Initial sped of the car, u = 0

Distance covered by the car, d = 100 m

Time taken, t = 4.5 s

Let a is the acceleration of the car. It can be calculated using second equation of kinematics as :

d=ut+\dfrac{1}{2}at^2

d=\dfrac{1}{2}at^2

a=\dfrac{2d}{t^2}

a=\dfrac{2\times 100}{(4.5)^2}

a=9.87\ m/s^2

So, the acceleration of the car is 9.87\ m/s^2. Hence, this is the required solution.

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A very long uniform line of charge has charge per unit length λ1 = 4.80 μC/m and lies along the x-axis. A second long uniform li
Elodia [21]

Answer:

a) E=228391.8 N/C

b) E=-59345.91N/C

Explanation:

You can use Gauss law to find the net electric field produced by both line of charges.

\int \vec{E_1}\cdot d\vec{r}=\frac{\lambda_1}{\epsilon_o}\\\\E_1(2\pi r)=\frac{\lambda_1}{\epsilon_o}\\\\E_1=\frac{\lambda_1}{2\pi \epsilon_o r_1}\\\\\int \vec{E_2}\cdot d\vec{r}=\frac{\lambda_2}{\epsilon_o}\\\\E_2=\frac{\lambda_2}{2\pi \epsilon_o r_2}

Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.

The net electric field at point r will be:

E=E_1+E_2=\frac{1}{2\pi \epsilon_o}(\frac{\lambda_1}{r_1}+\frac{\lambda_2}{r_2})

a) for y=0.200m, r1=0.200m and r2=0.200m:

E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C

b) for y=0.600m, r1=0.600m, r2=0.200m:

E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C

5 0
2 years ago
How and why does the distance between 2 electrodes affect the rate of electrolysis? ...?
Rasek [7]
 <span>Nothing, in terms of the chemistry. 

The distance between the electrodes affects the electrical resistance very slightly. Increasing the distance increases the resistance and reduces the current slightly, which reduces slightly the amount of product. 

For most practical applications, for electrolysis done in a beaker, varying the distance between the electrodes will make little difference. 

Increasing the concentration of the electrolyte will increase the current flow because there are more charged particles to carry charge, and increase the product yield.</span>
4 0
2 years ago
A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po
Umnica [9.8K]

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + 0.5mV1^{2} = mgh2 + 0.5mV2^{2}

where

  • speed at highest point = V2
  • speed at lowest point = V1
  • mass of the girl and swing = m
  • at the highest point, the  speed is minimum (V1 = 0)
  • at the lowest point the speed is maximum (V2 is the maximum speed)
  • therefore the equation becomes mgh1 + 0.5mV1^{2} = mgh2

      m(gh1 + 0.5V1^{2}) = m(gh2)

      gh1 + 0.5V1^{2} = gh2

      V1 = \sqrt{\frac{gh2 - gh1}{0.5}}

now we can substitute all required values into the equation above.

V1 = \sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}

V1 = \sqrt{\frac{32.34}{0.5}}

V1 =8.1 m/s

8 0
2 years ago
Motion maps for two objects, Y and Z, are shown.
Vesna [10]

Answer:

The answer is B) 3 seconds

Explanation:

I just took the test on 2020 edge and got it right

5 0
2 years ago
How far could a rabbit run if it ran 36km/h for 5.0min?
Korolek [52]

3 kilometers, it is just 5/60 or 1/12 multiplied by 36.

4 0
2 years ago
Read 2 more answers
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