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Tamiku [17]
2 years ago
6

Using the graph above determine the acceleration for the object represented between 6.0s and 8.0s

Physics
1 answer:
grin007 [14]2 years ago
4 0

The object was moving at constant velocity between the points on the graph that we marked 6 s to 8s.

<h3>What is acceleration?</h3>

The term acceleration has to do with the change of velocity with time. In this case, we saw that the velocity of the car was not changed between 6 s to 8s.

Now , the object was moving at constant velocity between the points on the graph that we marked 6 s to 8s.

Learn more about acceleration:brainly.com/question/12550364?

#SPJ1

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A potential problem is that you are willing to accept a <u>5% </u>chance of being wrong if you reject the null hypothesis.

The significance level is the probability of rejecting the null hypothesis if it is true. For example, a significance level of 0.05 indicates a 5% risk of concluding that there is a difference when there is actually no difference. Rejecting the true null hypothesis results in a Type I error.

The smaller the value of α the more difficult it is to reject the null hypothesis. Therefore, choosing a low value for α can reduce the likelihood of Type I errors. The result here is that if the null hypothesis is false, it may be more difficult to reject using a lower value for α. The alpha value or statistical significance threshold is arbitrary. Which value to use depends on your field of study.

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4 0
1 year ago
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Veseljchak [2.6K]

Answer:

The speed of q₂ is 4\sqrt{10}\ m/s

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

E_{i}=E_{f}

\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2

\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})

Put the value into the formula

\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})

0.00075(400-v_{f}^2)=0.18&#10;

400-v_{f}^2=\dfrac{0.18}{0.00075}

-v_{f}^2=240-400

v_{f}^2=160

v_{f}=4\sqrt{10}\ m/s

Hence, The speed of q₂ is 4\sqrt{10}\ m/s

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A skateboarder starts from rest and maintains a constant acceleration of 0.50 m/s² for 8.4 s. What is the rider's displacement d
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