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3 years ago

What is the average velocity of the object from t=1s to t= 3 s?

Physics

1 answer:

galina1969 [7]3 years ago

7 0

Answer:

Explanation:

Since this is a distance v time graph, the slope of the line from 1s to 3s is the velocity. However, it looks like, at t=3, the velocity is 0, so getting the definite velocity is not going to happen. We can estimate it as closely as possible. Since the line is tending from the upper left to the lower right, the slope is negative, so the velocity is also negative. That leaves only C or D as our answers. And the slope is closer to -1 than to -5, so choice D. is the one you want.

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In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of t

muminat

<h2><em><u>⇒</u></em>Answer:</h2>

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Step-by-Step Solution:

Solution 35PE

This question discusses about the increased range. So, we shall assume that the angle of jumping will be as the horizontal range is maximum at this angle.

Step 1 of 3<

/p>

The legs have an extension of 0.600 m in the crouch position.

So, m

The person is at rest initially, so the initial velocity will be zero.

The acceleration is m/s2

Acceleration m/s2

Let the final velocity be .

Step 2 of 3<

/p>

Substitute the above given values in the kinematic equation ,

m/s

Therefore, the final velocity or jumping speed is m/s

Explanation:

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Answer:

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Explanation:

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3 years ago

What is the string theory.

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Which of the following is located in the temperature climate zone?

barxatty [35]

The correct answer is C. Taiga hope it helps ( :

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3 years ago

The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2

Vlada [557]

Answer:

Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]

Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the reaction with OH

Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]

Than, considering the concentration of Cl increases by a factor of 100 to about 4 × $10^{5}$ molecules $cm^{-3}$ , the result will be that the reaction with OH will be 535 + (100 to about 4 × $10^{5}$ molecules $cm^{-3}$ ) times faster than the reaction with Cl

Explanation:

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2 years ago