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sdas [7]
4 years ago
11

You walk with a velocity of 2 m/s north. You see a man approaching you, and from your frame of reference he has a speed of 3 m/s

to the south. What is the velocity of the man from the frame of reference of a stationary observer? (2 points)
Physics
1 answer:
kupik [55]4 years ago
5 0
The answer to this question would be: 1m/s

When you are walking to the north with 2m/s velocity, the stationary object(velocity=0m/s) will look like moving south at 2m/s velocity. That happens because the relative distance between you and the object is reduced by 2m/s in both conditions. In this question, the man seems like have 3m/s velocity. The real velocity should be:
3m/s - 2m/s = 1m/s
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You shake a bottle of soda and take off the cap. If the soda shoots out of the
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Answer:

A. 11.5 m

Explanation:

Given,

The initial velocity of the soda cap, u = 15 m/s

The soda bottle cap is projected vertically upwards,

Hence, the angle formed with the ground, Ф = 90°

The maximum height of the projectile is given by the formula,

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Substituting the given values in the above equation

                                   h_{max} =\frac{15^{2}sin^{2}90}{2X9.8}

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8 0
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 3x10^{-11} e^{-255/T}  and 2x10^{-12} e^{-940/T}  

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]  

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]

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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 3x10^{-11} e^{-255/T}  and 2x10^{-12} e^{-940/T}  

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]  

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x10^{-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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Answer:

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Explanation:

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Explanation:

Below is an attachment containing the solution.

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