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shutvik [7]
3 years ago
9

The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vert

ical circle. What is the maximum speed the ball can have at the top of the circle? 3.74 m/s 9.67 m/s 5.86 m/s None of the above
Physics
1 answer:
zimovet [89]3 years ago
3 0

Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

F=\dfrac{mv^2}{r}

v is the maximum speed

v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s

Hence, the maximum speed of the ball is 5.86 m/s.

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Please help quickly,the picture and choices are above.
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Answer:

C because it make sens

C the light wave traveled through ice and then through a Dimond.

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2 years ago
I got the correct answer, but I don't know what I did. Why is the answer E?
MariettaO [177]

Explanation:

Displacement is the straight line distance from the starting position to the final position.

Person X walks halfway around circle.  So her displacement is 100 m.

Person Y walks 3/4 of the way around.  So his displacement is 50√2 m ≈ 70.7 m.

Person Z walks completely around the circle, so their displacement is 0 m.

Therefore:

Z < Y < X

6 0
2 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]

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 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:

4 0
2 years ago
If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int
vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system on the surrounding


In this problem, the work done by the system is

W=-225 kcal=-941.4 kJ

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

Q=-5 \cdot 10^2 kJ=-500 kJ

with a negative sign as well because it is released by the system.


Therefore, by using the initial equation, we find

\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ

8 0
3 years ago
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IgorC [24]

Answer:

d) shortening the string

Explanation:

Time period of a pendulum clock is dependent on two factors namely:length and acceleration due to gravity.

When a clock loses time, the time period of the pendulum clock increases.

This however can be corrected by decreasing the length of the pendulum.The time period of the pendulum clock is not dependent on the mass of the bob. The time period of the pendulum clock can be corrected only by changing the length of the pendulum string.

5 0
3 years ago
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