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Dmitry_Shevchenko [17]
3 years ago
5

Zach runs a 100-meter race. 7 seconds after the race started Zach is 35 meters from the starting line and reaches his max speed;

he runs at this max speed for the rest of the race. Zach notices that he is 75 meters from the starting line 12 seconds after the race started. What is Zach's max speed? 7 Incorrect meters per second Suppose Zach runs for an additional z seconds after reaching his max speed... How far will Zach travel during those additional z seconds? 5z Incorrect meters What is Zach's distance from the starting line 7 + z seconds after the race started? 35+z Incorrect meters What is Zach's distance from the starting line x seconds after the race started (provided x ≥ 7 )? 35+7x Incorrect meters
Physics
1 answer:
Angelina_Jolie [31]3 years ago
7 0

Answer:MaxSpeed=8 m/s. Additional distance during z seconds after top speed is 8*z meters. Now the distance at 7+z seconds after the race starts is D=( 35+8*z) meters. For x ≥ 7 the time of the race, then  D=(35+(x-7)*8) meters is the distance traveled during that time.

Explanation:

First we are interested in calculating the top speed, that is constant. We know at 35 meters he reaches the top speed and this happens 7 seconds after the race started. Also he is at 75 meters from the starting line 12 seconds after the beginning, then we can use the definition of speed, in this case is constant, to get its value:  Speed=change in distance / change in time

Then Speed= (75-35)meters/(12-7)seconds = 40m/5s=8m/s. That is the topspeed=8 m/s  

Now from the 35 meters he runs at constant speed, then if we are told he runs z seconds and we are asked the distance he runs in that time, we know the distance at constant speed is D=V*t,

then D= 8m/s * z seconds=8z meters.

But here since the run is a 100 meters-race we have an upper bound for z, here he has left (100-35)=65 meters to run at 8m/s, then he can maximum run additional 8,125 second or 15,125 seconds in total for the race.

That is D=8*z meters, with z < 8,125 seconds, this is the additional distance he can run after reaching his top speed.

Now the distance at 7+z seconds after the race starts is D=( 35+8*z) meters, the first 35 m he runs until reaching his top speed and the rest of the distance at this speed.

for x the total time of the run at a given point, we are asked the distance he has traveled at that time, then we are also told x ≥ 7, then  D=(35+(x-7)*8) meters, we subtract the 7 initial seconds because they are already counted in the initial 35 meters.

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Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of
morpeh [17]

Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner(v_1)=3.1 m/s

speed of second runner(v_2)=4.65 m/s

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3 0
3 years ago
Identify the labeled parts in the figure.
Hitman42 [59]

Answer:

I. a, c, f and h

II. e

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IV. i

Explanation:

I. Chemical symbols are simple abbreviations used to represent various elements or compound. They consist entire of alphabet.

For the diagram given above, the labelled parts which represent chemical symbol are: a, c, f and h

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For the diagram given above, the labelled part which represent Coefficient is: e

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For the diagram given above, the labelled part which represent the number of atoms of the element are: b, d, g and i

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5 0
3 years ago
Assignments
sladkih [1.3K]

Hello there!

I hope you and your family are staying safe and healthy during this unprecendented time.

A) What is the work done?

Answer: We need to use the formula

w=-F_f(d)

w=-(35)(20)

w=-700J

B) What is the work done on the cart by the gravitational force?  

Alright, we know that the gravitional force is perpendicular to the diplacement. Therefore, we gonna use the following formula:

w=Fdcos90

w=0

C) What is the work done on the cart by the shopper?

This is the easier part, since we already know that the work done by the shopper is the same as the work done by the friction force

W shopper + W friction = 0\\W shopper = W-friction \\W shopper = 700J

D) Find the force the shopper exerts, using energy considerations.

F_f+Fcos25=0\\-35+Fcos25=0\\F=38.6N

E) What is the total work done?

You just need to add them:

w=wshopper+wfriction\\w=0

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