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strojnjashka [21]
3 years ago
14

a runner is training for a marathon.he follows a trail going 5m/s N, then 2.5m/s S, then 4 m/s N and finishes strong going 7m/s

W. what is the resultant velocity in km/h of the cyclist throughout the training session
Physics
1 answer:
marta [7]3 years ago
5 0

In order to calculate his velocity for the whole session, we need to know ...

(1).  the straight-line distance from the start-point to the end-point;

(2).  the direction from start-point to end-point

(3).  the elapsed time from start to finish

We could find (1) and (2) if we knew how much time he spent on each segment of the training run.  Then we could calculate the distance covered on each segment, and construct the straight-line displacement and direction of the whole thing.

For (3), no time information is given anywhere in the question, so it's pretty hard to come up with a total time.

We definitely have to mark this one "<em>INSUFFICIENT GIVEN INFORMATION</em>".  We just don't have what we need in order to solve it.

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Ferns that eject spores generally do so in pairs, with two spores flying off in opposite directions. The structure from which th
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It cancels recoil.

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For each action there is an equal an opposite reaction.

The principle of conservation of momentum tell us that if a single spore were ejected the fern would suffer a recoil from it. This recoil would take energy and speed from the spore. But if they are ejected in pairs the recoil is canceled and all the energy is transferred to the spores resulting in higher speeds.

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Iron and stone

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This is a two part question: What is the force of friction acting on a 100 kg steel slab at rest on a steel floor?
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2) F=ma
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3 years ago
The speed at which a light aircraft can take off is 120 km/h. (A) What is the minimum constant acceleration required for the pla
kondor19780726 [428]

Answer:

A) a = 2.31[m/s^2]; B) t = 14.4 [s]

Explanation:

We can solve this problem using the kinematic equations, but firts we must identify the data:

Vf= final velocity = take off velocity = 120[km/h]

Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\  a=2.31[m/s^2]\\

To find the time we must use another kinematic equation.

v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]

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3 years ago
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