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3 years ago

Write a short description of how the motion of the racers might change from the start of the race to the finish line

1 answer:

6 0

The motion of the racers might change from the start because the pressure goes up so all the racer wants is to speed up and win, so when the racer first starts he or she is calm because he's not driving yet and when he or she is on his/hers way to he finish line he/she just wants to win and gets under pressure so he speeds up even more and drifts. Your welcome

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A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav

gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

$v^2=v_o^2+2gy$

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}$

With this value you can find the spring constant k from the equation (1):

$mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}$

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

$W=F_e\\\\mg=kx$

you solve the last expression for x:

$x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m$

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}$

Next, you calculate x from the equation (1):

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m$

The net fire is stretched 2.72 m

5 0

3 years ago

A student on her way to school walks eastward in a straight line 20.0 meters towards the bus stop, but realizes she dropped her

larisa [96]

Answer:

Total displacement will be 47 meter

Total distance will be 83 meters

Explanation:

We have given that first the student go eastward towards bus stop 20 meters

But he realizes that she dropped his physics notebook and so h=she turns back along the same way up to 18 meters

So displacement = 20-18 = 2 meters

And he travel 45 meters in east along the bus stop so total displacement = 45+2 = 47 meters

Total distance traveled by the student = 20+18+45 = 83 meters

3 0

3 years ago

I NEED THIS A.S.A.P I WILL GIVE BRAINLIEST PLEASE HELP!!

VLD [36.1K]

Explanation:

change 0.5 g to kg so 0.005kg then change 100 ml to m so 0.001m so density=mass over volume so from there you can continue

8 0

3 years ago

Read 2 more answers

In 1 to 2 sentences explain how you would determine the constant K for a spring

Olin [163]

When spring times comes around you ever be like, "k." If you feel like that every year during spring then there you go

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3 years ago

Miguel's parents drove him to a track meet that was 305 km away from the school. It took them 3 hours, which included a 30 minut

OverLord2011 [107]

A., 101.7 km/h is the correct answer for this question

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3 years ago