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3 years ago

A runner accelerates to 4.2 m/s2 for 10 seconds before winning the race. How far did he/she run?

Physics

2 answers:

timofeeve [1]3 years ago

8 0

She run the distance of 42metres

hodyreva [135]3 years ago

6 0

Answer:

Runner will run for 210m

Step by Step Explanation:

According to Newton's first of motion,

Vf=Vi+at

Given is

the speed with which runner is running will be Vi, he is running with accelertion a=4.2m/s2 fot time t=10sec

Vf will be 0 because runner will stop after winning the race

we will find Vi from equation 1,

Vi=Vf-at

= 0- 4.2*10

=-42 m/s (negative sign is showing decceleration)

So, Vi=42 m/s

According to Newton's third of motion,

2aS= Vf^2-Vi^2

S is distance, he will cover before stopping,

2(4.2)S= 0- 42^2

S=1764/8.4

S=210m

Runner will run for 210m

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ELECTROMAGNETIC FIELDS PLAN AN INVESTIGATION LAB REPORT FOR EDG.!!!!! WORTH 100 POINTS????????

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2 years ago

Bill is farsighted and has a near point located 121 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm fro

Arada [10]

Answer:

Explanation:

The lens equation is

1 / f = 1 / di + 1 / do

Where

f is focal length

di is the image distance

do is the object distance

Both Annie and Billy use a glass whose near point is 25cm

Then, the object distance is

do = 25 - 2 = 23cm

The have the same object distance.

Let find the vocal length of bills eye

Given that,

Bill near point is 121cm and distance of the glass from the eye is 2cm

Then,

Image distance of bill is

di_B = -(121-2) = -119cm

object distance do = 23cm

Then,

1 / f_B = 1 / di_B + 1 / do

1 / f_B = -1 / 119 + 1 / 23

1 / f_B = -119^-1 + 23^-1

1 / f_B = 0.0351

Then, f_B = 28.51 cm

Also, let find Annie focal length

Given that,

Annie near point is 74 cm and distance of the glass from the eye is 2cm

Then,

Image distance of Annie is

di_A = -(74-2) = -72cm

object distance do = 23cm

Then,

1 / f_A = 1 / di_A + 1 / do

1 / f_A = -1 / 72 + 1 / 23

1 / f_A = -72^-1 + 23^-1

1 / f_A = 0.02959

Then, f_A = 33.8 cm

Distance of object from the lens when Annie uses Billy glass

Then,

1 / f_B = 1 / di_A + 1 / do

1 / 28.51 = -1 / 72 + 1 / do

28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

do^-1 = 0.048964

do = 20.42 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_A = 20.42 + 2 = 22.42cm

do_A = 22.42 cm

Distance of object from the lens when Billy uses Annie glass

Then,

1 / f_A = 1 / di_B + 1 / do

1 / 33.8 = -1 / 119 + 1 / do

33.8^-1 = -119^-1 + do^-1

do^-1 = 33.8^-1 + 119^-1

do^-1 = 0.03799

do = 26.32 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_B = 26.32 + 2 = 28.32 cm

do_B = 28.32 cm

7 0

3 years ago

5. The net external force on a rock of mass 4.2 kg is 8.0 N forward. Find the acceleration of the rock.

andreev551 [17]

Answer:

1.904

Explanation:

F= ma

8 = 4.2 a

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a = 1.904

6 0

2 years ago

À A car moves with an initial velocity of 18 m/s due north. Find the velocity of the car after 7 Os if

Nonamiya [84]

Answer:

(a) $v_f=28.5m/s$

(b) $v_f=7.5m/s$

Explanation:

Hello.

(a) In this case since the car is moving at an initial velocity of 18 m/s due north, the final velocity is computed considering the acceleration as positive since it is due north as well:

$v_f=v_0+at=18m/s+1.5m/s^2*7s\\\\v_f=28.5m/s$

(b) In this case, since the car is moving due north by the acceleration is due south it is undergoing a slowing down process, thereby the acceleration is negative therefore the final velocity turns out:

$v_f=v_0+at=18m/s-1.5m/s^2*7s\\\\v_f=7.5m/s$