Ask question

Ask question

All categories

1 month ago

6

f a car is speeding down a road at 40 miles/hour (mph), how long is the stopping distance D40 compared to the stopping distance

D25 if the driver were going at the posted speed limit of 25 mph? Express your answer as a multiple of the stopping distance at 25 mph. Note that D25 is already written for you, so just enter the number. View Available Hint(s)

1 answer:

kicyunya [3.2K]1 month ago

5 0

Answer:

D40 = 2.56 × D25

This means the stopping distance at 40 mph is 2.56 times that at 25 mph

Explanation:

provided data

speed = 40 miles/hour

distance = D40

speed limit = 25 miles/hour

distance = D25

to determine

express how the stopping distance relates to the 25 mph distance

solution

it is established that stopping distance varies directly with the square of the speed

thus speed ratio is

initial speed = $\frac{40}{25}$

thereby, initial speed = 1.6

so

stopping distance increase = (1.6)²

$\frac{D40}{D25}$ = (1.6)²

$\frac{D40}{D25}$ = 2.56

therefore

D40 = 2.56 × D25

This shows that the stopping distance at 40 mph is 2.56 times that at 25 mph

You might be interested in

Mrs. Gonzalez is about to give birth and Mr. Gonzalez is rushing her to the hospital at a speed of 30.0 m/s. Witnessing the spee

Ostrovityanka [3204]

Answer: The frequency is 1714.3 Hz

Explanation: The calculation is derived from the Doppler effect formula.

Since the source is approaching the observer, the observer's velocity is considered positive.

Refer to the attached document for the detailed derivation

3 0

2 months ago

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

inna [3103]

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ $\frac{du}{dy}$

therefore

= µ $\frac{v}{h}$ ............1

substituting the values

= 1.75 × $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

the area between the air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

now applying Newton's second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

solving for t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

the time needed after impact for the puck is 2.18 seconds

3 0

1 month ago

Your school science club has devised a special event for homecoming. You've attached a rocket to the rear of a small car that ha

inna [3103]

La force agissant pendant 9 s et la décélération pendant 12 - 9 = 3 s.

Distance totale parcourue = 990 m

vitesse initiale u = 0

Distance parcourue pendant l'accélération

s₁ = 1/2 a 9² où a est l'accélération

= 40.5 a

vitesse finale après 9 s

v = at = 9a

pendant la décélération

v² = u² - 5 x s₂

0 = (9a)² - 5 s₂

s₂ = 16.2 a²

Distance parcourue pendant la décélération = 16.2 a²

s₁ + s₂ = 990

40.5 a + 16.2 a² = 990

16.2 a² + 40.5 a - 990 = 0

a = 6.5

4 0

24 days ago

A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.

serg [3582]

A) 16.1 N

The force of electricity acting between the corks can be calculated using Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where

k represents Coulomb's constant

$q_1 = 6.0 \mu C=6.0 \cdot 10^{-6} C$ denotes the charge magnitude on the first cork

$q_2 = 4.3 \mu C = 4.3 \cdot 10^{-6}C$ indicates the charge magnitude on the second cork

r = 0.12 m is the distance separating the corks

By inserting the values into the formula, we arrive at

$F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(6.0\cdot 10^{-6}C)(4.3\cdot 10^{-6} C)}{(0.12 m)^2}=16.1 N$

B) Attractive

<pas per="" coulomb="" law="" the="" orientation="" of="" electric="" force="" between="" two="" charged="" entities="" relies="" on="" their="" charge="" signs.=""><pmore specifically="">

- when both are similarly charged (e.g. positive-positive or negative-negative), the force is repulsive

- when charges are of opposite signs (e.g. positive-negative), the resulting force is attractive

<pin this="" case="" we="" have="">

Cork 1 holds a positive charge

Cork 2 possesses a negative charge

<pthus the="" force="" acting="" between="" them="" is="" attractive.="">

C) $2.69\cdot 10^{13}$

The total charge of the negative cork is

$q_2 = -4.3 \cdot 10^{-6}C$

<pwe understand="" that="" a="" single="" electron="" has="" charge="" of="">

$e=-1.6\cdot 10^{-19}C$

<pthe total="" charge="" of="" the="" negative="" cork="" arises="" from="" having="" n="" extra="" electrons="" so="" we="" can="" express="" it="" as="">

$q_2 = Ne$

<pafter solving="" for="" n="" we="" can="" determine="" the="" count="" of="" excess="" electrons:="">

$N=\frac{q_2}{e}=\frac{-4.3\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=2.69\cdot 10^{13}$

D) $3.75\cdot 10^{13}$

The overall charge on the positive cork is

$q_1 = +6.0\cdot 10^{-6}C$

<pthe charge="" of="" a="" single="" electron="" is="" known="" to="" be="">

$e=-1.6\cdot 10^{-19}C$

<pthe total="" charge="" of="" the="" positive="" cork="" results="" from="" n="" excess="" electrons="" which="" can="" be="" depicted="" as="">

$q_1 = -Ne$

<pby calculating="" for="" n="" we="" derive="" the="" number="" of="" electrons="" cork="" has="" lost:="">

$N=-\frac{q_1}{e}=-\frac{+6.0\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=3.75\cdot 10^{13}$

</pby></pthe></pthe></pafter></pthe></pwe></pthus></pin></pmore></pas>

6 0

1 month ago

If the coefficient of static friction between a table and a uniform massive rope is μs, what fraction of the rope can hang over

inna [3103]

Here's the procedure explained: Assume F represents the portion of the rope that is extending over the table. In this scenario, the frictional force that holds the rope on the table can be calculated using the formula: Ff = u*(1-f)*m*g. Additionally, it is important to determine the gravitational force that attempts to pull the rope off the table, Fg, calculated through: Fg = f*m*g. You then need to set these two equations equal to each other and resolve for f: f*m*g = u*(1-f)*m*g leads to f = u*(1-f) = u - uf. Simplifying gives f + uf = u, which results in f = u/(1+u) representing the fraction of the rope. This will lead you to the final answer.

8 0

1 month ago

Read 2 more answers