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2 years ago

How is it technically correct to say that a car making a u-turn can have a constant speed but cannot have a constant velocity?

1 answer:

5 0

During the "U" part of the turn, the car would follow an approximately circular path, and if it's moving at a constant speed, it would have to accelerate toward the center of the circle in order to change its direction.

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you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

3 0

3 years ago

A racecar driver steps on the gas, and his racecar travels 20 meters in 2 seconds starting from rest. The acceleration of the ra

givi [52]

Answer:

2.5m/s^2

Explanation:

Step one:

given

distance = 20meters

time = 2 seconds

initial velocity u= 0m/s

let us solve for the final velocity

velocity = distance/time

velocity= 20/2

velocity= 10m/s

$v^2=u^2+2as\\\\10^2=0^2+2*a*20\\\\100=40a$

divide both sides by 40

$a= 100/40\\\\a=10/4\\\\a= 5/2\\\\a= 2.5m/s^2$

5 0

3 years ago

a horse moves a sleigh 1.00 kilometers by applying a horizontal 2000 Newton force on its harness for 45 minutes. what is the pow

tatyana61 [14]

-- If 2,000 newtons of force were applied through a distance of 1,000 meters,
then 2,000,000 newton-meters = 2,000,000 joules of work were done.

-- 45 minutes = (45 x 60) = 2,700 seconds

-- Power = (work) / (time) = (2,000,000 j) / (2,700 s) = <u>740.74 watts</u>

Interestingly, that's almost exactly 1 horsepower. (0.99295... of 746 watts)

8 0

3 years ago

We see a bolt of lightning and 4 s later we hear the thunderclap. If the speed of

Alex73 [517]

Answer:

i guess 0.8 miles away

Explanation:

mathematically:-

speed is given in question

Time is given

and we have to find distance

simply by using speed formula (s) = d/t

we get answer

6 0

2 years ago

Which of the following metals require ultraviolet light to exhibit the photoelectric effect?The options available: a. Cs, work f

Eddi Din [679]

Answer:

b. AG, work function=4.74eV

Explanation:

Ultraviolet light starts at the end of the visible light spectrum, where violet light ends:

$\lambda=380 nm =3.8\cdot 10^{-7}m$ (wavelength of lowest-energy ultraviolet light)

So, the lowest energy of ultraviolet light can be found by using the formula

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

Substituting,

$E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.8\cdot 10^{-7} m}=5.23\cdot 10^{-19}J$

And keeping in mind that

$1 eV = 1.6\cdot 10^{-19}J$

This energy converted into electronvolts is

$E=\frac{5.23\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=3.27 eV$

The work function of a metal is the minimum energy needed to extract a photoelectron from the surface of the metal. Therefore, the metals that exhibit photoelectric effect are the ones whose work function is larger than the energy we found previously, so:

b. AG, work function=4.74eV

Because for all the other metals, visible light will be enough to extract photoelectrons.

7 0

3 years ago