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cupoosta [38]
3 years ago
13

According to recent typical test data, a Ford Focus travels 0.240 mi in 19.3 s, starting from rest. The same car, when braking f

rom 62.5 mph on dry pavement, stops in 150 ft. Assume constant acceleration in each part of its motion, but not necessarily the same acceleration when slowing down as when speeding up.(a) Find the magnitude of this car's acceleration while braking.(b) Find the magnitude of this car's acceleration while speeding up.(c) If its acceleration is constant while speeding up, how fast (in mph) will the car be traveling after 0.250 mi of acceleration?(d) How long does it take the car to stop while braking from 59.0 mph?
Physics
1 answer:
Anit [1.1K]3 years ago
4 0

Answer:

Explanation:

a )

While breaking initial velocity u = 62.5 mph

= 62.5 x 1760 x 3 / (60 x 60 )  ft /s

= 91.66 ft / s

distance trvelled s = 150 ft

v² = u² - 2as

0 = 91.66²  - 2 a x 150

a = - 28 ft / s²

b ) While accelerating initial velocity u = 0

distance travelled s = .24 mi

time = 19.3 s

s = ut + 1/2 at²

s is distance travelled in time t with acceleration a ,

.24 = 0 + 1/2 a x 19.3²

a = .001288 mi/s²

= 2.06 m /s²

c )

If distance travelled s = .25 mi

final velocity v = ? a = .001288 mi / s²

v² = u² + 2as

= 0 + 2 x .001288 x .25

= .000644

v = .025 mi / s

= .0025 x 60 x 60 mi / h

= 91.35 mph .

d ) initial velocity u = 59 mph

= 86.53 ft / s

final velocity = 0

acceleration = - 28 ft /s²

v = u - at

0 = 86.53 - 28 t

t = 3 sec approx .

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storchak [24]

Answer:

v = \frac{kQ}{a}  

Explanation:

We define the linear density of charge as:

\lambda = \frac{Q}{L}

     Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

dv = \frac{kdq}{r}

          where k is the coulomb's constant

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Thus.

v = \int_{}^{}\frac{kdq}{a}  

v = \frac{k}{a}\int_{}^{}dq

v = \frac{kQ}{a}     Potential at the center of the semicircle

4 0
3 years ago
The speed of sound in room temperature (20°C) air is 343 m/s; in room temperature helium, it is 1010 m/s. The fundamental freque
Lera25 [3.4K]

Answer: f = 927.55Hz

Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below

L = λ/4 where λ = wavelength.

speed of sound in air = v = 343m/s.

fundamental frequency of open closed tube = 315Hz

λ = 4L.

v = fλ

343 = 315 * 4L

343 = 1260 * L

L = 343/ 1260

L = 0.27m

In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.

The length of tube and wavelength are related by the formulae below

L = λ/4, λ=4L

λ = 4 * 0.27

λ = 1.087m.

v = fλ

1010 = f * 1.087

f = 1010/1.807

f = 927.55Hz

4 0
3 years ago
A car travels 240km in 4h what’s the cars velocity
Tresset [83]

The car's speed is 240km/4hr= 60km/hr.

There's not enough information given in the question to determine its velocity.

6 0
3 years ago
Read 2 more answers
If the equation of an average velocity of a sport car was given by the equation V(t) = 3t^2 -6t +24. Determine its displacement
alisha [4.7K]

Answer:

72

Explanation:

The displacement of an object can be found from the velocity of the object by integrating the expression for the velocity.

In this problem, the velocity of the sport car is given by the expression

v(t)=3t^2-6t+24

In order to find the expression for the position of the car, we integrate this expression. We find:

x(t)=\int v(t) dt=t^3-3t^2+24t+C

where C is an arbitrary constant.

Here we want to find the displacement after 3 seconds. The position at t = 0 is

x(0)=0^3-0+0+C=C

While the position after t = 3 s is

x(3)=3^3-3(3)^2+24(3)+C=72+C

Therefore, the displacement of the car in 3 seconds is

d=x(3)-x(0)=72+C-C=72

7 0
3 years ago
What is 1113.28 in a sig fig?
Afina-wow [57]

Answer:

6 significant figure

Explanation:

The digits 111328 all are 6 figures with no figure being zero, neither zero after the other digits. In this case, all the numbers are significant and since they are only six numbers, then this is a six significant figure. In case we add another zero after digit 8, the zero is not significant but if added either infront of 8 or 2, the zero becomes significant.

4 0
3 years ago
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