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n200080 [17]
2 years ago
12

A car traveling at 40.4 m/s skids to a stop in 3.34 s. Determine the skidding distance of the car. (Assume uniform acceleration

) Be sure to show your work.
Physics
1 answer:
Misha Larkins [42]2 years ago
7 0
Okay so you have to find acceleration first. I used v=vo+at. Plug in the numbers and you get -11.9m/s^2 as your constant acceleration. Then plug acceleration into d=vi(t)+1/2(a)t^2
d=(40m/s)(3.34s)+1/2(-11.9m/s^2)(3.34s)^2
d=68.6m
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2. If a cyclist in the Tour de France traveled southwest a distance of 12,250 meters in one hour, what would the velocity of the
Luda [366]
  • Answer:

<em>12,25 km/h</em>

<em>≈ 3,4 m/s </em>

  • Explanation:

<em>v = d/t</em>

<em>= 12250m/h</em>

<em>= 12,25km/h</em>

<em>or</em>

<em>v = d/t</em>

<em>= 12250m/h</em>

<em>1h = 60m×60s = 3600s</em>

<em>= 12250m/3600s</em>

<em>≈ 3,4 m/s </em>

5 0
2 years ago
A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound
NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

f_n=\frac{nv_s}{4L}

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz

hence, the frequency of the third overtone is 102 Hz

8 0
2 years ago
A particle moves along a straight line with a velocity in meters per second given by v = 12 - 3t + 8t, where t is in seconds. Wh
elena-14-01-66 [18.8K]

Answer:

Explanation:

Given the velocity of a particle modeled by the equation

v = 13-3t+8t² where t is in seconds

Given t = 0 and position s = +5m

A) To get the position as a function of time, we will integrate the function with respect to t ad shown;

v = 13-3t+8t²

S = ∫13-3t+8t² dt

S = 13t-13t²/2+8t³/3 + C

at t = 0 and S = +5m

5 = 13(0)-13(0)²/2+8(0)³/3+C

5 = 0-0+0+C

C = 5

Substituting c = 5 into the displacement function

S = 13t-13t²/2+8t³/3 + C

S = 13t-13t²/2+8t³/3 + 5

B) acceleration is the change in velocity with respect to time.

a = dv/dt

Given v = 13-3t+8t²

a= dv/dt = -3+16t

a = 16-3t

C) acceleration at t = 6s is derived by plugging in t = 6 into the resulting equations in (B)

a = 16-3t

a = 16-3(6)

a = 16-18

a = -2m/s²

D) net displacement from t = 0 to t = 6s

At t = 0:

S(0) = 13(0)-13(0)²/2+8(0)³/3 + 5

S(0) = 0+5

S(0) = 5m

At t = 6s

S(6) = 13(6)-13(6)²/2+8(6)³/3 + 5

S(6) = 78-234+576+5

S(6) = 425m

Net displacement from t = 0s to t = 6s is s(6)-s(0)

= 425-5

= 420m

E) Total distance travelled D = S(6)+S(0)

= 425+5

= 430m

F) Average velocity = ∆S/∆t

Average velocity = S(6)-S(0)/6-0

Average velocity = 425-5/6

Average velocity = 420/6

Average velocity = 70m/s

4 0
2 years ago
Calculate the longest wavelength visible to the human eye. express the wavelength in nanometers to three significant figures.
slega [8]

NOTE: The given question is incomplete.

<u>The complete question is given below.</u>

The human eye contains a molecule called 11-cis-retinal that changes conformation when struck with light of sufficient energy. The change in conformation triggers a series of events that results in an electrical signal being sent to the brain. The minimum energy required to change the conformation of 11-cis-retinal within the eye is about 164 kJ/mole. Calculate the longest wavelength visible to the human eye.

Solution:

Energy (E) = 164 kJ/mole

             E = 164 kJ/mole = 164 kJ /6.023 x 10²³

                = 2.72 x 10⁻²² kJ = 2.72 x 10⁻¹⁹J

Planck's constant = 6.6 x 10⁻³⁴ J s,

Speed of light = 3.00 x 10⁸ m/s

Let the required wavelength be λ.

Formula Used: E = hc / λ

or,                  λ = hc / E

or,                  λ = (6.6 x 10⁻³⁴ J s)× (3.00 x 10⁸ m/s) / (2.72 x 10⁻¹⁹J)

or,                  λ = 7.28 x 10⁻⁷ m

or,                  λ = (7.28 x 10⁻⁷ m) ×( 1.0 x 10⁹ nm / 1.0 m)

or,                  λ = (7.28 x 10² nm)

or,                  λ = 728 nm

Hence, the required wavelength will be 728 nm.

6 0
2 years ago
A car is traveling at 20.0 m/s on tires with a diameter of 70.0 cm. The car slows down to a rest after traveling 300.0 m. If the
cupoosta [38]

Answer: deceleration of 1.904\ rad/s^2

Explanation:

Given

Car is traveling at a speed of u=20 m/s

The diameter of the car is d=70 cm

It slows down to rest in 300 m

If the car rolls without slipping, then it must be experiencing pure rolling i.e. a=\alpha \cdot r

Using the equation of motion

v^2-u^2=2as\\

Insert v=0,u=20,s=300

0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2

Write acceleration as a=\alpha \cdot r

-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2

So, the car must be experiencing the deceleration of 1.904\ rad/s^2.

4 0
2 years ago
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