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iris [78.8K]
3 years ago
6

Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a

ccelerations. (a) a = -4.0 m/s^2 (b) a = -8.0 m/s^2
Physics
1 answer:
seropon [69]3 years ago
4 0

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, d=24.44\times 2=48.88\ m

(a) Acceleration, a=-4\ m/s^2

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{-(24.44)^2}{2\times -4}

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, a=-8\ m/s^2

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{-(24.44)^2}{2\times -8}

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

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A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.4 ft/s,
Art [367]

Answer:

\dfrac{d\theta}{dt} =-0.233\ rad/s

Explanation:

given,

length of ladder = 10 ft

let x be the distance of the bottom and y be the distance of the top of ladder.

x² + y² = 100

differentiating with respect to time we get

2 x\dfrac{dx}{dt}+2y\dfrac{dy}{dt} = 0..............(1)

when x = 8 and y = 6 and when \dfrac{dx}{dt} = 1.4ft/s

from equation (1)

now,

16\times 1.4 + 12\dfrac{dy}{dt} = 0

\dfrac{dy}{dt} = -\dfrac{5.6}{3}

let the angle between the ladders be θ

tan\theta = \dfrac{y}{x}

y = xtan θ

\dfrac{dy}{dt} =\dfrac{dy}{dt} tan\theta + x sec^2\theta\dfrac{d\theta}{dt}

-\dfrac{5.6}{3} =1.4\times \dfrac{6}{8} + 8 (1+\dfrac{9}{16})\dfrac{d\theta}{dt}

\dfrac{25}{2} \dfrac{d\theta}{dt} =\dfrac{-17.5}{6}

\dfrac{d\theta}{dt} =-0.233\ rad/s

6 0
3 years ago
A wooden block has a mass of 20.0 kg and a specific heat of 1700 J/kg °C. Find the change in thermal energy of the block as it w
andrew-mc [135]
M = 20.0kg, the mass of the block.
c = 1700 J/(kg-°C), the specific heat
ΔT = 25 - 15 = 10 °C = 10 K, the change in temperature.

The change in thermal energy is
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Answer: 340 kJ (or 340,000 J)

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3 years ago
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MrRissso [65]

Answer:

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Explanation:

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3 years ago
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An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft
sergeinik [125]

Answer:

a) 80 V

b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

<u>Given :</u>

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( K_{A}= 0) to the point B at distance l = 0.500m where its kinetic energy is (  K_{B}= 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

<u>Required :</u>

<em>(a) We are asked to find the electric potential VB </em>

<em>(b) We want to determine the magnitude and the direction of the electric field E. </em>

<u> Solution </u>

(a) We are given the values for VA,K_{B} and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

                                   K_{A} +U_{A} =K_{B} +U_{B} .........................................(1)                                          

Where K_{A}= 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential.  So, equation (1) will be in the form :

                                  0+qVA=K_{B} +qVB                      (Divide by q)

                                         VA=K_{B} /q + VB                  (solve for VB)

                                         VB=VA- K_{B}/q .......................................(2)

We get the relation between VB, VA and K_{B}, now we can plug our values for VA, K_{B} and q into equation (2) to get VB

                                         VB=VA- K_{B}/q

                                              =30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

                                              =80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

                                   VA-VB =\int\limits^1_0 {E} \, dl...................................(a)

                                               =E\int\limits^1_0 {} \, dl

                                   VA-VB=El                      (solve for E)

                                            E= VA-VB/l..................................(3)

Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E

                                            E= VA-VB/l

                                              =-100 N/C

The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

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Answer:

110 yds

Explanation:

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