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Fittoniya [83]
3 years ago
7

You are the juror of a case involving a drunken driver whose 1041 kg sports car ran into a stationary 1928 kg station wagon stop

ped at a red traffic light. the cars stuck together and slid with locked wheels for 12.0 m before coming to rest. the coefficient of sliding friction on the dry road was 0.60. estimate the speed of the sports car when it hit the station wagon.

Physics
1 answer:
mina [271]3 years ago
7 0
1. Find the force of friction between the sports car and the station wagon stuck together and the road. The total mass m = 1928kg + 1041kg = 2969kg. The only force in the x-direction is friction: F = μ*N = μ * m * g 
2. Find the acceleration due to friction: 
F = m*a =  μ * m * g => a = μ * g = 0.6 * 9.81
3. Find the time it took the two cars stuck together to slide 12m:
x = 0.5*a*t² 
t = sqrt(2*x / a) = sqrt(2 * x / (μ * g) )
4. Find the initial velocity of the two cars:
v = a*t = μ * g * sqrt(2 * x / (μ * g) ) = sqrt( 2 * x * μ * g)
5. Use the initial velocity of the two cars combined to find the velocity of the sports car. Momentum must be conserved:

m₁ mass of sports car
v₁ velocity of sports car before the crash
m₂ mass of station wagon
v₂ velocity of station wagon before the crash = 0
v velocity after the crash

m₁*v₁ + m₂*v₂ = (m₁+m₂) * v = m₁*v₁ 
v₁ = (m₁+m₂) * v / m₁ = (m₁+m₂) * sqrt( 2 * x * μ * g) / m₁
v₁ = 33.9 m/s


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Lemur [1.5K]

Answer:

a)   v = 0.9167 m / s,  b)  A = 0.350 m,  c)  v = 0.9167 m / s, d)  A = 0.250 m

Explanation:

a) to find the velocity of the wave let us use the relation

          v = λ f

the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength

           λ = x

           λ = x

the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period

          T / 2 = t

           T = 2t

period and frequency are related

           f = 1 / T

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we substitute

           v = x / 2t

           v = 5.50 / 2 3

           v = 0.9167 m / s

b) the amplitude is the distance from a maximum to zero

          2A = y

           A = y / 2

           A = 0.700 / 2

           A = 0.350 m

c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same

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d) the amplitude is

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4 0
3 years ago
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

We know that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

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3 0
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v = \frac{D}{t}

here

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t = total time of the motion

so here we will say that there is no such relation between initial or final speed or we can say maximum or minimum speed of object with average speed of object.

We only need to find the total distance and total time of motion in order to find the average speed

here we can see many examples like let say an object moves with speed v1 for time t1 and then with speed v2 for time t2 then here average speed is given as

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since we know that distance covered is product of speed and time

that's why we used above equation for finding total distance

now the average speed will be

v_{avg} = \frac{v_1t_1 + v_2 t_2}{t_1 + t_2}

so this is how we can find the average speed for above motion

so average speed is always between maximum and minimum speed any value in-between.

It is neither the maximum value nor it is minimum value

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3 years ago
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