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2 years ago

A car moving 12 m/s stops in 3 seconds vf=

Physics

2 answers:

Alla [95]2 years ago

4 0

Your question is not complete
For this I can answer only this
That it can stop in 3 seconds

allsm [11]2 years ago

4 0

Answer:

The question seems to be looking for vf, which may be the final velocity.

We know that the car stops, so the car is at rest, this means that the final velocity is equal to zero.

vf = 0 m/s

Now, we also can find the acceleration, the velocity equation can be written as:

v(t) = a*t + 12m/s

We know that in 3 seconds the car stops, so we have:

v(3s) = 0 = a*3s + 12m/s

a= -12m/s*(1/3s) = -4m/s^2

So in those 3 seconds where the car is stoping, the acceleration of the car is -4m/s^2, and the velocity of the car in those 3 seconds can be described by the equation:

v(t) = -4m/s^2*t + 12m/s

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Answer:

D an B

Explanation:

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2 years ago

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Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm sli

Orlov [11]

Answer:

a.3.20m

b.0.45cm

Explanation:

a. Equation for minima is defined as: $sin \theta=\frac{m\lambda}{\alpha}$

Given $m=3$ , $\lambda=6.33\times 10^-^7$ and $\alpha=0.00015$ :

#Substitute our variable values in the minima equation to obtain $\theta$ :

$\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad$

#draw a triangle to find the relationship between $\theta, y \$ and $L$ .

$tan(\theta)=y/L$ #where $y=4.05cm$

$L=y/tan(\theta)=3.20$

Hence the screen is 3.20m from the split.

b. To find the closest minima for green(the fourth min will give you the smallest distance)

#Like with a above, the minima equation will be defined as:

$sin \theta=\frac{m\lambda}{\alpha}$ , where $m=4$ given that it's the minima with the smallest distance.

$sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad$

#we then use $tan(\theta)=y/L$ to calculate $L$ =4.5cm

Then from the equation subtract $y_3$ from $y$ :

$4.50cm-4.05cm=0.45cm$

Hence, the distance $\bigtriangleup y$ is 0.45cm

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2 years ago

What phenomenon does this image demonstrate? *

Ostrovityanka [42]

Answer:

Constructive Interference

Explanation:

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2 years ago

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Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

schepotkina [342]

If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!

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2 years ago

Water at 20°C flows by gravity through a smooth pipe from one reservoir to a lower one. The elevation difference is 60 m. The pi

Serga [27]

Answer:

Flow Rate = 80 m^3 /hours (Rounded to the nearest whole number)

Explanation:

Given

Hf = head loss
f = friction factor
L = Length of the pipe = 360 m
V = Flow velocity, m/s
D = Pipe diameter = 0.12 m
g = Gravitational acceleration, m/s^2
Re = Reynolds's Number
rho = Density =998 kg/m^3
μ = Viscosity = 0.001 kg/m-s
Z = Elevation Difference = 60 m

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Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)

The energy equation for this system will be,

Hp = Z + Hf

The other three equations to solve the above equations are:

Re = (rho*V*D)/ μ

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Power = 15000 W = rho*g*Q*Hp

1/f^0.5 = 2*log ((Re*f^0.5)/2.51)

We can iterate the 5 equations to find f and solve them to find the values of:

Re = 235000

f = 0.015

V = 1.97 m/s

And use them to find the flow rate,

Q = V*(pi/4)*D^2

Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours

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2 years ago