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Vanyuwa [196]
3 years ago
13

Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m

/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s. How much time does the driver of the car measure for his trip between the poles?
Physics
1 answer:
Oksi-84 [34.3K]3 years ago
4 0

Answer:

Observed time, t = 5.58 s  

Explanation:

Given that,

Speed of light in a vacuum has the hypothetical value of, c = 18 m/s

Speed of car, v = 14 m/s along a straight road.

A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s.

We need to find the time the driver of the car measure for his trip between the poles. The relation between real and observed time is given by :

T=\dfrac{t}{\sqrt{1-\dfrac{v^2}{c^2}} }

t is observed time.

t=T\times \sqrt{1-\dfrac{v^2}{c^2}} \\\\t=8.89\times \sqrt{1-\dfrac{14^2}{18^2}} \\\\t=5.58\ s

So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.

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In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 2.2 rev/
Hunter-Best [27]

Answer:

n_{T} = 31.68\,rev

Explanation:

The angular acceleration is:

\ddot n_{1} = \frac{2.2\,\frac{rev}{s} -0\,\frac{rev}{s} }{8.8\,s}

\ddot n_{1} = 0.25\,\frac{rev}{s^{2}}

And the angular deceleration is:

\ddot n_{2} = \frac{0\,\frac{rev}{s}-2.2\,\frac{rev}{s} }{20\,s}

\ddot n_{2} = -0.11\,\frac{rev}{s^{2}}

The total number of revolutions is:

n_{T} = n_{1} + n_{2}

n_{T} = \frac{\left(2.2\,\frac{rev}{s} \right)^{2}-\left(0\,\frac{rev}{s} \right)^{2}}{2\cdot \left(0.25\,\frac{rev}{s^{2}} \right)} + \frac{\left(0\,\frac{rev}{s} \right)^{2}-\left(2.2\,\frac{rev}{s} \right)^{2}}{2\cdot \left(-0.11\,\frac{rev}{s^{2}} \right)}

n_{T} = 31.68\,rev

4 0
3 years ago
You have two vectors, which are 2.59 m at 30.0° north of east and 4.18 m at 60.0° north of west. What is the magnitude in meters
andrew11 [14]

Answer:\ec{r}=0.153\hat{i}+4.914\hat{j}v

Explanation:

Given

Vector 1

\vec{a}=2.59\left ( cos30\hat{i}+sin30\hat{j}\right )

Vector 2

\vec{b}=4.18\left ( -cos60\hat{i}+sin60\hat{j}\right )

Resultant \vec{r}=\vec{a}+\vec{b}

\vec{r}=2.59\left ( cos30\hat{i}+sin30\hat{j}\right )+4.18\left ( -cos60\hat{i}+sin60\hat{j}\right )

\vec{r}=0.153\hat{i}+4.914\hat{j}

|r|=4.916

5 0
3 years ago
A car is driven 13 mi east and then certain distance due north, ending up at the position 25 degrees north of east of its initia
Kobotan [32]

Answer: Hello mate!

lets define the north as the y-axis and east as the x-axis.

Using the notation (x,y) we can define the initial position of the car as (0,0)

then the car travells 13 mi east, so now the position is (13,0)

then the car travels Y miles to the north, so the position now is (13, Y)

and we know that the final position is 25° degrees north of east of the initial position. This angle says that the distance traveled to the north is less than 13 mi because this angle is closer to the x-axis (or east in this case).

This angle is measured from east to north, then the adjacent cathetus is on the x-axis, in this case, 13mi

And we want to find the distance Y, so we can use the tangent:

Tan(25°) = Y/13  

tan(25°)*13 mi = Y = 6.06 mi.

3 0
3 years ago
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