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3 years ago

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An automobile is traveling on a long, straight highway at a steady 80.0 mi/h when the driver sees a wreck 190 m ahead. At that i

nstant, she applies the brakes (ignore reaction time). Between her and the wreck are two different surfaces. First there is 100 m of ice, where the deceleration is only 1.20 m/s2 . From then on, it is dry concrete, where the deceleration is a more normal 6.80 m/s2 .
a) What was the car’s speed just after leaving the icy portion of the road?
b)What is the total distance her car travels before it comes to a stop?
c)What is the total time it took the car to stop?

1 answer:

kozerog [31]3 years ago

6 0

Answer:

a) The car’s speed just after leaving the icy portion of the road is the first part

Explanation:

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A motor is the devise

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1- A 30 gram bullet travels at 300 m/s. How much kinetic energy does it have?

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Answer:

1.35 kJ

Explanation:

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Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

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Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

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2 years ago

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Answer:

The waves will increase in frequency

Explanation:

As the young girl moves her hand back and forth faster, it will be observed that number of back and forth motions increase every second. Also the distance between crest and trough of the wave (wavelength) will be reduced as she moves her hand back and forth faster.

Frequency = number of turns (moves) per second

The waves will increase in frequency since there will be more number of back and forth motions in every second.

Also,

The distance between crest and trough will be reduced, which implies that there will be decrease in waves wavelength.

This can also be verified using wave equation;

V = Fλ

At constant velocity,

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The right answer is : The waves will increase in frequency

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3 years ago