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Natasha_Volkova [10]
3 years ago
13

A driver is traveling along a straight road at the speed limit of 60 mph. After two minutes, she slows at a constant rate to a s

top at a stop light. Two minutes later, the light turns green and she accelerates at a constant rate back up to the speed limit. Three and a half minutes later, she again slows at a constant rate to a stop. After three minutes, she performs a U-turn, then accelerates at a constant rate back up to 60 mph. Two minutes later, she reaches her destination and slows at a constant rate to a stop. Assume that each period of slowing down and speeding up lasts 30 s and that the driver is initially moving in the +x- direction. Create a graph of the driver's velocity versus time that represents her trip.

Physics
1 answer:
Yuri [45]3 years ago
7 0

Explanation :

From the given information, the graph is plotted.

It is given that, a driver is traveling along a straight road at the speed limit of 60 mph. Initially, he was at point A.

After two minutes, she slows at a constant rate to a stop at a stop light. BC shows this part.

Two minutes later, the light turns green and she accelerates at a constant rate back up to the speed limit. CD shows this part.

Three and a half minutes later, she again slows at a constant rate to a stop. DE shows this part.

Hence, this is the required solution.

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A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d
I am Lyosha [343]

Answer:

2.09\ \text{m/s}

22298.4\ \text{J}

Explanation:

m = Mass of each the cars = 1.6\times 10^4\ \text{kg}

u_1 = Initial velocity of first car = 3.46 m/s

u_2 = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}

Speed of the three coupled cars after the collision is 2.09\ \text{m/s}.

As energy in the system is conserved we have

K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}

The kinetic energy lost during the collision is 22298.4\ \text{J}.

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The part of earth where all living things are found is called the
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Answer:

Hey Dude....

Explanation:

This is ur answer.....

<em>The biosphere is made up of the parts of Earth where life exists. The biosphere extends from the deepest root systems of trees to the dark environment of ocean trenches, to lush rain forests and high mountaintops.</em>

Hope it helps!

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Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00
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Answer:

2.62898\times 10^{-6}\ C/m^3

1979.99974\ N/C

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C

Volume charge density is given by

\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3

The volume charge density for the sphere is 2.62898\times 10^{-6}\ C/m^3

E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C

The magnitude of the electric field is 1979.99974\ N/C

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careful observation, curiosity, logic, creativity, skepticism, and objectivity

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