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2 years ago

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A rocket carrying a satellite is accelerating straight up from the earth’s surface. At 1.15 s after liftoff, the rocket clears t

he top of its launch platform, 63 m above the ground. After an additional 4.75 s, it is 1.00 km above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75-s part of its flight and (b) the first 5.90 s of its flight.

1 answer:

yarga [219]2 years ago

8 0

Answer:

197.263157895 m/s

169.491525424 m/s

Explanation:

x Denotes position

t Denotes time

Average velocity is given by

$v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-63}{4.75}\\\Rightarrow v_a=197.263157895\ m/s$

The average velocity is 197.263157895 m/s

$v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-0}{5.9}\\\Rightarrow v_a=169.491525424\ m/s$

The average velocity is 169.491525424 m/s

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Read 2 more answers

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

The force between charges separated 2d is

$\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}$

$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

a.

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

b.

The ratio of the greatest to the smallest net force is

$\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9$

The greatest force is 9 times the smallest net force

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The carnot cycle attempts to model the most efficient possible process by avoiding irreversible processes.

In essence, the Carnot cycle is a reversible cycle made up of four other reversible processes. A reversible process is one that can be thought of as consisting of a sequence of equilibrium stages because it is carried out endlessly slowly.

Essentially, this means that any reversible cycle can be performed in reverse and that the amount of work or heat exchanged along the forward and backward pathways is the same.

It goes without saying that such reversible processes are not possible because they would take an unlimited amount of time. Therefore, the Carnot Engine is described as an idealized heat engine that uses the Carnot Cycle, a reversible cycle.

Learn more about carnot cycle here;

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9 months ago

The temperature inside a vehicle can rise ____ degrees higher than the outside temperature.

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Answer:

Generally, 40 to 50 degrees

Explanation:

About the heat-up over time, whether the windows of a vehicle are locked or partially open makes very little difference. In both situations, in an internal temperature of a vehicle, even at an outside temperature of only 72 ° F, it may exceed approximately 40 ° F within one hour. This happens mainly due to the greenhouse effect that is the heat inside the car is trapped and not allowed to escape. Thus, raising the temperature of the vehicle.

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3 years ago