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3 years ago

Calculate the period (T) of uniform circular motion if the velocity is 40.0 m/s and centripetal acceleration is 20.0 m/s2.

1 answer:

3 0

T is the time for a whole round.

centripetal acceleration = V^2/R,

20 = 40^2 / R, find R = 40^2/20 = 40*40/20 = 80 m, right?

Now, one round is L = 2*pi*R = 2*pi*80 = 160*pi

And T = L/v (distance/speed) = 160*pi/40 = 4*pi seconds, or ~ 12.57 s

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Write down the condition required for the thermonuclear fusion.

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Explanation:

The temperature must be hot enough to allow the ions of deuterium and tritium to have enough kinetic energy to overcome the Coulomb barrier and fuse together. The ions must be confined with a high ion density to achieve a suitable fusion reaction rate.

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2 years ago

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Based on the free-body diagram, the net force acting<br> on this firework is

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0N. The net force acting on this firework is 0.

The key to solve this problem is using the net force formula based on the diagram shown in the image. Fnet = F1 + F2.....Fn.

Based on the free-body diagram, we have:

The force of gases is Fgases = 9,452N

The force of the rocket Frocket = -9452

Then, the net force acting is:

Fnet = Fgases + Frocket

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3 years ago

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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

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

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

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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3 years ago

A non-conducting sphere of radius R = 3.0 cm carries a charge Q = 2.0 mC distributed uniformly throughout its volume. At what di

BlackZzzverrR [31]

Answer:

$r =3 *\sqrt{2} = 4.24 cm$

Explanation:

given data

Radius of sphere 3.0 cm

charge Q = 2.0 m C

We know that maximum electric field is given as

$E_{MAX}= \frac{KQ}{r^{2}}$

electric field inside the sphere can be determine by using below relation

$\frac{KQ}{r^{2}}= \frac{1}{2}*\frac{KQ}{R^{2}}$

$r = \sqrt{2}R$

$r =3 *\sqrt{2} = 4.24 cm$

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UHHH WHAT? I DONT GET THAT AT ALLOW

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3 years ago