Ask question

Ask question

All categories

Katyanochek1 [597]

3 years ago

9

According to the Guinness Book of World Records (1999) the highest rotary speed ever attained was 2010 m/s (4500 mph) The rotati

ng rod was 15.3 cm long. Assume that the speed quoted is that of the end of the rod. a. What is the centripetal acceleration of the end of the rod? (2.64 x 107 m/s2) b. If you were to attach a 1.0 g object to the end of the rod, what force would be needed to hold it on the rod?

1 answer:

Zinaida [17]3 years ago

7 0

Answer:

a. 2.645 * 10^7 m/s^2

b. 2.645 * 10^4 N

Explanation:

Parameters given:

Velocity of rod = 2010m/s

Length of rod = 15.3cm = 0.153m

Mass of object placed at the end of the rod = 1g = 0.001kg

a. Centripetal acceleration is given as:

a = (v*v)/r

Where v = velocity

r = radius of curvature.

The radius of curvature in this case is equal to the length of the rod, since the rod makes the circular path of the motion.

Hence, centripetal acceleration at the end of the rod:

a = (2010*2010)/(0.153)

a = 26432156.86 m/s^2 = 2.64 * 10^7 m/s

b. The force needed to hold the object at the end of the rod is equal to the centripetal force at the end of the rod. Centripetal force is given as:

F = ma = (m*v*v)/r

Where a = centripetal acceleration

F = 0.001 * 2.64 * 10^7

F = 2.64 * 10^4N

You might be interested in

Find the force necessary to pull a 6 kg object 3 m/s2

mars1129 [50]

Answer:

<h2>18 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 6 × 3

We have the final answer as

<h3>18 N</h3>

Hope this helps you

6 0

3 years ago

Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q

scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

a E = $1.685*10^3 N/C$

b E = $36.69*10^3 N/C$

c E = 0 N/C

d $V = 6.7*10^3 V$

e $V = 26.79*10^3V$

f V = $34.67 *10^3 V$

g $V= 44.95*10^3 V$

h $V= 44.95*10^3 V$

i $V= 44.95*10^3 V$

Explanation:

From the question we are given that

The first charge $q_1 = 2.00 \mu C = 2.00*10^{-6} C$

The second charge $q_2 =1.00 \muC = 1.00*10^{-6}$

The first radius $R_1 = 0.500m$

The second radius $R_2 = 1.00m$

$Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}$

And $Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]$

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

r = 4.00 m

$E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}$

$= 1.685*10^3 N/C$

Considering b

$r = 0.700 m \ , R_2 > r > R_1$

This implies that the electric field would be

$E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}$

This because it the electric filed of the charge which is below it in distance that it would feel

$E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}$

= $36.69*10^3 N/C$

Considering c

r = 0.200 m

=> $r$

The electric field = 0

This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

Considering d

r = 4.00 m

=> $r > R_1 >r>R_2$

Now the potential difference is

$V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V$

This so because the distance between the charge we are considering is further than the two charges given

Considering e

r = 1.00 m $R_2 = r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V$

Considering f

$r = 0.700 m \ , R_2 > r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V$

Considering g

$r =0.500\m , R_1 >r =R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering h

$r =0.200\m , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering i

$r =0\ m \ , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

8 0

3 years ago

If you add 1 proton to Carbon, it will no longer be Carbon, it will be ________________.

Whitepunk [10]

Nitrogen

Explanation:

Adding one proton to a carbon atom makes Nitrogen.

A quick introspection on atoms:

An atom is made up of three fundamental particles.
They are protons, neutrons and electrons.
The protons are positively charged and the neutrons do not carry any charges.
Electrons are negatively charged.

The difference between an atom and another is the number of protons in them. This is the atomic number.

The periodic table of element is a list of elements arranged based on the number of protons they have. Every element on the table has unique number of protons which makes it differ from another.

Atoms do not readily lose their protons because they are held by nuclear forces in the nucleus of an atom.

When an element gains a proton, it becomes another element.

Carbon has proton number of 6

If a proton is added to it, it becomes 7

This is the proton or atomic number of nitrogen.

Learn more:

Atomic number brainly.com/question/5425825

#learnwithBrainly

4 0

3 years ago

Which of the following is not true of a form?

Ludmilka [50]

Answer:

It can only display one record at a time

Explanation:

Form ;

1. This is a document with spaces (also called placeholders or fields ) in which a series of documents with similar content can be written or selected.

2.This is the most popular method of data entry

3.It may contain images in the background.

4.This can be sorted data regardless of its source of information.

Only option C is wrong.

Therefore the answer C is correct.

4 0

3 years ago

Question 6 (1 point)

jarptica [38.1K]

Explanation:

decrease,then increase

8 0

3 years ago