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3 years ago

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There are very, very few times that scientists make estimations because they don’t have enough time or data to test all the poss

ible answers to the question they have. TRUE OR FALSE

1 answer:

Zanzabum3 years ago

6 0

True, it’s limited to when they use estimations at most

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Which is not true about the variables in the relationship F = ma?

tatuchka [14]

Answer:

A

Explanation:

Because D is definitely true and there is only one false sentence what means if that non of B or C is false because if one is false so other one needs to be too.

8 0

3 years ago

How is it possible for a person watching a baseball game on TV to hear sounds before someone who is actually watching the same g

Andreyy89

Answer:

It is possible because, the TV broadcast audio and video signals in radio frequency which travels at the speed of light while the audio signals travel to those present in the stadium at the speed of sound which is over eight hundred thousand times slower than the speed of light

Explanation:

It is possible because of the following;

1. TV signals from the camera (including the captured sound) very close to the field of play are transmitted through the radio frequency bands and as such are a form of electromagnetic radiation that travels at the speed of light which is about 300,000 km/second

It will therefore, take 1 second for a sound of the game to reach someone located at 300,000,000 meters watching a live televised game

2. The speed of sound is about 343 m/second and it therefore takes up to 2 seconds for a sound to reach someone 686 meters away from the ball in the stadium.

5 0

3 years ago

Nuclearissionoccursthroughmanydifferent pathways. For the ission of U-235 induced by a neutron, write a nuclear equation to form

ruslelena [56]

Answer: a) $^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n$

b) $^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}$

Explanation:

A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.

a) The given reaction is $^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + x^{1}_{0}n$

Now, as the mass on both reactant and product side must be equal:

$235+1=87+146+x$

$x=3$

Thus three neutrons are produced and nuclear equation will be: $^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n$

b) For the another fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{A}_{56}\textrm{Ba}+^{94}_{Z}\textrm{X}+2^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 2

A = 140

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = 56 + Z + 0

Z = 36

As Krypton has atomic number of 36,Thus the nuclear equation will be :

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}$

8 0

3 years ago

The Sun orbits the center of the Milky Way galaxy once each 2.60 × 108 years, with a roughly circular orbit averaging 3.00 × 104

Mamont248 [21]

To solve this problem it is necessary to apply the kinematic equations of linear and angular motion, as well as the given definitions of the period.

Centripetal acceleration can be found through the relationship

$a_c = \frac{v^2}{R}$

Where

v = Tangential Velocity

R = Radius

At the same time linear velocity can be expressed in terms of angular velocity as

$v = R\omega$

Where,

R = Radius

$\omega =$ Angular Velocity

PART A) From this point on, we can use the values used for the period given in the exercise because the angular velocity by definition is described as

$\omega = \frac{2\pi}{T}$

T = Period

So replacing we have to

$\omega = \frac{2\pi}{2.6*10^8years}\\\omega = 2.4166*10^{-8}rad/years\\\omega = 2.4166*10^{-8}rad/years(\frac{1years}{365days})(\frac{1day}{86400s})\\\omega = 7.663*10^{-16}rad/s$

Since $1 Light year = 9.48*10^{15}m$

Then the radius in meters would be

$R = (3*10^4ly)(\frac{9.48*10^{15}m}{1ly})$

$R = 2.844*10^{20}m$

Then the centripetal acceleration would be

$a_c = \frac{v^2}{R}\\a_c = \frac{(R\omega)^2}{R}\\a_c = R\omega^2 \\a_c = 2.844*10^{20}(7.663*10^{-16})^2\\a_c = 1.67*10^{-10}m/s^2$

From the result obtained, considering that it is an unimaginably low value of an order of less than $10^{-10}$ it is possible to conclude that it supports the assertion on the inertial reference frame.

8 0

3 years ago

You are a project manager for a manufacturing company. One of the machine parts on the assembly line is a thin, uniform rod that

ra1l [238]

Answer:

a) $I=0.012\ kg.m^2$

b) $I=0.012\ kg.m^2$

Explanation:

Given that:

mass of rod, $m=0.4\ kg$

length of the rod, $l=0.6\ kg$

<u>(a)</u>

<u>Moment of inertia of rod about its center and perpendicular to the rod is given as:</u>

$I=\frac{1}{12} m.l^2$

$I=\frac{1}{12} 0.4\times 0.6^2$

$I=0.012\ kg.m^2$

(b)

<u>Moment of inertia on bending the rod to V-shape of 60 degree angle and axis being perpendicular to the plane of V at the vertex.</u>

<em>We treat it as two rod with axis of rotation at the end and perpendicular to the plane of rotation. </em>

<em>So, the mass and the length of the rod will become half of initial value.</em>

$I=I_1+I_2$

$I=\frac{1}{3} \frac{m}{2} (\frac{l}{2})^2 +\frac{1}{3} \frac{m}{2} (\frac{l}{2})^2$

$I=2[ \frac{1}{3}\times 0.2\times 0.3^2]$

$I=0.012\ kg.m^2$

7 0

3 years ago