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icang [17]
3 years ago
14

Upon completing an interview, it is important that you send a follow-up thank you

Physics
2 answers:
Schach [20]3 years ago
8 0
A, it makes you seem like your a better person and grateful for the opportunity you have.
fomenos3 years ago
4 0
A True it shows that you are very grateful for the opportunity and it shows that you really would like the job and you’re not just doing it for the money even if you are. Also it just shows professionalism
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PLEASE HLEP I WILL GIVE BRAINLIEST TO CORRECT ANSWER!!!!!!
Hoochie [10]

Answer:

75 km/h

Explanation:

Speed = Distance divided by Time.

The train went a distance of 300 km in 4 hours. Therefore, the speed of the train is: 300 divided by 4 = 75 km/h

8 0
3 years ago
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2. Neutrons have a ____<br> charge.<br> a. positive<br> b. negative<br> c. neutral
AnnyKZ [126]

Answer:

b. negative

Explanation:

neutrons have a negative charge and protons have a proton has a positive charge

3 0
2 years ago
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Which term best describes the basic unit that makes up all matter?
neonofarm [45]

Answer:

C. Atoms

Explanation:

The basic unit of matter and the smallest, indivisible unit of a chemical element. It comprises a nucleus (neutrons + protons) that is surrounded by a cloud of electrons.

6 0
2 years ago
Name three methods that can be used to change the overall density of an object.
Gnesinka [82]
<span> changing its shape
changing its mass
or changing its volume</span>
6 0
3 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
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