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3 years ago

What type of relationship exists between acceleration and mass?

1 answer:

7 0

Here, you can derive that by numerical method, as follows:
F = m.a
m = F/a

So, here we can see when we decrease one, other increase by same effect; we can say they are "Indirectly Proportional" to each other!

Hope this helps!

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How do you do this? I know that the Density equation is D=m/v but these are in kilograms and Liters...so do I convert it to gram

BARSIC [14]

Explanation:

The problem doesn't specify that the units have to be g/mL, so you can calculate the density in kg/L without converting the mass or volume.

Just make sure that either way, you write the units.

8 0

3 years ago

To distinguish between properties of the two major types of supernovae: massive star supernovae and white dwarf supernovae. all

const2013 [10]

In my opinion, I myself believe that there are only two supernovae. The first is the white dwarf. It makes sense because if something is too big for its size, it will "explode". Just like a basketball with too much air. Massive star supernovae is like something has reached it's full potential and cannot get any bigger or better.

3 0

3 years ago

A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are

slava [35]

Answer:

The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

Explanation:

Given that,

Mass $m = 1.81\times10^{-3}\ kg$

Velocity $v = (3.00\times10^{4}\ m/s)j$

Charge $q = 1.22\times10^{-8}\ C$

Magnetic field $B= (1.63\hat{i}+0.980\hat{j})\ T$

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

$F = ma=q(v\times B)$

$a =\dfrac{q(v\times B)}{m}$

We need to calculate the value of $v\times B$

$v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})$

$v\times B=4.89\times10^{4}$

Now, put the all values into the acceleration 's formula

$a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}$

$a= -0.3296\ \hat{k}\ m/s^2$

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

7 0

3 years ago

Read 2 more answers

How do quantum numbers relate to electrons? please explain?

alukav5142 [94]

Quantum numbers allow us to both simplify and dig deeper into electron configurations. Electron configurations allow us to identify energy level, subshell, and the number of electrons in those locations. If you choose to go a bit further, you can also add in x,y, or z subscripts to describe the exact orbital of those subshells (for example 2px). Simply put, electron configurations are more focused on location of electrons then anything else.

Quantum numbers allow us to dig deeper into the electron configurations by allowing us to focus on electrons' quantum nature. This includes such properties as principle energy (size) (n), magnitude of angular momentum (shape) (l), orientation in space (m), and the spinning nature of the electron. In terms of connecting quantum numbers back to electron configurations, n is related to the energy level, l is related to the subshell, m is related to the orbital, and s is due to Pauli Exclusion Principle.

7 0

3 years ago

A projectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above the horizontal. It strikes

Alchen [17]

<h2>Answer: x=125m, y=48.308m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

$x=V_{o}cos\theta t$ (1)