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Marrrta [24]
3 years ago
8

Whats the formula of mass? example Density= mass/volume

Physics
2 answers:
DochEvi [55]3 years ago
8 0
There are a few ways to calculate mass. You can do Density * Volume. But you can also do Weight/Gravitational Constant (9.81 m/s2 on earth)
DochEvi [55]3 years ago
4 0
The formula for mass you can find from the density equation. Multiply the volume by both sides of the equation.

density = mass/volume

volume*density = mass 

And there you are :)
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A 2 kg ball is released from rest at the top of a track and
xz_007 [3.2K]
A) The formula for kinetic energy is E = 1/2 mv^2, so the energy of the ball is 1/2 * 2 * 10^2 = 100J. b) Energy is always conserved, and so if no energy is lost to resistive forces then all 100J of kinetic energy came from its potential energy at the top of the track. c) The formula for potential energy is E = mgh, which we can rearrange for h = E/mg. We know the energy, the mass and the strength of gravity, so we can find h = 100 / (2*9.81) = 5.10m.
7 0
3 years ago
Can someone help me?​
Leviafan [203]

Car X traveled 3d distance in t time.  Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t,  the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.

In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

V_y.t = \frac{2d}{t} .t = 2d

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.

4 0
3 years ago
I mostly know how to do these vector problems i’m just confused on how many decimal places i use
Degger [83]
I believe you use 4? I’m really unsure about that but I like your Joy Division picture
5 0
4 years ago
Read 2 more answers
Write the abbreviation for each of the following units: meter, kilometer, centimeter, millimeter,
nikdorinn [45]
M=meter, km=kilometer, mm=millimeter, mg=micrometer, cm=centimeter
3 0
3 years ago
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A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk
Murrr4er [49]

Answer:

\omega_2=5.1rad/s

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is L=I\omega. If we call 1 the situation when the student is at the rim and 2 the situation when the student is at r_2=1.39m from the center, then we have:

L_1=L_2

Or:

I_1\omega_1=I_2\omega_2

And we want to calculate:

\omega_2=\frac{I_1\omega_1}{I_2}

The total moment of inertia will be the sum of the moment of intertia of the disk of mass m_D=119.1 kg and radius r_D=3.23m, which is I_D=\frac{m_Dr_D^2}{2}, and the moment of intertia of the student of mass m_S=54.3kg at position r (which will be r_1=r=3.23m or r_2=1.39m) will be I_{S}=m_Sr_S^2, so we will have:

\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}

or:

\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}

which for our values is:

\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s

6 0
3 years ago
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