Answer: Here this will help you..
Explanation:
1 kg-m/s to kilogram-force meter/second = 1 kilogram-force meter/second
5 kg-m/s to kilogram-force meter/second = 5 kilogram-force meter/second
10 kg-m/s to kilogram-force meter/second = 10 kilogram-force meter/second
20 kg-m/s to kilogram-force meter/second = 20 kilogram-force meter/second
30 kg-m/s to kilogram-force meter/second = 30 kilogram-force meter/second
40 kg-m/s to kilogram-force meter/second = 40 kilogram-force meter/second
50 kg-m/s to kilogram-force meter/second = 50 kilogram-force meter/second
75 kg-m/s to kilogram-force meter/second = 75 kilogram-force meter/second
100 kg-m/s to kilogram-force meter/second = 100 kilogram-force meter/second
Answer:
Heat can travel from one place to another in three ways: Conduction, Convection and Radiation. Both conduction and convection require matter to transfer heat. If there is a temperature difference between two systems heat will always find a way to transfer from the higher to lower system.
Explanation:
Hope this helps
There is no image!?...was there meant to be something attached?
Answer:
Explanation:
Given that,
Force is downward I.e negative y-axis
F = -2 × 10^-14 •j N
Magnetic field is westward, +x direction
B = 8.3 × 10^-2 •i T
Charge of an electron
q = 1.6 × 10^-19C
Velocity and it direction?
Force in a magnetic field is given as
F = q(V×B)
Angle between V and B is 270, check attachment
The cross product of velocity and magnetic field
F =qVB•Sin270
2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2
Then,
v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)
v = 1.51 × 10^6 m/s
Direction of the force
Let x be the direction of v
-F•j = v•x × B•i
From cross product
We know that
i×j = k, j×i = -k
j×k =i, k×j = -i
k×i = j, i×k = -j OR -k×i = -j
Comparing -k×i = -j to given problem
We notice that
-F•j = q ( -V•k × B×i)
So, the direction of V is negative z- direction
V = -1.51 × 10^6 •k m/s
Answer: hope it helps you...❤❤❤❤
Explanation: If your values have dimensions like time, length, temperature, etc, then if the dimensions are not the same then the values are not the same. So a “dimensionally wrong equation” is always false and cannot represent a correct physical relation.
No, not necessarily.
For instance, Newton’s 2nd law is F=p˙ , or the sum of the applied forces on a body is equal to its time rate of change of its momentum. This is dimensionally correct, and a correct physical relation. It’s fine.
But take a look at this (incorrect) equation for the force of gravity:
F=−G(m+M)Mm√|r|3r
It has all the nice properties you’d expect: It’s dimensionally correct (assuming the standard traditional value for G ), it’s attractive, it’s symmetric in the masses, it’s inverse-square, etc. But it doesn’t correspond to a real, physical force.
It’s a counter-example to the claim that a dimensionally correct equation is necessarily a correct physical relation.
A simpler counter example is 1=2 . It is stating the equality of two dimensionless numbers. It is trivially dimensionally correct. But it is false.
