Answer:
Explanation:
There are 4 forces. These are 1) Gravity, 2) Weak Nuclear Force, 3) Electromagnetism, and 4) Strong Nuclear Force.
Order of strength from weakest to strongest: Gravity, Weak Nuclear Force, Electromagnetism, Strong Nuclear Force
Type of Range:
Gravity - Unlimited range
Weak Nuclear Force - Limited range
Electromagnetism - Infinite range
Strong Nuclear Force - Limited Range
Found in:
Gravity - Exists between all objects with mass
Weak Nuclear Force - Governs over beta decays like the emission of electron or positron
Electromagnetism - the attraction found between particles that are electrically charged
Strong Nuclear Force - Found in atoms and subatomic particles. It is responsible for holding the atoms' nucleus together.
Using K.E=1/2MV^2
answer is 125joules
<span>Answer:
The moments of inertia are listed on p. 223, and a uniform cylinder through its center is:
I = 1/2mr2
so
I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2
Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration:
t = Ia
-1.20 Nm = (0.0120984 kgm2)a
a = -99.19 rad/s/s
Now we have a kinematics question to solve:
wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s
w = 0
a = -99.19 rad/s/s
Let's find the time first:
w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s
t = 10.558 s = 10.6 s
And the displacement (Angular)
Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form
s = (u+v)t/2
Which is
q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s
q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians
But the problem wanted revolutions, so let's change the units:
q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>
Answer: 321 J
Explanation:
Given
Mass of the box 
Force applied is 
Displacement of the box is 
Velocity acquired by the box is 
acceleration associated with it is 
Work done by force is 
change in kinetic energy is 
According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy
![\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J](https://tex.z-dn.net/?f=%5CRightarrow%20W%2BW_f%3D%5CDelta%20K%5Cquad%20%5BW_f%3D%5Ctext%7BWork%20done%20by%20friction%7D%5D%5C%5C%5C%5C%5CRightarrow%20375%2BW_f%3D54%5C%5C%5CRightarrow%20W_f%3D-321%5C%20J)
Therefore, the magnitude of work done by friction is 
Answer:
Mass is constant everywhere,
But weight is different,
If earth g = 10 then moon's is 1.6666667
Now billie's weight in moon is 41.6667
