Specific heat. The definition of specific heat is the amount of energy required to raise the temperature of 1g of a substance by 1K or 1°C.
Answer:
288N
Explanation:
Given parameters:
Mass of Cheetah = 12kg
Acceleration = 24m/s²
Unknown:
Force needed by the cheetah to run = ?
Solution:
The force needed by the Cheetah to run is the net force.
According to Newton's law;
Force = mass x acceleration
Insert the given parameters and solve;
Force = 12 x 24 = 288N
The mass of an object has no effect whatsoever on the object's
acceleration during free-fall. If there is no air resistance to interfere
with the natural effects of gravity, then a feather and a battleship ...
dropped at the same time ... fall together, and hit the ground at the
same time.
Answer:
Explanation:
Given that
The mass of the body is 0.04kg
M=0.04kg
The radius of the paths is 0.6m
r=0.6m
The normal force exerted at A is 3.9N
Fa=3.9N
The normal force exerted at B is 0.69N
Fb=0.69N
Then work done by friction from point A to B will be the change in K.E
W=∆K.E+P.E
So we need to know the velocity at both point A and B
Then since the centripetal force is given as
Ft=mv²/r
Then,
For point A
Fa=mv²/r
3.9=0.04v²/0.6
3.9=0.0667v²
v²=3.9/0.0667
v²=58.5
v=√58.5
v=7.65m/s
Va=7.65m/s
Now at point B
Fb=mv²/r
0.69=0.04v²/0.6
0.69=0.0667v²
v²=0.69/0.0667
v²=10.35
v=√10.35
v=3.22m/s
Vb=3.22m/s
Then, the work done is
W=∆K.E+P.E
P.E is given as mgh
The height will be 2R =1.2m
P.E=mgh
P.E=0.04×9.81×1.2
P.E=0.471J
Final kinetic energy at B minus initial kinetic energy at A
W=K.Eb-K.Ea
K.E is given as 1/2mv²
W=1/2m(Vb²-Va²) +P.E
W=0.5×0.04(3.22²-7.65²) +0.471
W=0.5×0.04×(-48.1541) +0.471
W=-0.96+0.471
W=-0.49J
work was done on the block by friction during the motion of the block from point A to point B is 0.49J.
Friction opposes motions and that is why the work done is negative
F = ma, where m = mass in kg, a = acceleration in m/s², F = Force in Newton
F = 1 * 2
F = 2 N
Force needed is 2 Newtons.
