Max height occurs when v = 0.
v(t) = ds(t)/dt
v(t) = 80 - 32t
0 = 80 - 32t
t = 5/2
s(5/2) = 80(5/2) - 16(5/2)^2
s(5/2) = 100
Answer: 100 ft
96 = 80t - 16t²
t = 3, 2
(80 ± √256) / 32 using the quadratic equation.
v(2) = 16
v(3) = -16
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
Two vectors have magnitudes of 10 and 15. The angle between them when they are drawn with their tails at the same point is 65. The component of the longer vector along the line of the shorter is 6.33 .
A vector is a quantity or phenomenon that has two independent properties: magnitude and direction. The term also denotes the mathematical or geometrical representation of such a quantity. Examples of vectors in nature are velocity, momentum, force, electromagnetic fields, and weight.
The taller component will be 15 . There will be two components taller component , one in the direction of shorter component and other perpendicular to the shorter wavelength .
The component of longer wavelength in the direction of shorter will be
= 15 cos (theta ) = 15 cos (65) = 6.33
where theta is the angle between both the vectors
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The Mercury's mass for the given acceleration due to gravity is 0.3152 x 10²⁴ kg.
The ratio of the calculated and accepted value of the Mercury's mass is 0.95.
<h3>What is mass?</h3>
Mass is the amount of matter present in the object.
The mass of the object is always constant, anywhere it is on the Earth or Moon or any other planet.
Given is the acceleration due to gravity of Mercury planet at North pole is g = 3.698 m/s² and the radius of Mercury planet is 2440 km.
The acceleration due to gravity is related with mass as
g = GM/R²
Substitute the values, we have
3.698 = 6.67 x 10⁻¹¹ x M/(2440 x1000)³
M = 2.2016 x 10¹³ / 6.67 x 10⁻¹¹
M = 0.3152 x 10²⁴ kg
Thus, the mercury's mass is 0.3152 x 10²⁴ kg.
(b) Accepted value of Mercury's mass is 3.301 x 10²³ kg
Ratio of the value of mass calculated and accepted is
Mcalc/M accep = 0.3152 x 10²⁴ kg / 3.301 x 10²³ kg
= 0.95
Thus, the ratio is 0.95
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