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stich3 [128]
3 years ago
9

A bullet with a mass 2.25g is fired up into the air with a velocity of 187.5 m/s. What is the maximum height of the bullet

Physics
1 answer:
Ksivusya [100]3 years ago
3 0

Answer:

1793.7m

Explanation:

From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

Now the kinetic energy; is

K.E = 1/2 × m × v2

Where m is mass

v is velocity

Hence.

K.E = 1/2 × 2.25 × (187.5)^2

Now this should be same with the potential energy which is given as;

P.E = m× g× h

Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P.E = 2.25 × 9.8 × h

From the foregoing analysis of energy conversation it implies;

1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h

=> 1/2 × (187.5)^2 = 9.8 × h

=>1/2 × (187.5)^2 / 9.8 = h

=> 1793.69m = h

h= 1793.69m

h =1793.7m to 1 decimal place

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15. A locomotive moved 18.0 m [W] in a time of 6.00 s and stopped. After stopping, the
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Answer:

Distance = 30m

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Explanation:

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Therefore, total distance traveled equals :

Movement 1 + movement 2

18m + 12m = 30m

B) Displacement also measures the movement made by an object. However, Displacement is a vector quantity and therefore, considers both magnitude and direction of travel of the object. Therefore, it measures the overall change in position of the object from its starting position.

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Although these quantities vary from one type of cell to another, a cell can be 1.9 μm in diameter with a cell wall 60 nm thick.
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Answer: 2(10)^{-9} mg

Explanation:

We know the total diameter of the cell (assumed spherical) is:

d=1.9\mu m=1.9(10)^{-6} m

Then its total radius r=\frac{d}{2}=\frac{1.9(10)^{-6} m}{2}=9.5(10)^{-7} m

On the other hand, we know the thickness of the cell wall is r_{t}=60 nm= 60(10)^{-9} m and its density is the same as water (\rho=997 kg/m^{3}).

Since density is the relation between the mass m and the volume V:

\rho=\frac{m}{V}

The mass is: m=\rho V (1)

Now if we are talking about this cell as a thin spherical shell, its volume will be:

V=\frac{4}{3}\pi R^{3} (2)

Where  R=r-r_{w}=9.5(10)^{-7} m - 60(10)^{-9} m

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V=2.952(10)^{-18} m^{3} (4)

Substituting (4) in (1):

m=(997 kg/m^{3})(2.952(10)^{-18} m^{3}) (5)

m=2.94(10)^{-15} kg (6)

Knowing 1 kg=1000 g and 1 mg=0.001 g:

m=2.94(10)^{-15} kg=2(10)^{-9} mg

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