Answer:
Explanation:
The cannonball goes a horizontal distance of 275 m . It travels a vertical distance of 100 m
Time taken to cover vertical distance = t ,
Initial velocity u = 0
distance s = 100 m
acceleration a = 9.8 m /s²
s = ut + 1/2 g t²
100 = .5 x 9.8 x t²
t = 4.51 s
During this time it travels horizontally also uniformly so
horizontal velocity Vx = horizontal displacement / time
= 275 / 4.51 = 60.97 m /s
Vertical velocity Vy
Vy = u + gt
= 0 + 9.8 x 4.51
= 44.2 m /s
Resultant velocity
V = √ ( 44.2² + 60.97² )
= √ ( 1953.64 + 3717.34 )
= 75.3 m /s
Angle with horizontal Ф
TanФ = Vy / Vx
= 44.2 / 60.97
= .725
Ф = 36⁰ .
Answer:
Explanation:
When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that
Work done by all the forces = change in kinetic energy of the system
here we know that
also we know that the length of the incline is given as
now we have
so we have
We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"
- it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
From the question we are told
the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?
Generally the equation for the Force is mathematically given as
F=\frac{F}{dx}
Therefore
F=-kdx
k=600Nm^{-1}
now
K.E=0.5x ds^2
K.E=600*(-0.1^2)
K.E=3J
Therefore
the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
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Initial velocity = Vo= 25 m/s
Final velocity = V = x
Acceleration= a = 6 m/s^2
time= t = 4 seconds
Appy the equation:
V = Vo + at
Replacing:
V = 25 + 6(4) = 25 + 24 = 49 m/s
Density = (mass) divided by (volume)
We know the mass (2.5 g). We need to find the volume.
The penny is a very short cylinder.
The volume of a cylinder is (π · radius² · height).
The penny's radius is 1/2 of its diameter = 9.775 mm.
The 'height' of the cylinder is the penny's thickness = 1.55 mm.
Volume = (π) (9.775 mm)² (1.55 mm)
= (π) (95.55 mm²) (1.55 mm)
= (π) (148.1 mm³)
= 465.3 mm³
We know the volume now. So we could state the density of the penny,
but nobody will understand what we have. Here it is:
mass/volume = 2.5 g / 465.3 mm³ = 0.0054 g/mm³ .
Nobody every talks about density in units of ' gram/(millimeter)³ ' .
It's always ' gram / (centimeter)³ '.
So we have to convert our number for the volume.
(0.0054 g/mm³) x (10 mm / cm)³
= (0.0054 x 1,000) g/cm³
= 5.37 g/cm³ .
This isn't actually very close to what the US mint says for the density
of a penny, but it's in a much better ball park than 0.0054 was.
