Answer:
Explanation:
If the initial velocity is U
Then the horizontal component of the velocity is
Ux= Ucosθ
Then the range for a projectile is give as
R=Ux.t
Where t is the time of flight
The time of flight is given as
t=2USinθ/g
Therefore,
R=Ux.t
R=UCosθ.2USinθ/g
R=U^2×2SinθCosθ/g
Then, from trigonometric ratio
2SinθCosθ= Sin2θ
R=U^2Sin2θ/g
Given that θ=32° and g=9.81m/s^2
Then
R=U^2Sin2×32/9.81
R=U^2Sin64/9.81
R=0.0916U^2
Then, range is given by R=0.0916U^2
A=0.0916U^2.
T
The box is at a distance A from the point of projection. Then the range R=A
R=0.0916U^2
A=0.0916U^2
Then,
U^2=A/0.0916
U^2=10.915A
Then the initial velocity should be
U=√10.915A
U=3.3√A
The speed of the object increases
Explanation:
We can answer this question by applying the work-energy theorem, which states that the work done on an object is equal to the change in kinetic energy of the object. Mathematically:

where
W is the work done on the object
are the final and initial kinetic energy of the object, respectively
m is the mass of the object
v is its final speed
u is its initial speed
In this case, the force does a positive amount of work on the object, so

This also implies that

And so

And therefore

which means that the speed of the object increases.
Learn more about work:
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Answer:
Both Thomson and Rutherford used charged particles in their experiments.
Explanation:
The thermal energy is proportional to the movement of the particles in every state.
Decreasing the thermal energy will decrease the movement.
Find the volume of the bottom and top separately and then add them.
Cylinder volume is the area of the bottom times the height
(22/7)(5^2)•175=13750 ft^3
The volume of a sphere is
V=(4/3)(22/7)r^3
where r is the radius. Here that's also 5 since it fits on the cylinder.
Also we only want half the sphere so use
V=(2/3)(22/7)•5^3=261.9 ft^3
Which we round upto 262.
Now add the parts together
13750+262=14,012 ft^3