Answer:
The velocity at the ground is <u>63.25 m/s</u> and time taken is <u>6.325 s.</u>
Explanation:
Given:
As the object is released, the initial velocity is, 
Displacement of the object is, 
To find:
Velocity at the bottom, 
Time to reach the bottom, 
The acceleration of the object is due to gravity and hence equal to 
Now, using the following equation of motion:
Now, using the equation of motion relating time and velocity:
Therefore, the velocity at the ground is 63.25 m/s and time taken is 6.325 s.
It is the acceleration of an object in free fall
Explanation:
When an object is in free fall, it is subjected only to one force: the force of gravity, which pulls the object downward, with a magnitude (near the Earth's surface) which is given by
where
m is the mass of the object
is the acceleration due to gravity
We can apply Newton's second law to the object in free fall:
where
F is the net force on the object
a is the acceleration of the object
m is the mass
However, since there is only the force of gravity acting on the object, the net force is equal to the force of gravity: so we can equate the two equations, obtaining that
Which means that the acceleration of an object in free fall (acted upon the force of gravity only) is equal to the acceleration due to gravity, .
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Answer:
Explanation:
100 kph = 27.8 m/s
80 kph = 22.2 m/s
v² = u² + 2as
a = (v² - u²) / 2s
a = (22.2² - 27.8²) / (2(88))
a = -1.578282828...
a = -1.58 m/s²
t negative sign means the acceleration opposes the initial velocity.
t = Δv/a
t = (22.2 - 27.8) / -1.57828
t = 3.52 s
Answer: 1.5 N/C
Explanation:
The equation is derived from Coulomb's Law, but is not exactly Coulomb's Law:
F = qE
q is the test charge
E is the electrical field produced by the source charge
F is the force in Newtons
In this problem, we were given the value of the test charge q, and the force acting on it due to the electric field from the source charge.
E = F/q
E = 6.5N/4.3C
E = 1.5 N/C
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