Consider a day in the physics lab room with room temperature of 20 0C and pressure of 1.0 atm. A container of 3.5 L is left open
1 answer:
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1) 88 ft
2) 4.09 s
3) 1.38 s
4) 118.2 m
Explanation:
1)
For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:
(1)
where
is the height at time t = 0
is the initial vertical velocity
is the acceleration due to gravity
The function that describes the height of the rock above the surface at a time t in this problem is
(2)
By comparing the terms with same degree of eq(1) and eq(2), we observe that
which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.
2)
The rock will hit the water when its height becomes zero, so when
which means when
First of all, we can simplify the equation by dividing each term by 4:
This is a second-order equation, so we solve it using the usual formula and we find:
Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):
t = 4.09 s
So, the rock hits the water after 4.09 seconds.
3)
Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.
For an object in free fall motion, the vertical velocity is given by the expression
where
u is the initial velocity
g is the acceleration due to gravity
t is the time
The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:
which means
Here we have
(initial velocity)
(acceleration due to gravity)
Solving for t, we find the time at which this occurs:
4)
The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when
t = 1.38 s
The expression that gives the height of the rock at time t is
Substituting t = 1.38 s, we find:
So, the maximum height reached by the rock during its motion is
Which means 118.2 m above the water.
Answer:
a) F = 2.25 10³ N, b) REPULSIVE.
Explanation:
a) The electric force is given by coulomb's law
F = k q1q2 / r2
in this case nso indicate that the two charges have the same value
q₁ = q₂ = 2.5 10⁻⁶ C
Let's reduce the magnitudes to the SI system
r = 0.5 cm (1m / 100cm) = 5 10⁻³ m
let's calculate
F =
F = 2.25 10³ N
b) In electricity, electric charges of the same sign repel and those of the opposite sign attract
In this exercise, the balls are equal and are rubbed with the same material, for which it acquires charges of the same type, consequently, as the charges are of the same type, they indicate that the negative force is REPULSIVE.