4) A cannon shoots a cannonball of mass 5kg vertically upward from the mouth of a cannon with muzzle velocity of 7 m/s. At a hei
ght of 25 m above the mouth of the cannon, the cannonball attaches to another mass (1kg) and they continue to travel up. What is the maximum height that both masses will reach?
1 answer:
Answer:
h =220 m
Explanation:
Given that
u = 7 m/s
Even mass will attach but this will not produce any effect on the maximum height of the ball.Because in energy conservation the effect of mass does not present.
So the final speed of the ball will be zero at the maximum height.
v² = u² - 2 g (25 + h)
0 = 7² - 2 x 10 (25 +h)
49 = 20 ( 25 +h)
49 = 500 +20 h
Here h comes out negative that is why we are taking the 70 m/s in place of 7 m/s.
0 = 70² - 2 x 10 (25 +h) ( take g =10 m/s²)
4900 = 20 ( 25 +h)
4900 = 500 +20 h
4900- 500 = 20 h
4400 = 20 h
440 = 2 h
h =220 m
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Let's explain the following electrical terms.
• Electric current, can be said to be the flow of electric charge. These charges are through a conductor by moving electrons.
• Ammeter, is an instrument used to measure the amount of electric current in a circuit.
The name ammeter was derived from the unit of electric current (Amperes).
• Voltmeter, can be said to be an instrument used to measure the voltage(potential difference) in a circuit. It measures the electric potential between two points in an electrical circuit. It can also be called voltage meter.
• Electric circuit, can be defined as the conductive path for the flow of electric current.
It allows electric charge carriers to flow continuously.
• Resistance ,can be said to be the property of an electrical conductor which resists the flow of electric current. It is measured in ohms.
• Ohms law, states that the potential difference (V) between two points is directly proportional to the electric current across two points.
It is deonted as: V = I x R
Voltage = Current x Resistance
• Electric potential ,can be said to be the amount of electric potential energy at a point.
a) See free-body diagram in attachment
b) The book is stationary in the vertical direction
c) The net horizontal force is 35 N in the forward direction
d) The net force on the book is 35 N in the forward horizontal direction
e) The acceleration is in the forward direction
Explanation:
a)
The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.
From the diagram of this book, we see there are 4 forces acting on the book:
- The applied force, F = 50 N, pushing forward in the horizontal direction
- The frictional force, , pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)
- The weight of the book, , where m is the mass of the book and
is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:
- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book
b)
The book is in equilibrium in the vertical direction, therefore there is no motion.
In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is
where a is the acceleration; however, since N = W, this becomes
And since the book is initially at rest on the desk, this means that there is no motion.
c)
We said there are two forces acting in the horizontal direction:
- The applied force, F = 50 N, forward
- The frictional force, , backward
Since they act along the same line, we can calculate their resultant as
and therefore the net force is 35 N in the forward direction.
d)
The net force is obtained as the resultant of the net forces in the horizontal and vertical direction. However, we have:
- The net force in the horizontal direction is 35 N
- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction
Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.
e)
The acceleration of the book can be calculated by using Newton's second law:
where
is the net force
m is the mass
a is the acceleration
Here we have:
(in the forward direction)
m = 4 kg
Therefore, the acceleration is
(forward)
Learn more about forces, weight and Newton's second law:
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