Answer: Both cannonballs will hit the ground at the same time.
Explanation:
Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.
then the acceleration equation is only on the vertical axis, and can be written as:
a(t) = -(9.8 m/s^2)
Now, to get the vertical velocity equation, we need to integrate over time.
v(t) = -(9.8 m/s^2)*t + v0
Where v0 is the initial velocity of the object in the vertical axis.
if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s
and:
v(t) = -(9.8 m/s^2)*t
Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)
And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.
You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)
<h2>The voltmeter reading will be 35.7 volt </h2>
Explanation:
The resistor 1000 ohm and 4000 ohm are connected in parallel .
Their combined resistance is supposed R₁
Thus 
= 
+ 
or R₁ = 800 ohm
Therefore the total resistance in circuit is = 2000 + 800 = 2800 ohm
Because these are in series .
We can find current flowing through the circuit I = 
= 
= 
here R is total resistance .
The potential difference across 1000 ohm = x 1000 = 35.7 volt
Thus voltmeter reading will be 35.7 volt
Answer:
m1/m2 = 0.51
Explanation:
First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:
V = √F/u
This is the equation that describes the relation between speed of a pulse and a force exerted on it.
the value of "u" is:
u = m/L
Where m is the mass of the rod, and L the length.
Now, for the rod 1:
V1 = √F/u1 (1)
rod 2:
V2 = √F/u2 (2)
Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:
1.4V2 = √F/u1 (3)
Replacing (2) in (3):
1.4(√F/u2) = √F/u1 (4)
Now, let's solve the equation 4:
[1.4(√F/u2)]² = F/u1
1.96(F/u2) =F/u1
1.96F = F*u2/u1
1.96 = u2/u1 (5)
Now, replacing the expression of u into (5) we have the following:
1.96 = m2/L / m1/L
1.96 = m2/m1 (6)
But we need m1/m2 so:
1.96m1 = m2
m1/m2 = 1/1.96
m1/m2 = 0.51
Answer:
The gravitational potential energy of the ball is 13.23 J.
Explanation:
Given;
mass of the ball, m = 0.5 kg
height of the shelf, h = 2.7 m
The gravitational potential energy is given by;
P.E = mgh
where;
m is mass of the ball
g is acceleration due to gravity = 9.8 m/s²
h is height of the ball
Substitute the givens and solve for gravitational potential energy;
PE = (0.5 x 9.8 x 2.7)
P.E = 13.23 J
Therefore, the gravitational potential energy of the ball is 13.23 J.
There's nothing mysterious about it at all. "Frequency" simply means
"often-ness" ... how often or how frequently something happens.
-- The frequency of traditional meals is 3 per day.
-- The frequency of an equinox is 2 per year.
-- The frequency of my sleeping really late is 1 per week.
-- The frequency of my intense desire to sleep late is 30 per month.
etc.
-- The standard unit of frequency in the SI system is "per second".
The special name for that unit is "Hertz". (Hz)
