a wave travels at a speed of 793 m/s and has a wavelength of 2.3 meters. calculate the frequency of the wave
1 answer:
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Answer:
Five: B
Ten: A
Explanation:
Five
An alpha decay looks like this.
So whatever is produced must have a mass of 4 less than 234 and a number on the periodic table of 2 less than 92. In other words, B has a mass of 230 and a number on the periodic table of 90.
The answer should be Thorium which is B.
Ten
There is actually not enough information to do 10. You get it by making an assumption and seeing if it works.
In general a beta decay can (the most common one ) look like this.
Is there anything that looks like that?
The weight stays the same (234) and the atomic number of the mother element (on the left) is 1 lower than the given chemical on the right.
The answer is Thorium A
Beta decays are very tricky. Be very careful how you handle them. One of three items can be what is decayed. I have assumed it was an electron, but there are two other possibilities. I won't confuse you by adding them. Just be aware that they exist.
Answer:
(a) L = 0·73 m
(b) 4·39 × J
Explanation:
(a) From the figure, consider the torque about the point where the rod is attached because if we consider another point then there will be hinge forces acting on the rod at the point of attachment
Let L m be the length of the rod and β be the angle between the rod and the vertical
Let α be the angular acceleration of the rod
As the force of gravity acts at the centre so from the figure, the torque about the point of attachment will be 0·53 × g ×(L ÷ 2) ×sinβ
Assuming that the value of amplitude of this oscillation to be small
As torque = moment of inertia × angular acceleration
0·53 × g ×(L ÷ 2) ×sinβ = ((0·53 × L²) ÷ 3) × α (∵ moment of inertia of the rod from the point of attachment)
<h3>For small oscillations, α = ω² × β</h3>After substituting the value of α and solving we get
ω = √((3 × g) ÷ (2 × L))
Time period = (2 × π) ÷ ω = (2 × π) ÷ √((3 × g) ÷ (2 × L))
∴ (2 × π) ÷ √((3 × g) ÷ (2 × L)) = 1·4
Substituting the value of g as 9·8 m/s² and solving we get
L = 0·73 m
(b) At the maximum amplitude condition the velocity will be 0 and potential energy will be maximum and maximum kinetic energy will be attained at the lowest point and hinge forces will not do work as the point of attachment is not moving
∴ Taking the reference for finding the potential energy as the lowest point
<h3>Maximum potential energy = Maximum kinetic energy </h3><h3>As total energy is constant, since there is no dissipative force</h3>Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 (∵ increment in height is (L × (1 - cosβ)) ÷ 2
∴ Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 After substituting the value we get
Maximum potential energy = 4·39 × J
∴ Maximum kinetic energy = 4·39 × J
Answer:
<h3><em>28.01m/s.</em></h3>Explanation:
Given maximum height reached by the ball as H = 40 metres
Since the ball rises straight up when hit by a ball, then the angle of launch will be perpendicular to the ground and that is 90°.
To determine the upward speed of the ball in meters per second after it got struck by the bat, we will use the formula for calculating the maximum height according to projectile motion;
Maximum Height H = where;
u is the speed of the ball
is the angle of launch
g is the acceleration due to gravity = 9.81m/s²
Substituting the given parameters into the formula;
<em>Hence the upward speed of the ball in meters per second after it got struck by the bat is 28.01m/s.</em>