Answer:
1.67m/s
Explanation:
Total Distance to be travelled by a Runner=100m
Time Taken=10*6s
Speed=Distance/Time
=100/10*6=10/6=1.67m/s
To find work, you use the equation: W = Force X Distance X Cos (0 degrees)
Following the Law of Conservation of Energy, energy cannot be destroyed nor created.
So you would do 75 N x 10m x Cos (0 degrees)= 750 J
Answer:
374.39 J/K
Explanation:
Entropy: This can be defined as the degree of disorder or randomness of a substance.
The S.I unit of entropy is J/K
ΔS = ΔH/T ..................................... Equation 1
Where ΔS = entropy change, ΔH = Heat change, T = temperature.
ΔH = cm................................... Equation 2
Where,
c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.
Substitute into equation 2
ΔH = 333000×0.3071
ΔH = 102264.3 J.
Also, T = 273.15 K
Substitute into equation 1
ΔS = 102264.3/273.15
ΔS = 374.39 J/K
Thus, The change in entropy = 374.39 J/K
Answer:
required distance is 233.35 m
Explanation:
Given the data in the question;
Sound intensity 
= 1.62 × 10⁻⁶ W/m²
distance r = 165 m
at what distance from the explosion is the sound intensity half this value?
we know that;
Sound intensity is proportional to 1/(distance)²
i.e
∝ 1/r²
Now, let r² be the distance where sound intensity is half, i.e ₂ = ₁/2
Hence,
₂/₁ = r₁²/r₂²
1/2 = (165)²/ r₂²
r₂² = 2 × (165)²
r₂² = 2 × 27225
r₂² = 54450
r₂ = √54450
r₂ = 233.35 m
Therefore, required distance is 233.35 m
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)
