Complete question:
The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?
Answer:
The recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s
Explanation:
Given;
combined mass of the shotgun and arm–shoulder, m₁ = 15 kg
mass of the projectile, m₂ = 0.04 kg
speed of the projectile, u₂ = 380 m/s
let the recoil velocity of the shotgun and arm–shoulder combination = u₁
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = 0
m₁u₁ = - m₂u₂
Therefore, the recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s
Answer:
The Angle between the vector A and B is 12.32°
Explanation:
Given data
|A×B|=32
|A|=10
|B|=15
To find
Angle α
Solution
From Cross product properties we know that
The Angle between the vector A and B is 12.32°
All objects in orbit must follow the path of an ellipse (one of Keplers laws)
Answer:
Q2: a) make sure the sugar sample used is constant
b) conduct the experiment twice and use the average
c) the type of container used
Q3: a) out of the three, icing sugar dissolves the fastest
b) substances dissolves faster in warmer waters
Q4: molecules in warmer waters moves at a greater speed than in colder waters, having more kinetic energy and colliding with the solute more frequently. More energy is then transferred to the solute molecules to break their bonds, making them dissolve faster.
The field lines spread apart as we move away from the charge, and they point away from the charge
Explanation:
The electric field produced by a single-point positive charge is a radial field, whose strength is given by the equation
where
k is the Coulomb's constant
Q is the magnitude of the charge
r is the distance from the charge at which the field is calculated
There are two pieces of information given by the field lines shown in the graph:
- The spacing between the lines gives an indication of the strength of the field: the closer to each other they are, the stronger the field. In this case, as we move away from the charge, the spacing between the lines increases, and this means that the field becomes weaker (in fact, it follows an inverse square law,
- The direction of the lines gives the direction of the electric field, which points away from the central charge. This is because the direction of the electric field corresponds to the direction of the force that a positive test charge would feel when immersed in the electric field: in this case, if we place a positive test charge in this field, then it would get repelled away from the central charge (remember that the electric force between two positive charges is repulsive), and therefore, the direction of the electric field is away from the central charge.
Learn more about electric field:
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