Normal atmospheric pressure at sea level is ?
1 answer:
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Answer:
The distance is 0.53 m.
Explanation:
Given that,
Target distance = 100.0 m
Speed of bullet = 300 m/s
We need to calculate the total time
Using formula of time
Put the value into the formula
Now, consider vertical motion of bullet.
Initial velocity of bullet in vertical direction = 0 m/s
We need to calculate the vertically distance
Using equation of motion
Put the value in the equation
Hence, The distance is 0.53 m.
Answer : The correct option is, (C) 17 m/s
Explanation :
Formula used :
where,
K.E = kinetic energy = 6.8 J
m = mass of object = 46 g = 0.046 kg (1 kg = 1000 g)
v = velocity
Now put all the given values in the above formula, we get:
Therefore, the ball's velocity be as it leaves the cannon is, 17 m/s
Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
