Which object is least likely to allow light to pass through it?
2 answers:
You might be interested in
Answer:
The recoil velocity is 0.354 m/s.
Explanation:
Given that,
Mass of hunter = 70 kg
Mass of bullet = 42 g = 0.042 kg
Speed of bullet = 590 m/s
We need to calculate the recoil speed of hunter
Using conservation of momentum
Where, = mass of hunter
= mass of bullet
u = initial velocity
v = recoil velocity
Put the value in the equation
Hence, The recoil velocity is 0.354 m/s.
Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s
Answer:
The maximum height reached is 4.63 m.
Explanation:
Given:
Initial speed of the man (u) = 31.0 m/s
Speed at the top of trajectory () = 29.5 m/s
Acceleration due to gravity (g) = 9.8 m/s²
When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.
Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.
So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.
Now, initial velocity can be rewritten in terms of its components as:
Where, are the initial horizontal and vertical velocities of the man.
Now, plug in the given values and simplify. This gives,
Now, we know that, for a projectile motion, the maximum height is given as:
Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,
Therefore, the maximum height reached is 4.63 m.