A 15 g bullet traveling horizontally at 865 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 53
1 answer:
Answer:
0.0613°C
Explanation:
the given parameters are m=15gm=15×10⁻³ V₁=865m/s V₂=534m/s
the bullet moves with different kinetic energies before and after the penetration, therefore
Kinetic energy before - kinetic energy after = 1/2 × m × ( V₁² - V₂²)
= × 15×10⁻³ × (865² - 534²)
= 3.47 × 10⁻³J
this loss in energy is transferred to the water, therefore
change in temperature =
where c = heat capacity of water = 4.19 x 10^3
m = mass of water = 13.5 kg
= {3.47 × 10⁻³} / {13.5 x 4.19 x 10^3 }
=0.0613°C
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To solve this problem we will apply the linear motion kinematic equations, which describe the change in velocity, depending on the acceleration and the distance traveled, that is,
Where,
= Final Velocity
= Initial Velocity
a = Acceleration
h = height
Our values are given as,
Replacing we have,
Therefore the height of the cliff is 121ft
Answer:
The answer is below
Explanation:
The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm
radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m
a) Initial angular velocity () = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s
θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad
angular acceleration (α) is:
b)
c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad
a) When it stops, the final angular velocity is 0. Hence:
θ = 323 rad