Let F be the magnitude of the frictional force. This force performs an amount of work W on the bullet such that
W = -Fx
where x is the distance over which F is acting. This is the only force acting on the bullet as it penetrates the tree. The work-energy theorem says the total work performed on a body is equal to the change in that body's kinetic energy, so we have
W = ∆K
-Fx = 0 - 1/2 mv²
where m is the body's mass and v is its speed.
Solve for F and plug in the given information:
F = mv²/(2x)
F = (0.00426 kg) (881 m/s)² / (2 (0.0444 m))
F = 37,234.8 N ≈ 37.2 kN
Answer:
The answer to your question is: C. -9.81 m/s²
Explanation:
A. 9.81 m/s² acceleration is considered positive when it goes to the center of the earth, so this option is incorrect.
B. 0 m/s² This option is incorrect because acceleration is 0 for a linear motion without acceleration.
C. -9.81 m/s² If a projectile goes to the sky, then the acceleration will be negative.
D. It is not constant. Acceleration is constant.
Is D
is D because the inner layers are the Core, Radiative Zone and Convection Zone.
Answer:
H_w = 2.129 m
Explanation:
given,
Width of the weir, B = 1.2 m
Depth of the upstream weir, y = 2.5 m
Discharge, Q = 0.5 m³/s
Weir coefficient, C_w = 1.84 m
Now, calculating the water head over the weir
now, level of weir on the channel
H_w = y - H
H_w = 2.5 - 0.371
H_w = 2.129 m
Height at which weir should place is equal to 2.129 m.
Answer:
Average force is F = mass times change in V/ change in time so..
1 366.07143 N
Explanation:
51 kg x 15 m/s / 0.56
1 366.07143 m kg / s
1 366.07143 N
1 kilogram 1 meter per second per second = 1 N
