The 61.0 kg object<span> ... F = (300kg)(6.673×10−11 </span>N m<span>^2 </span>kg<span>^−2)(61kg)/(.225m)^2. F = 2.412e-5 </span>N<span> towards the 495 </span>kg<span> block. </span>b. [195kg] ===.45m ... (b<span>) You cannot achieve this </span>position<span>. For the </span>net force<span> to become zero, one or both of the </span>masses<span> must ...</span>
Answer:
2.28
Explanation:
From mirror formula,
1/f = 1/u+1/v .......... Equation 1
Where f = focal length of the mirror, v = image distance, u = object distance.
Note: The focal length mirror is positive.
make v the subject of the equation,
v = fu/(u-f)............ Equation 2
Given: f = 2.5 cm, u = 1.4 cm
Substitute into equation 2
v = 2.5(1.4)/(1.4-2.5)
v = 3.5/-1.1
v = -3.2 cm.
Note: v is negative because it is a virtual image.
But,
Magnification = image distance/object distance
M = v/u
Where M = magnification.
Given: v = 3.2 cm, u = 1.4 cm
M = 3.2/1.4
M = 2.28.
Thus the magnification of the tooth = 2.28.
Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Explanation:
Given that;
mass of vehicle m = 1000 kg
for a low speed test; V = 2.5 m/s
bumper maximum deflection = 4 cm = 0.04 m
First we determine the energy of the vehicle just prior to impact;
W_v = 1/2mv²
we substitute
W_v = 1/2 × 1000 × (2.5)²
W_v = 3125 J
now, the the effective design stiffness k will be:
at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;
hence;
W_v = 1/2kx²
we substitute
3125 = 1/2 × k (0.04)²
3125 = 0.0008k
k = 3125 / 0.0008
k = 3906250 N/m
Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Increasing its velocity will add to the kinetic energy more as the formula for kinetic energy is 0.5*m*v^2. (The speed will be squared making it greater)
