1) True 2) True 3)True I hope I helped
Answer:
a) t = 1.47 h b) t = 1.32 h
Explanation:
a) In this problem the plane and the wind are in the same North-South direction, whereby the vector sum is reduced to the scalar sum (ordinary). Let's calculate the total speed
v =
f -
v = 585 -32.1
v = 552.9 km / h
We use the speed ratio in uniform motion
v = x / t
t = x / v
t = 815 /552.9
t = 1.47 h
b) We repeat the calculation, but this time the wind is going in the direction of the plane
v=
f -
v 585 + 32.1
v = 617.1 km / h
t = 815 /617.1
t = 1.32 h
Answer:
The minimum coefficient of friction is 0.544
Solution:
As per the question:
Radius of the curve, R = 48 m
Speed of the car, v = 16 m/s
To calculate the minimum coefficient of static friction:
The centrifugal force on the box is in the outward direction and is given by:

where
= coefficient of static friction
The net force on the box is zero, since, the box is stationary and is given by:
Answer:
pretty sure its motor 2, with body 1
Explanation:
Answer:
a)W=8.333lbf.ft
b)W=0.0107 Btu.
Explanation:
<u>Complete question</u>
The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.
Solution
Preload = F₀=0 lbf
Spring constant k= 200 lbf/in
Initial length of spring x₁=0
Final length of spring x₂= 1 in
At any point, the force during deflection of a spring is given by;
F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

Change to lbf.ft by dividing the value by 12 because 1ft=12 in
100/12 = 8.333 lbf.ft
work required to compress the spring, W=8.333lbf.ft
The work required to compress the spring in Btu will be;
1 Btu= 778 lbf.ft
?= 8.333 lbf.ft----------------cross multiply
(8.333*1)/ 778 =0.0107 Btu.