Answer:
We can use 2 g H = v2^2 - v1^2 or
v2^2 = 2 g H + v1^2
Since 88 ft/sec = 60mph we have 30 mph = 44 ft/sec
The object will return with the same speed that it had initially so the object
starts out with a downward speed of 44 ft/sec
Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2
v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2
v2 = 110 ft/sec
Answer:
The maximum mass the bar can support without yielding = 32408.26 kg
Explanation:
Yield stress of the material (
) = 200 M Pa
Diameter of the bar = 4.5 cm = 45 mm
We know that yield stress of the bar is given by the formula
Yield Stress = 
⇒
=
---------------- (1)
⇒ Area of the bar (A) =
×
⇒ A =
× 
⇒ A = 1589.625 
Put all the values in equation (1) we get
⇒
= 200 × 1589.625
⇒
= 317925 N
In this bar the
is equal to the weight of the bar.
⇒
=
× g
Where
is the maximum mass the bar can support.
⇒
= 
Put all the values in the above formula we get
⇒
= 
⇒
= 32408.26 Kg
There fore the maximum mass the bar can support without yielding = 32408.26 kg
Work is done when spring is extended or compressed. Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.
Answer:
carbon + oxygen → carbon dioxide