I’m pretty sure it’s speed=distance/time
Answer:
a.)
I.) Open system
II.) Closed system
b.)
The total momentum of the object and earth system stay the same as the object fall toward the earth
The kinetic energy of the object increases as the object falls toward the earth.
Explanation:
a.)
I.) The system containing only the object is an open system because of the influence of external forces and presence of matters. External forces in this scenario means gravitational force acting on the object. And air is the matter that has influence on the object.
II.) The system containing only the object and earth is a closed system. Because in a closed system, there are no external dissipative forces acting on it. Universe is a closed system. And the mechanical energy of a close system is conserved. The mechanical energy will remain constant. In other words, it will not change (become more or less). This is called the Law of Conservation of Mechanical Energy.
b.) The total momentum of the object and earth system stay the same as the object fall toward the earth. Because momentum will always be conserved.
The kinetic energy of the object increases as the object falls toward the earth. When the object was initially at rest, kinetic energy equals zero
Answer:
The speed is 173 m/s.
Explanation:
Given that,
A = 47
B = 14
Length 1 urk = 58.0 m
An hour is divided into 125 time units named dorts.
3600 s = 125 dots
dorts = 28.8 s
Speed v= (25.0+A+B) urks/dort
We need to convert the speed into meters per second
Put the value of A and B into the speed
Hence, The speed is 173 m/s.
Answer:
yes
Explanation:
this is true don't believe other people whether right or wrong
Answer:
a)Yes will deform plastically
b) Will NOT experience necking
Explanation:
Given:
- Applied Force F = 850 lb
- Diameter of wire D = 0.15 in
- Yield Strength Y=45,000 psi
- Ultimate Tensile strength U = 55,000 psi
Find:
a) Whether there will be plastic deformation
b) Whether there will be necking.
Solution:
Assuming a constant Force F, the stress in the wire will be:
stress = F / Area
Area = pi*D^2 / 4
Area = pi*0.15^2 / 4 = 0.0176715 in^2
stress = 850 / 0.0176715
stress = 48,100.16 psi
Yield Strength < Applied stress > Ultimate Tensile strength
45,000 < 48,100 < 55,000
Hence, stress applied is greater than Yield strength beyond which the wire will deform plasticly but insufficient enough to reach UTS responsible for the necking to initiate. Hence, wire deforms plastically but does not experience necking.
