Hello!
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.
Data:
Hooke represented mathematically his theory with the equation:
F = K * Δx
On what:
F (elastic force) = 2 N
K (elastic constant) = 4 N/cm
Δx (deformation or elongation of the elastic medium or distance from a spring) = ?
Solving:
simplify by 2
Answer:
B.) 1/2 cm
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Answer:
-10C= Solid
10C= Liquid
50C= Liquid
90C= Liquid
110C= Gas
120C= Gas
Explanation:
Below 0 degree C (Celsius), water is frozen means it is in the form of ice. After 0 degree, once we keep it in room temperature, the ice starts becoming liquid (water), and once we heat water, after 100 degree C (Celsius) water starts boiling and thus starts entering gaseous state.
Answer:
The force ratio of a machine is 5 and it velocity ratio is 5 means that the load moved is five times the effort applied and the distance moved by the effort is five times the distance moved by the load at the same time interval.
Explanation:
Neither source is a renewable one.
Answer: v= 160ft/s
a=32ft/s^2 constant
Explanation:
s(t)=400-16t^2 derivative of position is velocity v(t) and derivative of velocity is acceleration a(t) so let s(t)=0 to find the time of flight to reach the ground and take the two derivatives and use the time found and solve. Also acceleration is a constant as it’s gravity.
0=400-16t^2
400=16t^2
25=t^2
t=5s
ds/dt=v(t)=0-32t
dv/dt=a(t)=-32 constant(gravity)
v(t)=-32(5s)= -160ft/s negative sign is only showing direction
