Answer:
a) 1.20227 seconds
b) 0.98674 m
c) 7.3942875 m/s
Explanation:
t = Time taken
u = Initial velocity = 4.4 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
b) Her highest height above the board is 0.98674 m
Total height she would fall is 0.98674+1.8 = 2.78674 m
a) Her feet are in the air for 0.75375+0.44852 = 1.20227 seconds
c) Her velocity when her feet hit the water is 7.3942875 m/s
86.4×10^6 joule is energy does one house use during each 24 hr day.
20 MJ of light energy
Consumption of electricity is 1 kW.
The energy consumption lasts for 24 hours.
energy=power×time
energy=10^3×24×3600
energy=86.4×10^6 joule
Energy in physics is the ability to perform work. Different shapes, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc., may be assumed by it. Other examples of energy being transferred from one body to another include heat and work. Energy is always distributed after it has been transported in accordance with its type. Thus, heat transfer could result in thermal energy, whereas work could result in mechanical energy.
Motion is a trait shared by all forms of energy. For instance, if a body is moving, it has kinetic energy. Due to the object's design, which incorporates potential energy, a tensioned object, like a spring or bow, has the ability to move even when at rest.
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Solid elements are rigid elements. For example an element iron is in solid form. You can touch it and it’s hard.
Answer:
4NaBr + 2CaF2 ---------- 4 NaF + 2CaBrz
Explanation:
Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis
![F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}](https://tex.z-dn.net/?f=F_N%3Dk_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D)
Explanation:
Let the force on +Q charge y-axis due to +2Q charge be 
and force on +Q charge y axis due to -Q charge on x-axis be 
.
Distance between the +2Q charge and +Q charge = d units
Distance between the -Q charge and +Q charge = 
units
= Coulomb constant
Net force on +Q charge on y-axis is:
![|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|](https://tex.z-dn.net/?f=%7CF_N%7C%3D%7Ck_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D%7C)
The net froce on the +Q charge on y-axis is
