Momentum = mv
where m is the mass of an electron and v is the velocity of the electron.
v = momentum ÷ m
= (1.05×10∧-24)÷(9.1×10∧-31) = 1,153,846.154 m/s
kinetic energy = (mv∧2)÷2
= (9.1×10∧-31 × 1,153,846.154∧2) ÷2
= (1.21154×10∧-18) ÷ 2
= 6.05769×10∧-19 J
For E = 200 gpa and i = 65. 0(106) mm4, the slope of end a of the cantilevered beam is mathematically given as
A=0.0048rads
<h3>What is the slope of end a of the cantilevered beam?</h3>
Generally, the equation for the is mathematically given as
Therefore
A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}
A=0.00288+0.00192=0.0048rads
A=0.0048rads
In conclusion, the slope is
A=0.0048rads
Read more about Graph
brainly.com/question/14375099
Answer:
a) 5.851× 10¹⁰m/s²
b) 2.411×10⁻¹¹s
c) 1.70×10⁻¹¹m
d) 1.661×10⁻²⁷KJ
Explanation:
A proton in the field experience a downward force of magnitude,
F = eE. The force of gravity on the proton will be negligible compared to the electric force
F = eE
a= eE/m
= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷
= 5.851× 10¹⁰m/s²
b)
V = u + at
u= 0
v= 1.4106m/s
v= (0)t + at
t= v/a
= 1.4106m/s/5.851 ×10¹⁰
= 2.411×10⁻¹¹s
c)
S = ut + at²
= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²
= 1.70×10⁻¹¹m
d)
Ke = 1/2mv²
= (1.67×10⁻²⁷×)(1.4106)²/2
= 1.661×10⁻²⁷KJ
The cup is acted upon by an unbalanced force which is the cars acceleration, but before it was an object at rest that stayed at rest. This jet propels their body forward.
Use of lubricant
Use of ball bearers
Use of streamlined body
Use of graphite
