Answer:
Explanation:
GIVEN
diameter = 15 fm =
m
we use here energy conservation
there will be some initial kinetic energy but after collision kinetic energy will zero
on solving these equations we get kinetic energy initial
J ..............(i)
That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e
and gains kinetic energy K =e∆V ..........(ii)
by accelerating through a potential difference ∆V
Thus the alpha particle will
just reach the 
nucleus after being accelerated through a potential difference ∆V
equating (i) and second equation we get
e∆V = 35.33 Me V
Mary and her younger brother Alex decide to ride the carousel at the State Fair, Mary's and Alex's angular speed M and tangential speed vM is mathematically given as
Mary's and Alex's angular speed=1.43
Tangential speed mary=3.22 m/s
Tangential speed alex =2.260m/s
<h3>What is Mary's and Alex's angular
speed M and tangential speed vM?</h3>
Generally, the equation for angular speed is mathematically given as
w = 1.61 rev/see 3.9
Centripetal acc mary = v^2/r
Centripetal acc mary = w^2r
Centripetal acc mary = w^2x 2m
Centripetal acc. of Alex = w²x L.u
Therefore
Hence
tang. speed V=Wr
tang. speed of mary = 1.61x2 = 3.22 m/s
tang. speed of Alex: 1.61X1·4 =2.260m/s
Read more about Speed
brainly.com/question/4931057
#SPJ1
Answer:
(a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
Explanation:
Given that,
Power factor = 0.6
Power = 600 kVA
(a). We need to calculate the reactive power
Using formula of reactive power
...(I)
We need to calculate the 
Using formula of
Put the value into the formula
Put the value of Φ in equation (I)
(b). We draw the power triangle
(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95
Using formula of reactive power
We need to calculate the difference between Q and Q'
Put the value into the formula
Hence, (a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
Answer:
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
Explanation:
Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:
(1)
Where:
- Explosive force, measured in newtons.
- Barrel length, measured in meters.
- Mass of the shell, measured in kilograms.
, 
- Initial and final speeds of the shell, measured in meters per second.
If we know that 
, 
, 
and 
, then the explosive force experienced by the shell inside the barrel is:
![F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%281250%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%5Cleft%28750%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5E%7B2%7D-%5Cleft%280%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5E%7B2%7D%5Cright%5D%7D%7B2%5Ccdot%20%2815%5C%2Cm%29%7D)
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
Applicable linear expansion equation:
ΔL = αΔTL
In which
ΔL = change in length, α = Linear expansion coefficient of steel, ΔT = change in temperature, L = original length
Therefore,
ΔL = 12*10^-6*(18.5-(-3))*1410 = 0.36378 m
