Answer:
The pressure is the same on points that are at the same level but on opposite sides.
Answer:
amount of work done is
Explanation:
Formula for work done by force field
where
r(t) is parametrization of line
as it is straight line so
thus,
r(t) = (1-t)(-1,1) + t(3,-3)
= (-1+t,1-t) + (3t - 3t)
= (-1+t +3t, 1-t-3t)
r(t) = (4t -1, 1- 4t)
r'(t) = (4,-4)
putting value in above integral
![= 4[ -16 \frac{t^3}{3} + 16\frac{t^2}{2} - 3t]_{0}^{1}](https://tex.z-dn.net/?f=%3D%204%5B%20-16%20%5Cfrac%7Bt%5E3%7D%7B3%7D%20%2B%2016%5Cfrac%7Bt%5E2%7D%7B2%7D%20-%203t%5D_%7B0%7D%5E%7B1%7D)
Answer:
a) K = 3 MeV b) K= 1.5 MeV
Explanation:
We can solve this experiment using the equation of the magnetic force with Newton's second law, where the acceleration is centripetal.
F = q v x B
We can also write this equation based on the modules of the vectors
F = qv B sin θ
With Newton's second law
F = ma
F = m v² / r
q v B = m v² / r
v = q B r / m
The kinetic energy is
K = ½ m v²
Substituting
K = ½ m (q B r/ m)²
K = ½ B² r² q² / m
K = (½ B² R²) q²/m
The amount in brackets does not change during the experiment
K = A q² / m
For the proton
K = 3.0 10⁶eV (1.6 10⁻¹⁹ J / 1eV) = 4.8 10⁻¹³ J
With this data we can find the amount we call A
A = K m/q²
A = 4.8 10⁻¹³ 1.67 10⁻²⁷ /(1.6 10⁻¹⁹)²
A = 3.13 10⁻²
With this value we can write the equation
K = 3.13 10⁻² q² / m
Alpha particle
m = 4 uma = 4 1.66 10⁻²⁷ kg
K = 3.13 10⁻² (2 1.6 10⁻¹⁹)² / 4.0 1.66 10⁻²⁷
K = 4.82 10⁻¹³ J ((1 eV / 1.6 10⁻¹⁹ J) = 3 10⁶ eV
K = 3 MeV
Deuteron
K = 3.13 10⁻² (1.6 10⁻¹⁹)²/2 1.66 10⁻²⁷
K = 2.4 10⁻¹³ J (1eV / 1.6 10⁻¹⁹J)
K = 1.5 10⁶ eV
K= 1.5 MeV
Momentum before collision must be equal to momentum after collision
(m1<span> + m</span>2)v = m1v1<span> + m</span>2v<span>2</span>
3.62g*270 m/s=2.30kg* (x)
convert the units to be the same that is convert the kg mass to grams
1kg=1000g
2.30kg=y
230/100*1000=2300g
3.62*270=2300X
977.4g/m/s=2300X
977.4/2300=0.4250M/S
Finding a velocity after the gun is embedded on the block of wood
3.62*270+2300g*0.425=(3.62+2300)*V
977.5+977.5=2303V
1955=2303V
V=1955/2303
V=0.8489M/S
