Answer:
a) v = √ g x
, b) W = 2 m g d
, c) a = ½ g
Explanation:
a) For this exercise we use Newton's second law, suppose that the block of mass m moves up
T-W₁ = m a
W₃ - T = M a
w₃ - w₁ = (m + M) a
a = (3m - m) / (m + 3m) g
a = 2/4 g
a = ½ g
the speed of the blocks is
v² = v₀² + 2 ½ g x
v = √ g x
b) Work is a scalar, therefore an additive quantity
light block s
W₁ = -W d = - mg d
3m heavy block
W₂ = W d = 3m g d
the total work is
W = W₁ + W₂
W = 2 m g d
c) in the center of mass all external forces are applied, they relate it is
a = ½ g
Answer:
The separation between the first two minima on either side is 0.63 degrees.
Explanation:
A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:
with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:
for the first minimum
solving for θ1:


for the second minimum:



So, the angular separation between them is the rest:


Answer:
you must throw 3 snowballs
Explanation:
We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment
Initial. Before bumping that refrigerator
p₀ = n m v₀
Where n is the snowball number
Final. When the refrigerator moves
pf = (n m + M) v
The moment is preserved because the forces during the crash are internal
n m v₀ = (n m + M) v
n m (v₀ - v) = M v
n = M/m v/(vo-v)
Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton
F = m a
a = F / m
The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm
v₁² = v₀² + 2 a x
v₁² = 0+ 2 (100/1) 1
v₁ = 14.14 m / s
This is the initial speed for the crash
v₀ = v = 14.14 m / s
Let's calculate
n = M/m v/ (v₀-v)
n = 10/1 3 / (14.14 -3)
n = 2.7 balls
you must throw 3 snowballs
Given data:
mass of the bullet (m) = 25 g = 0.025 kg,
mass of the gun (M) = 0.9 kg,
speed of the bullet (v) =230 m/s,
speed of the bullet (V) = ?
From the given data it is clear that, the momentum is conserved. According to "<em>law of conservation of momentum" </em>the total momentum before and after the collision is equal.
In this problem the momentum before collision (bullet+gun) is zero.
Therefore, after the gun fires a bullet, the momentum must be zero.
Mathematically,
M × V + m × v = 0
where,
M × V = momentum of the gun
m × v = momentum of the bullet
(0. 9 × V) + (0.025 × 230) = 0
0.9 V = -5.75
V = -5.75/0.9
= -6.39 m/s
<em>The gun recoils with a speed of 6.39 m/s</em>
Option (A) is correct.
A noble gas is different from other elements because its highest electron energy level is completely filled.The examples of noble gases are helium, neon, Argon , krypton,Xenon , radon.
All the noble gases have completely filled outermost shell. for example, Helium has two electrons and both of them are present in first shell. Neon has 10 electrons, so its electronic configuration is 2,8.It has two electrons in the first shell and eight electrons in the second shell. Thus the outermost shell of both Helium and Neon is completely filled.
This property of having completely filled outermost shells makes noble gases different from the rest of the elements.These noble gases are very less reactive .