Need help but the subject is science will give brainleist
2 answers:
You might be interested in
Answer:
λ = 482.05 nm
Explanation:
The diffraction phenomenon and the diffraction grating is described by the expression
d sin θ = m λ
where d is the distance between two consecutive slits, λ the wavelength and m an integer representing the order of diffraction
in this case they indicate the distance between slits, the angle and the order of diffraction
λ = d sin θ / m
let's calculate
λ = 1.00 10⁻⁶ sin 74.6 / 2
λ = 4.82048 10⁻⁷ m
Let's reduce to nm
λ = 4.82048 10⁻⁷ m (10⁹ nm / 1 m)
λ = 482.05 nm
The coefficient of friction is missing and it has a value of μ = 0.4
Answer:
a = 3.924 m/s²
Explanation:
I've attached the kinematic free body diagram.
Taking the sum of all upward and downward forces,
EFy = 0;
N1 + N2 - m_p•g - m_v•g = 0
N1 + N2 = m_p•g + m_v•g
Where;
N1 and N2 are the normal reactions at the wheels
m_p is the mass of the driver
m_v is the mass of the vehicle
g is the acceleration due to gravity.
Plugging in the relevant values in the question,we obtain;
N1 + N2 = (385 + 75) x 9.81
N1 + N2 = 4512.6N - - - (eq1)
Now, taking sum of all horizontal forces;
EFx = (m_p + m_v) x a
So,
μ(N1 + N2) = (mp + mv) x a
Thus,
0.4(N1 + N2) = (385 + 75)a
0.4(N1 + N2) = 460a
N1 + N2 = 1150a
From eq(1),N1 + N2 = 4512.6N
Thus,
1150a = 4512.6N
a = 4512.6/1150
a = 3.924 m/s²
Therefore, the acceleration, a = 3.924 m/s²
Answer:
<em>The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>
<em></em>
Explanation:
Let us first consider the initial characteristics of the angular motion of the disk
moment of inertia =
angular speed = ω
For the second case, we consider the characteristics to now be
moment of inertia = (five times larger)
angular speed = ω/5 (five times smaller)
Recall that the kinetic energy of a spinning body is given as
therefore,
for the first case, the K.E. is given as
and for the second case, the K.E. is given as
<em>this is one-tenth the kinetic energy before its spinning characteristics were changed.</em>
<em>This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>
Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²