Answer:
Explanation:
A charge within an electric field E experiences a force proportional to the field whose module is F = qE, whose direction is the same, if the charge is negative, it experiences a force in the opposite direction to the field and if the charge is positive, experience a force in the same direction of the field.
In our case we are interested in the magnitude of the force, therefore the sign of the charge has no relevance
N·m
(b) Find the angular acceleration of the airplane when it is inlevel flight.
rad/s2
(c) Find the linear acceleration of the airplane tangent to itsflight path.
m/s2
(a) 24.6 Nm
The torque produced by the net thrust about the center of the circle is given by:
where
F is the magnitude of the thrust
r is the radius of the wire
Here we have
F = 0.795 N
r = 30.9 m
Therefore, the torque produced is
(b)
The equivalent of Newton's second law for a rotational motion is
where
is the torque
I is the moment of inertia
is the angular acceleration
If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is
And so we can solve the previous equation to find the angular acceleration:
(c)
The linear acceleration (tangential acceleration) in a rotational motion is given by
where in this problem we have
is the angular acceleration
r = 30.9 m is the radius
Substituting the values, we find
Answer:
b
Explanation:
Answer:
3.44 rad
Explanation:
The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk
Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is ΔK = 21 J/s × 3.3 s = 69.3 J
So, ΔK = 1/2I(ω² - ω₀²)
Since ω₀ = 0 rad/s
ΔK = 1/2I(ω² - 0)
ΔK = 1/2Iω²
ΔK = 1/2(MR²/2)ω²
ΔK = MR²ω²/4
ω² = (4ΔK/MR²)
ω = √(4ΔK/MR²)
ω = 2√(ΔK/MR²)
Substituting the values of the variables into the equation, we have
ω = 2√(ΔK/MR²)
ω = 2√(69.3 J/( 4 kg × (4 m)²))
ω = 2√(69.3 J/[ 4 kg × 16 m²])
ω = 2√(69.3 J/64 kgm²)
ω = 2√(1.083 J/kgm²)
ω = 2 × 1.041 rad/s
ω = 2.082 rad/s
The angular displacement θ is gotten from
θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²
Substituting the values of the variables into the equation, we have
θ = ω₀t + 1/2αt²
θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²
θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²
θ = 1/2 × 6.87159 rad
θ = 3.436 rad
θ ≅ 3.44 rad
"The equation can be used to calculate the power absorbed by any surface" statement concerning the Stefan-Boltzmann equation is correct.
Answer: Option A
Explanation:
According to Stefan Boltzmann equation, the power radiated by black body radiation source is directly proportionate to the fourth power of temperature of the source. So the radiation transferred is absorbed by another surface and that absorbed power will also be equal to the fourth power of the temperature. So the equation describes the relation of net radiation loss with the change in temperature from hotter temperature to cooler temperature surface.
So this law is application for calculating power absorbed by any surface.
Answer:
we see it is a linear relationship.
Explanation:
The magnetic flux is u solenoid is
B = μ₀ N/L I
where N is the number of loops, L the length and I the current
By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops
B = (μ₀ I / L) N
the amount between paracentesis constant, in the case of 4 loop the field is worth
B = cte 4
N B
4 4 cte
3 3 cte
2 2 cte
1 1 cte
as we see it is a linear relationship.
In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,
Answer:
h≅ 58 m
Explanation:
GIVEN:
mass of rocket M= 62,000 kg
fuel consumption rate = 150 kg/s
velocity of exhaust gases v= 6000 m/s
Now thrust = rate of fuel consumption×velocity of exhaust gases
=6000 × 150 = 900000 N
now to need calculate time t = amount of fuel consumed÷ rate
= 744/150= 4.96 sec
applying newton's law
M×a= thrust - Mg
62000 a=900000- 62000×9.8
acceleration a= 4.71 m/s^2
its height after 744 kg of its total fuel load has been consumed
h= 58.012 m
h≅ 58 m