The velocity of the aircraft relative to the ground is 240 km/h North
Explanation:
We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.
Mathematically:
where
v' is the velocity of the aircraft relative to the ground
v is the velocity of the aircraft relative to the air
is the velocity of the air relative to the ground.
Taking north as positive direction, we have:
v = +320 km/h
(since the air is moving from North)
Therefore, we find
(north)
Learn more about vector addition:
#LearnwithBrainly
Answer:
B_o = 1.013μT
Explanation:
To find B_o you take into account the formula for the emf:
where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.
By applying the derivative you obtain:
when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:
hence, B_o = 1.013μT
Answer:
q₁= +0.5nC
Explanation:
Theory of electrical forces
Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
To solve this problem we apply Coulomb's law:
Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
o solve this problem we apply Coulomb's law:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
F=K*q₁*q₂/d² Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁,q₂:Charges in Coulombs (C)
d: distance between the charges in meters
Data:
Equivalences
1nC= 10⁻⁹ C
1cm= 10⁻² m
Data
q₃=+5.00 nC =+5* 10⁻⁹ C
q₂= -2.00 nC =-2* 10⁻⁹ C
d₂= 5.00 cm= 5*10⁻² m
d₁= 2.50 cm= 2.5*10⁻² m
k = 8.99*10⁹ N*m²/C²
Calculation of magnitude and sign of q1
Fn₃=0 : net force on q3 equals zero
F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.
F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.
We propose the algebraic sum of the forces on q₃
F₂₃ - F₁₃=0
We eliminate k*q₃ of the equation
q₁= +0.5*10⁻⁹ C
q₁= +0.5nC
Answer:
Check Explanation
Explanation:
This is a question that is as a result of an experimental procedure.
The Phet simulation is put on the settings given in the question;
- select Oscillate
- Select No End
- Use the parameters in parentheses by sliding the bars for Amplitude (1.00 cm), Frequency (1.40 Hz)
- Damping (none)
- Tension (highest)
I'll attach an interface of the Phet simulation to this solution.
Once all of these settings have been fixed, the simulation gives a wave pattern whose wavelength can be read from the ruler attached to the background of the simulation.
The wavelength is the distance from crest to crest or from trough to trough.
From the simulation example I have attached, the distance from crest to crest is from the green indicator on one crest to the green indicator on the next crest, that is about 5 to 5.1 cm
The velocity of a wave, v, is related to the frequency, f, and wavelength, λ of the wave through
v = fλ
For the photon, the velocity of the wave is the speed of light,
v = c = (3.00 × 10⁸) m/s
The wavelength computed from the simulation = λ = 5.1 cm = 0.051 m
c = fλ
frequency = (c/λ) = (3.00 × 10⁸) ÷ 0.051 = (5.88 × 10⁹) Hz
So, this step can be used to obtaim rhe required frequency of the photon, just follow these steps and use the calculation method too. You should be able to obtain the frequency of the photon in your experiment.
Hope this Helps!!!
