Answer:
Frequency, 
Explanation:
Visible red light has a wavelength of 680 nanometers (6.8 x 10⁻⁷ m). The speed of light is 3.0 x 10 ⁸ m / s. What is the frequency of visible red light?
It is given that,
Wavelength of a visible red light is, 
Speed of light is, 
We need to find the frequency of visible red light. It can be calculated using below relation.
So, the frequency of visible red light is 
.
1.velocity and acceleration
2.
3.inertia
4.
5.speed
Answer: Wavelength is the measure of the length of a complete wave cycle. The velocity of a wave is the distance traveled by a point on the wave. In general, for any wave the relation between Velocity and Wavelength is proportionate. It is expressed through the wave velocity formula.
Explanation: For any given wave, the product of wavelength and frequency gives the velocity. It is mathematically given by wave velocity formula written as-
V=f×λ
Where,
V is the velocity of the wave measure using m/s.
f is the frequency of the wave measured using Hz.
λ is the wavelength of the wave measured using m. Velocity and Wavelength Relation
Amplitude, Frequency, wavelength, and velocity are the characteristic of a wave. For a constant frequency, the wavelength is directly proportional to velocity.
Given by:
V∝λ
Example:
For a constant frequency, If the wavelength is doubled. The velocity of the wave will also double.
For a constant frequency, If the wavelength is made four times. The velocity of the wave will also be increased by four times.
Hope you understood the relation between wavelength and velocity of a wave. I truely hope this helps you out tho! Goodluck!
Answer:
D. only briefly while being connected or disconnected.
Explanation:
As we know that transformer works on the principle of mutual inductance
here we know that as per the principle of mutual inductance when flux linked with the primary coil charges then it will induce EMF in secondary coil
So here when AC source is connected with primary coil then it will give output across secondary coil because AC source will have change in flux with time.
Now when we connect DC source across primary coil then it will not induce any EMF across secondary coil because DC source is a constant voltage source in which flux will remain constant always
So here in DC source the EMF will only induce at the time of connection or disconnection when flux will change in it while rest of the time it will give ZERO output
so correct answer will be
D. only briefly while being connected or disconnected.
Answer:
1.758820×10^11(-2.5i-0.8j) m/s^2
Explanation:
From the question, the parameters given are; E=(2.80i+ 5.20j) v/m, a uniform magnetic field,B= 0.400K T, acceleration, a= ??? and velocity vector, v= 11.0i metre per seconds (m/s)...
We can solve this problem using the formula below;
Ma= q[E+V × B] ---------------(1).
Note: q is negative, m= mass of electron.
Making acceleration,a the subject of the formula and substituting the parameters into equation (1);
a= -e/m × (2.5i + 5.2j +11.0i × 0.400K)
a= -e/m × (2.5i+5.2j-4.4j)
a= e/m × (-2.5i - 0.8j)
e/m= 1.758820×10^11 c/kg
Therefore, slotting in the value of charge to mass(e/m) ratio;
a= 1.7588×10^11×(-2.5i-0.8j) m/s^2
