Answer:
Frequency, ![f=4.41\times 10^{14}\ Hz](https://tex.z-dn.net/?f=f%3D4.41%5Ctimes%2010%5E%7B14%7D%5C%20Hz)
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, ![\lambda=6.8\times 10^{-7}\ m](https://tex.z-dn.net/?f=%5Clambda%3D6.8%5Ctimes%2010%5E%7B-7%7D%5C%20m)
Speed of light is, ![c=3\times 10^8\ m/s](https://tex.z-dn.net/?f=c%3D3%5Ctimes%2010%5E8%5C%20m%2Fs)
We need to find the frequency of visible red light. It can be calculated using below relation.
![c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.8\times 10^{-7}}\\\\f=4.41\times 10^{14}\ Hz](https://tex.z-dn.net/?f=c%3Df%5Clambda%5C%5C%5C%5Cf%3D%5Cdfrac%7Bc%7D%7B%5Clambda%7D%5C%5C%5C%5Cf%3D%5Cdfrac%7B3%5Ctimes%2010%5E8%7D%7B6.8%5Ctimes%2010%5E%7B-7%7D%7D%5C%5C%5C%5Cf%3D4.41%5Ctimes%2010%5E%7B14%7D%5C%20Hz)
So, the frequency of visible red light is
.
Answer:
Highly doubt something that requires great focus and mental strength exists
We can apply the law of conservation of energy here. The total energy of the proton must remain constant, so the sum of the variation of electric potential energy and of kinetic energy of the proton must be zero:
![\Delta U + \Delta K=0](https://tex.z-dn.net/?f=%5CDelta%20U%20%2B%20%5CDelta%20K%3D0)
which means
![\Delta K = - \Delta U](https://tex.z-dn.net/?f=%5CDelta%20K%20%3D%20-%20%5CDelta%20U)
The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference (
![\Delta V=-1000 V](https://tex.z-dn.net/?f=%5CDelta%20V%3D-1000%20V)
):
![\Delta U = q \Delta V=(1 eV)(-1000 V)=-1000 eV](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20q%20%5CDelta%20V%3D%281%20eV%29%28-1000%20V%29%3D-1000%20eV)
Therefore, the kinetic energy gained by the proton is
![\Delta K = -(-1000 eV)=1000 eV](https://tex.z-dn.net/?f=%5CDelta%20K%20%3D%20-%28-1000%20eV%29%3D1000%20eV)
<span>And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.</span>
The statement “Impulse is a vector quantity” is true about Impulse.
Answer: Option B
<u>Explanation:
</u>
The object’s action by applied force in a particular time interval, there happens changing in momentum called impulse. It is denoted by a symbol ‘J’ or ‘imp’ and expressed in a unit ‘Ns’. As impulse depends on the acted force, when a collision arises from front, behind or side, the force’s direction would be differed.
![\text {Impulse }=\text {Force } \times \text {time}=\vec{F} \Delta t](https://tex.z-dn.net/?f=%5Ctext%20%7BImpulse%20%7D%3D%5Ctext%20%7BForce%20%7D%20%5Ctimes%20%5Ctext%20%7Btime%7D%3D%5Cvec%7BF%7D%20%5CDelta%20t)
So, from this option A is false as impulse is not a force but changing momentum. The unit is not Newton, it is Newton second (Ns). The force direction differs (impulse direction) for each cases of collision, so option D also false. Hence, option B seems to be correct. Vector quantity deals with both direction and magnitude and important in motion study.
Answer:
The value of resistance when power is 1100 watts =
= 50 ohms
Explanation:
Power
= 2200 Watts
Resistance
= 25 ohms
Power
= 1100 Watts
Resistance
= we have to calculate
Given that the power in an electric circuit varies inversely with the resistance
⇒ P ∝ ![\frac{1}{R}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR%7D)
⇒
= ![\frac{R_{1} }{R_{2} }](https://tex.z-dn.net/?f=%5Cfrac%7BR_%7B1%7D%20%7D%7BR_%7B2%7D%20%7D)
⇒
= ![\frac{25}{R_{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B25%7D%7BR_%7B2%7D%20%7D)
⇒
= 50 ohms
This is the value of resistance when power is 1100 watts.