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Readme [11.4K]
2 years ago
9

if A and B are non zero vectors, is it possible for vector A×vector B and vector A.vector B both to be zero? Justify your answer

​
Physics
1 answer:
DedPeter [7]2 years ago
3 0

Answer:

not able to understand the question

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Calculate the wavelength of a photon having 3.26 x 10^-19 joules of energy
bekas [8.4K]
The energy of a photon is given by:
E=hf
where h is the Planck constant and f is the photon frequency.
We know the energy of the photon, E=3.26 \cdot 10^{-19} J, so we can rearrange the equation to calculate the frequency of the photon:
f= \frac{E}{h}= \frac{3.26 \cdot 10^{-19}J}{6.6 \cdot 10^{-34}Js}=4.94 \cdot 10^{14}Hz

And now we can use the following relationship between frequency f, wavelength \lambda and speed of light c to find the wavelength of the photon:
\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{4.94 \cdot 10^{14} Hz}=6.07\cdot 10^{-7} m=607 nm
8 0
2 years ago
An atom with seven valence electrons would most likely lose an electron to become stable true or fasle
mel-nik [20]

Answer:

false

Explanation:

since it has seven valence electrons it means it's a non metal and non metals always gain electrons.

5 0
2 years ago
Two point charges each have a value of 3.0 c and are separated by a distance of 4.0 m. what is the electric field at a point mid
swat32
 <span>Place a test charge in the middle. It is 2cm away from each charge. 
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point. 
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out. 
THIS IS A TRICK QUESTION. 
THe electric field exactly midway between them = 0/Q = 0. 
But if the point moves even slightly you need the following formula 
F= (1/4Piε)(Q1Q2/D^2) 
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
4 0
3 years ago
What is the force needed to give a .25 kg arrow an acceleration of 196m/s2?
Rzqust [24]
<span>49N is the force needed to give a .25 kg arrow an acceleration of 196m/s2. F =ma ⇒ =( 0.25kg)(196m/s2) = 49N if the arrow is shot horizontally where the applied force is entirely in the x-direction.</span>
5 0
2 years ago
Read 2 more answers
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