To solve this question, we use the wave equation which is:
C=f*λ
where:
C is the speed;
f is the frequency;
λ is the wavelength
So in this case, plugging in our values in the problem. This will give us:
C = 261.6Hz × 1.31m
= 342.696 m/s is the answer.
no BECQUSE POSUM BROOB SHSHSJ
Answer : 413.44N
Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .
- From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
- The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
- When a elevator accelerates down , the weight recorded is less than the actual weight .
From the Free body diagram ,
- Mass of the man = 64.2 kg
Answer:
C. 2 N forwards
Explanation:
Net Force in y direction = 0 since the lift force and gravity cancel out. Net force in x direction = 6 N forward - 4 N backwards = 2 N forwards.
The direction of electric field by the charge in and on the metal block will be along the direction line 5 as given in question.
<h3>
How to determine electric field direction in a metal block?</h3>
The charge always remain on outer surface of metal and inside the metal block, the net electric field is zero. But due to dipole there is an electric field at the center of metal block i.e. at point R along direction line 1.
Now, to make make the net electric field zero at center, the electric field by the charge in and on the metal block must be equal in magnitude to that of electric field due to dipole at point R and in opposite direction to that of the net electric field at at R due to dipole.
The electric field by the charge in and on the metal block will be making 180° angle to the electric field due to dipole at point R.
Hence the direction of electric field by the charge in and on the metal block will be along the direction line 5 as given in question.
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