a.
The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:
where F is the magnitude of the force, theta is the angle between them and d is the distance.
The problen gives the following data:
The magnitude of the force 750 N.
The angle between the force and the displacement which is 25°
The distance, 26 m.
Plugging this in the formula we have:
Therefore the work done is 17673 J.
b)
The power is given by:
the problem states that the time it takes is 6 s. Then:
Therefore the power is 2945.5 W
Answer:
A) Impulse is the same for both the objects
B) The higher is the speed, the greater will be the height.
Explanation:
Part a)
The time of interaction of the two bodies i.e the hanging mass and the stick is same. Thus, force caused by dart on the block = force caused by block on the dart. Hence, impulse is the same for both the objects.
Part B
The energy will be conserved in the entire reaction process
Hence, Kinetic energy = potential energy
0.5Mv^2 = gh(md+mb)
H is directly proportional to the square of speed.
Hence, the higher is the speed, the greater will be the height.
determining how data will he gathered
Explanation: Apex
Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.
Answer:
The thickness is 
Explanation:
From the question we are told that
The wavelength is 
The first order of the dark fringe is 
The second order of dark fringe considered is 
Generally the condition for destructive interference is mathematically represented as
Here y is the path difference between the central maxima(i.e the origin) and any dark fringe
So the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as
=> 
=> 
=>
