42.9°
Explanation:
Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:


Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at
Solving for the angle, we get

or

![\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]](https://tex.z-dn.net/?f=%5C%3B%5C%3B%5C%3B%3D%20%20%5Csin%5E%7B-1%7D%5Cleft%5B%5Cdfrac%7B34%5C%3A%5Ctext%7BN%7D%7D%7B%285.1%5C%3A%5Ctext%7Bkg%7D%29%289.8%5C%3A%5Ctext%7Bm%2Fs%7D%5E2%29%7D%5Cright%5D)

Answer:Reducing mass i.e. water
Explanation:
Frequency For given mass in glass is given by

where k =stiffness of the glass
m=mass of water in glass
from the above expression we can see that if mass is inversely Proportional to frequency
thus reducing mass we can increase frequency
Answer:
Explanation:
spring constant k = 425 N/m
a ) At the point of equilibrium
restoring force = frictional force
= kx = 10 N
425 x = 10
x = 2.35 cm
b )
Work done by frictional force
= -10 x 2.35 x 10⁻² x 2 J ( Distance is twice of 2.35 cm )
= - 0.47 J
= Kinetic energy remaining with the cookie as it slides back through the position where the spring is unstretched .
= 425 - 0.47
= 424.53 J
=
Answer:
The distance is
Explanation:
From the question we are told that
The initial speed of the electron is 
The mass of electron is 
Let
be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value
Let
be the initial kinetic energy of the electron \
Let
be the kinetic energy of the electron at the distance
from the proton
Considering that energy is conserved,
The energy at the initial position of the electron = The energy at the final position of the electron
i.e

are the potential energy at the initial position of the electron and at distance d of the electron to the proton
Here 
So the equation becomes

Here
are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton
is electrostatic constant with value 
i.e
is the velocity at distance d from the proton = 2
So the equation becomes

![\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20mv_i%5E2%20%20%3D%204%20%5B%5Cfrac%7B1%7D%7B2%7Dmv_i%5E2%20%5D-%20%5Cfrac%7Bk%28q%29%5E2%7D%7Bd%7D)
![3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}](https://tex.z-dn.net/?f=3%5B%5Cfrac%7B1%7D%7B2%7Dmv_i%5E2%20%5D%20%3D%20%5Cfrac%7Bk%28q%29%5E2%7D%7Bd%7D)
Making d the subject of the formula


