Answer:
The spring constant of this spring is 200 N/m.
Explanation:
Given:
Original unstretched length of the spring (x₀) = 10 cm =0.10 m [1 cm =0.01 m]
Stretched length of the spring (x₁) = 18 cm = 0.18 cm
Force acting on the spring (F) = 16 N
Spring constant of the spring (k) = ?
First let us find the change in length of the spring or the elongation caused in the spring due to the applied force.
So, Change in length = Final length - Initial length
Now, restoring force acting on the spring is directly related to its elongation or compression as:
Rewriting in terms of 'k', we get:
Now, plug in the given values and solve for 'k'. This gives,
Therefore, the spring constant of this spring is 200 N/m.
Answer:
a) 2.4 mm
b) 1.2 mm
c) 1.2 mm
Explanation:
To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:
a: width of the slit
λ: wavelength
m: order of the minimum
for little angles you have:
y: height of the mth minimum
a) the width of the central maximum is 2*y for m=1:
b) the width of first maximum is y2-y1:
![w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm](https://tex.z-dn.net/?f=w%3Dy_2-y_1%3D%5Cfrac%7B%28500%2A10%5E%7B-9%7Dm%29%281.2m%29%7D%7B0.50%2A10%5E%7B-3%7Dm%7D%5B2-1%5D%3D1.2mm)
c) and for the second maximum:
![w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm](https://tex.z-dn.net/?f=w%3Dy_3-y_2%3D%5Cfrac%7B%28500%2A10%5E%7B-9%7Dm%29%281.2m%29%7D%7B0.50%2A10%5E%7B-3%7Dm%7D%5B3-2%5D%3D1.2mm)
Answer:
Explanation:
Given
mass of ship =1240 kg
height=37 m
(a)Work done by Tension in cable+work done by gravity =0
Work done by gravity =change in potential energy of mass
Thus work done by tension=449.624 kJ
(b)work done by force of gravity
=-449.624 kJ
Answer:
D). energy resulting from the attraction between two masses.
Explanation:
