Answer:
A: 4 times as much
B: 200 N/m
C: 5000 N
D: 84,8 J
Explanation:
A.
In the first question, we have to caculate the constant of the spring with this equation:

Getting the k:
![k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bm%2Ag%7D%7Bx%7D%20%3D%5Cfrac%7B0%2C2%5Bkg%5D%2A9%2C81%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%5D%7D%7B0%2C05%5Bm%5D%7D%20%3D39%2C24%5B%5Cfrac%7BN%7D%7Bm%7D%5D)
Then we can calculate how much the spring stretch whith the another mass of 0,2kg:
![x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bm%2Ag%7D%7Bk%7D%20%3D%5Cfrac%7B0%2C4%5Bkg%5D%2A9%2C81%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%5D%7D%7B39%2C24%5B%5Cfrac%7BN%7D%7Bm%7D%5D%7D%20%3D0%2C1%5Bm%5D%5C%5C)
The energy of a spring:

For the first case:
![E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7D%20%2A39%2C24%5B%5Cfrac%7BN%7D%7Bm%7D%5D%2A%280%2C05%5Bm%5D%29%5E%7B2%7D%20%3D0%2C049%20%5BJ%5D)
For the second case:
![E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7D%20%2A39%2C24%5B%5Cfrac%7BN%7D%7Bm%7D%5D%2A%280%2C1%5Bm%5D%29%5E%7B2%7D%20%3D0%2C0196%20%5BJ%5D)
If you take the relation E2/E1 = 4.
B.
We have the next facts:
x=0,005 m
E = 0,0025 J
Using the energy equation for a spring:
⇒![k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]](https://tex.z-dn.net/?f=k%3D%5Cfrac%7BE%2A2%7D%7Bx%5E%7B2%7D%20%7D%20%3D%5Cfrac%7B0%2C0025%5BJ%5D%2A2%7D%7B%280%2C005%5Bm%5D%29%5E%7B2%7D%20%7D%20%3D200%5B%5Cfrac%7BN%7D%7Bm%7D%20%5D)
C.
The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.
![E=W*h=500[N]*10[m]=5000[J]](https://tex.z-dn.net/?f=E%3DW%2Ah%3D500%5BN%5D%2A10%5Bm%5D%3D5000%5BJ%5D)
Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.
D.
The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.
The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:
W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J
Based on the given details with this question, I can say that the direction of motion is not conserved. This is because the boat is subjected to an external force because of water resistance. So, the answer for this question would be NO.
Answer:
d= 7.32 mm
Explanation:
Given that
E= 110 GPa
σ = 240 MPa
P= 6640 N
L= 370 mm
ΔL = 0.53
Area A= πr²
We know that elongation due to load given as



A= 42.14 mm²
πr² = 42.14 mm²
r=3.66 mm
diameter ,d= 2r
d= 7.32 mm
Answer:
0.01631 T
Explanation:
current, i = 28 A
mass per unit length, m/l = 46.6 g/m = 0.0466 kg/m
Let the magnetic field is B.
the weight of the wire is balanced by the magnetic force .
mg = i l B
B = mg / i l
B = (m/l) x g/i
B = 0.0466 x 9.8 / 28
B = 0.01631 T
Thus, the magnetic field is 0.01631 T.
20.3 divided by 3.0 will get u velocity and v times 3.0s