The volumes are 200cm3 and 0.0002m3
Answer:
1) The magnetic field outside the loop is zero.
In region III the magnetic fields due to the two wire loops point in the opposite direction andhence cancel each other. Therefore the magnetic field is zero in region I, III and V
The diagram is attached
Answer:
Explanation:
For the problem, we should have same reynolds number
ρvd/mu = constant
1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600
d = 25.66 cm
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
Answer:
a) a= 8.33 m/s², T = 12.495 N
, b) a = 2.45 m / s²
Explanation:
a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.
We apply Newton's second law to the lower charge
fr₁ + fr₂ = ma
The equation for the force of friction is
fr = μ N
Y Axis
N - W₁ –W₂ = 0
N = W₁ + W₂
N = (m₁ + m₂) g
Since the beams are the same, it has the same mass
N = 2 m g
We replace
μ₁ 2mg + μ₂ mg = m a
a = (2μ₁ + μ₂) g
a = (2 0.30 + 0.25) 9.8
a= 8.33 m/s²
Let's look for cable tension with beam 2
T = m₂ a
T = 1500 8.33
T = 12.495 N
b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal
fr = m₂ a₂
N- w₂ = 0
N = W₂ = mg
μ₂ mg = m a₂
a = μ₂ g
a = 0.25 9.8
a = 2.45 m / s²