Answer:
(C) line integral along a closed loop
This integral is evaluated by dividing the path into infinitesimal segments dL, calculating the scalar product B•dL for each segment, and sum these products. The component of B parallel to dL at each point is always used in evaluating this integral.
The line integral is equal to mu_o × I.
Explanation:
Explanation:
Given
Velocity v = 23.0m/s
Distance S = 3.45m
Required
Time it will take the skier to reach the ground;
Using the equation of motion;
S = ut + 1/2gt²
3.45 = 23t + 1/2(9.8)t²
3.45 = 23t + 4.9t²
4.9t²+23t-3.45 = 0
Factorize;
t = -23 ±√23²-4(4.9)(-3.45)/2(4.9)
t = -23 ±√529+67.62/9.8
t = -23±√596.62/9.8
t = -23±24.43/9.8
t = 1.43/9.8
t = 0.146 secs
Hence take the skier 0.146 secs to reach the ground.
b) Horizontal distance covered is the range;
Range = U√2H/g
Range = 23√2(3.45)/9.8
Range = 23√6.9/9.8
Range = 23√0.7041
Range = 23(0.8391)
Range = 19.29m
Hence the horizontal distance travelled in air is 19.29m
Electric field is proportional to the charge and inversely proportional to the square of distance:

With charge 2q and distance 2r, the electric field is proportional to:

The new electric field is half of that of the original measured field.
They give us the cable tension t = 1500 newton. We assume that the mass of the cable is negligible compared to that of the tower.
We have a force t of 1500 Newton. This force has a vertical component on the y axis and a horizontal component on the x axis.
Of these two components of force, we are especially interested in calculating is the magnitude of the vertical component.
If the angle that it forms with the ground is 50 °, then the vertical component of the force is:
Fy = 1500sin (50)
Fy = 1149 N. <1200 N
The wire will not be loose
Answer:
350 m/s
31°C
Explanation:
Speed of sound is given as the product of frequency and wavelength
S=fw
Where s represent the speed in m/s, f is frequency and w is wavelength
Conversion
Taking 1m to be 100 cm then 35 cm will be 35/100=0.35m
Substituting 0.35 m for w and 1000 Hz for f then
S=1000*0.35=350 m/s
speed of sound (m/s) = 331.5 + 0.60 T(°C)
350=331.5+0.6T
T=30.833333333333
The temperature is approximately 31°C