Answer:
3.83 m/s
Explanation:
Given that,
Distance covered by Jan, d = 4 miles
1 mile = 1609.34 m
4 miles = 6437.38 m
Time, t = 28 minutes = 1680 s
Jan's average speed,
v = d/t

Hence, the average velocity of Jan is 3.83 m/s.
<span>Yes, there are! r1 and r2 are numbers. The volume of the hollow shell is 4 π 3 ( r 3 1 − r 3 2 ) 4π3(r13−r23). Now multiply by ρ to get the mass.</span>
Answer:
i. 0.34
ii. 0.4
iii. 1700 w/m²
iv. 2211.36 w/m²
Explanation:
Given that
Irradiation of the plate, G = 2500 w/m²
Reflected rays, p = 500 w/m²
Emissive power, E = 1200 w/m²
See attachment for calculations
Nearly equal the output work is greater than the input work because of friction.All machines use some amount of input work to overcome friction.The only way to increase the work output is to increase the work you put into the machine.You cannot get more work out of a machine than you put into it.
Answer:
FB = 0.187 N
Explanation:
To find the magnetic force FB in the wire you use the following formula:

the angle between B and L is given by:

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:
![|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N](https://tex.z-dn.net/?f=%7C%5Cvec%7BF_B%7D%7C%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7DB%28y%29dy%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7D%280.5y%29dy%5C%5C%5C%5C%7C%5Cvec%7BF_B%7D%7C%3D%282.0%2A10%5E%7B-3%7DA%29%28sin36.86%5C%C2%B0%29%280.5T%29%5B%5Cfrac%7B0.25%5E2%7D%7B2%7Dm%5D%3D0.187%5C%20N)
hence, the magnitude of the magnetic force is 0.187N