Answer:
Here ball and rod will repel each other as they are of similar charges
Explanation:
As we know that the two charges attract or repel each other by electrostatic force
This force is given as

so we know if two charges are similar in nature then they will repel each other and if the two charges are opposite in nature then they will attract each other
So here when rod touch the ball then it transfer its charge to the ball and due to similar charges in ball and rod they both repel each other
Not sure but i will say D
Answer:
Here's what I got:
Let's assume that N and E are + directions while S and W are - directions.
Wind is blowing from SW; thus, it is blowing towards NE (or at 45 deg N of E).
Dividing the wind's speed into components:y-component: +70.71 km/h; x-component: +70.71 km/h
Dividing the airplane's speed into components:y-component: -600 km/h; x-component: 0 km/h
Adding the components to get the resulting components:y-component: -529.29 km/h; x-component: +70.71
Using the Pythagorean Theorem to find the resulting speed:v^2 = y^2 + x^2 so v = 533.99 km/h
To find the angle of direction, use arctan (y/x):arctan (529.29/70.71) = 82.39 deg
ANSWER: velocity = 533.99 km/h at 82.39 deg S of E
Explanation:
Answer: 58.8235 km/h
speed = distance/time
the distance is 10 km
the time is 10 minutes
the unit is not correct, so we first change minute to hour
so 10/60 is 0.166667, rounded to 0.17.
10 km/ 0.17 hours =
Answer:
solution:
dT/dx =T2-T1/L
&
q_x = -k*(dT/dx)
<u>Case (1) </u>
dT/dx= (-20-50)/0.35==> -280 K/m
q_x =-50*(-280)*10^3==>14 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (3)
q_x =-50*(160)*10^3==>-8 kW
T2=T1+dT/dx*L=70+160*0.25==> 110° C
Case (4)
q_x =-50*(-80)*10^3==>4 kW
T1=T2-dT/dx*L=40+80*0.25==> 60° C
Case (5)
q_x =-50*(200)*10^3==>-10 kW
T1=T2-dT/dx*L=30-200*0.25==> -20° C
note:
all graph are attached