Answer:
The magnitude of the net electric field is:
Explanation:
The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).
On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).
Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>
Let's find first E1 and E2.
The electric field equation is given by:
Where:
- k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
- q1 is the first charge
- d1 is the distance from q1 to P
And E2 will be:
Finally, we need to use the Pythagoras theorem to find the magnitude of the net electric field.
I hope it helps you!
Answer:
59.4 meters
Explanation:
The correct question statement is :
A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?
Solution:
We know for a circle of radius r and θ angle by an arc of length S at the center,
S=rθ
This gives
θ=S/r
also we know angular velocity
ω=θ/t where t is time
or
θ=ωt
and we know
1 revolution =2π radians
From this we have
angular velocity ω = 1.4 revolutions per sec = 1.4×2π radians /sec = 1.4×3.14×2×= 8.8 radians / sec
Putting values of ω and time t in
θ=ωt
we have
θ= 8.8 rad / sec × 4.5 sec
θ= 396 radians
We are given radius r = 15 cm = 15 ×0.01 m=0.15 m (because 1 m= 100 cm and hence, 1 cm = 0.01 m)
put this value of θ and r in
S=rθ
we have
S= 396 radians ×0.15 m=59.4 m
It's true, when we lift an object we add energy to it.
because, when we lift an object by applying force , the object attains a height and hence the energy gets stored in it, in the form gravitational potential energy .
The answer is B for sure !