Answer:
First calculate the deceleration making use of the formulation: V(very last speed= U(initial speed) -a(acceleration) t(ime) so 10=20- 4a. unfavourable a as decelerating. making a the acceleration the challenge by using rearranging the elements 4a= 20-10 = 10 so a=10/4 making use of Newton's 2d regulation P(stress) = M (mass) x A (acceleration). so P = 800 x 10/4 = 8000 /4 = 2000 newtons
Explanation:
Answer:
Explanation:
a ) V( primary ) = 100 V
V( secondary ) = 10 V
No of turns ( secondary ) / No of turns ( primary ) = 10 / 100
= 1 / 10
b ) current in secondary
= volt ( secondary ) / resistance
= 10 /6 = 1.67 A
c )
Average power to secondary
= V ( secondary ) x current ( secondary )
= 10 x 10 / 6
= 16.67 W
d )
Power in primary = power in secondary = 16.67 W
e ) current drawn by ac line ( primary )
Volt ( primary ) x current ( primary ) = power in primary
= 16.67
current ( primary )
= 16.67 / 100
= 0.167 A
Answer:
Temporary? You have two options
If you have a positively charged object, bring it close to the LEFT side of the ball (not touching it!).
If you have a negatively charged object, bring it close to the RIGHT side of the ball (again, not in contact).
Either way, the electron cloud will migrate towards the right side of the of your sphere.
Another option is to have a capacitor large enough to house the sphere inside it, and polarize it by attaching it to a DC source making sure that the negative plate of it is to the right of the sphere.
The answer is a bc electric is fast
Answer:
<em>the centripetal acceleration will now be four times its initial value</em>
Explanation:
v=velocity
a=centripetal acceleration
r=radius to the center
take note tat a centripetal force is the force tat is needed to keep a body moving along a circular path
centripetal force is given by f=mv^2/r
a=v^2/r
v=v
wen v is doubled , we ave
2v
substituting into the previous equation of centripetal force
a=(
/r
a=4^{2}/(r)
from the equation, is can be discovered that the centripetal acceleration will now be four times its initial value
