Answer:
t = 2.2 s
Explanation:
Given that,
Height of the roof, h = 24.15 m
The initial velocity of the pumpkin, u = 0
We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

Here, u = 0 and a = g

So, it will take 2.22 s for the pumpkin to hit the ground.
Explanation:
Formula depicting relation between total flux and total charge Q is as follows.
(Gauss's Law)
Putting the given values into the above formula as follows.
Q =
= 
= 
= -8.4 nC
Therefore, when the unknown charge is q then,
-14.0 nC + 33.0 nC + q = -8.4 nC
q = -27.4 nC
Thus, we can conclude that charge on the third object is -27.4 nC.
v = average speed of movement of the Southwest Indian Ridge = 20 mm/year
d = distance moved by the Southwest Indian Ridge = 100 mm
t = number of years required to move distance "d"
distance traveled is given as
d = v t
inserting the above values in the formula
100 mm = (20 mm/year) t
dividing both side by 20 mm/year
t = 100 mm/(20 mm/year)
t = 5 years
Well, one AU is 149,597,870 km. So, we would basically have to divide 4.5 billion km by 149,597,870, right?
4,500,000,000/149,597,870=30.080642 AU.
So, the correct answer would be 30 AU. Hoped this helped!
Answer:
q = - 93.334 nC
Explanation:
GIVEN DATA:
Radius of ring 73 cm
charge on ring 610 nC
ELECTRIC FIELD p FROM CENTRE IS AT 70 CM
E = 2000 N/C
Electric field due tor ring is guiven as
![E = \frac{KQx}{[x^2+ R^2]^{3/2}}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BKQx%7D%7B%5Bx%5E2%2B%20R%5E2%5D%5E%7B3%2F2%7D%7D)

E1 = 3714.672 N/C
electric field due to point charge q



now the eelctric charge at point P is
E = E1 + E2
solving for q
q = - 93.334 nC