Answer:
the electric field strength of this charge is two times the strength of the other charge
Explanation:
Using the relationship between electric field and the charge, which is inversely proportionality. Let the the magnitude of the first charge be Q and the respective electric field be E. It implies that;
E1/E2 = Q2/Q1
E2 = E1 x Q1/Q2
= E x Q/ (Q/2)
= 2E
Answer: Cumulus
Explanation: Most large cloud fronts are made up of cumulus clouds, large storm clouds are cumulonimbus clouds.
Answer:
Explanation:
We know that acceleration is change in velocity over time.
v is the final velocity and u is the initial velocity.
Solve for v.
Multiply both sides by t.
Add u to both sides.
Given :
Two forces act on a 6.00-kg object. One of the forces is 10.0 N.
Acceleration of object 2 m/s².
To Find :
The greatest possible magnitude of the other force.\
Solution :
Let, other force is f.
So, net force, F = 10 + f.
Now, acceleration is given by :
Therefore, the greatest possible magnitude of the other force is 2 N.
Hence, this is the required solution.
Answer:
Speed =0.283m/ s
Direction = 47.86°
Explanation:
Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane
MU1 =MU2cos38 + MV2cos y ...x plane
0 = MU2sin38 - MV2sin y .....y plane
Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2
Substitute into equation above
.46 = .34cos38 + V2cos y ...equ1
.34sin38 = V2sin y...equ2
.19=V2cos Y...x
.21=V2sin Y ...y
From x
V2 =0.19/cost
Sub V2 into y
0.21 = 0.19(Sin y/cos y)
1.1052 = tan y
y = 47.86°
Sub Y in to x plane equ
.19 = V2 cos 47.86°
V2=0.283m/s
