Answer:
The false statement is 'Electric field lines form closed loops'.
Explanation:
- Electric field lines originate from positive end and terminates at negative end,i.e., field lines are inward in direction to the negative charges and outward from the positive charges.
- These lines when close together represents high intensity and when far apart shows low intensity of the field.
- These lines do not intersect, as the tangent drawn on these lines provides us with the field direction and intersection of these lines means two field directions which is not possible.
- These lines unlike magnetic field lines do not form closed loops as they do not turn around but originate at positive end and terminates at negative end which ensures no loop formation.
Answer:
0.011 m.
Explanation:
Energy stored in the spring = Energy of the projectile.
1/2ke² = mgh ................ Equation 1
Where k = spring constant, e = extension or compression, m = mass of the projectile, g = acceleration due to gravity, h = height.
make e the subject of the equation
e = √(2mgh/k)............................. Equation 2
Given: k = 12 N/cm = 1200 N/m, m = 15 g = 0.015 kg, h = 5.0 m
Constant: g = 9.8 m/s²
Substitute into equation 2
e = √(2×0.015×5/1200)
e = √(0.15/1200)
e = √(0.000125)
e = 0.011 m.
Answer:
a)ΔV = 6.48 KV
b)ΔU =18.79 mJ
Explanation:
Given that
E= 1.8 KV/m
a)
We know that
Electric potential difference ΔV given as
ΔV = E .d
Here
E= 1.8 KV/m
d= 3.6 m
ΔV = E .d
ΔV = 1.8 x 3.6 KV
ΔV = 6.48 KV
b)
Given that
q=+2.90 µC
Change in electric potential energy ΔU given as
ΔU = q .ΔV
ΔU =18.79 mJ
Answer:
Maximum Tension=224N
Minimum tension= 64N
Explanation:
Given
mass =8 kg
constant speed = 6m/s .
g=10m/s^2
Maximum Tension= [(mv^2/ r) + (mg)]
Minimum tension= [(mv^2/ r) - (mg)]
Then substitute the values,
Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N
Minimum tension= [8 × 6^2)/2 -(8×9.8)]
=64N
Hence, Minimum tension and maximum Tension are =64N and 2224N respectively
Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m
