Answer: 1.5×10^10 N/C
Explanation:
E= F/q
Where E= magnitude of the electric field
F= force of attraction
q= charge of the given body
Given F= 6.5×10^-8 N
q= 4.3× 10^-18 C
Therefore, E = 6.5×10 ^-8/ 4.3×10^-18
E = 1.5×10^10 N/C
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
ANSWER

EXPLANATION
Parameters given:
Initial velocity, u = 26.2 m/s
When the vase reaches its maximum height, its velocity becomes 0 m/s. That is the final velocity.
We can now apply one of Newton's equations of motion to find the height:

where a = g = acceleration due to gravity = 9.8 m/s²
Therefore, we have that:

That is the height that the vase will reach.
Answer:

Explanation:
Let x be the distance to the shore
From trigonometry properties:

Answer:
Explanation:
Given
Initially Reading on the odometer is 
Final reading on the odometer is 
Time taken is 
average velocity 



Thus the average velocity of mail truck is 