Answer:
The normal strain along an axis oriented 45° from the positive x axis in the clockwise direction is -ε₀/2
Explanation:
Given that

From equation of normal strain in x direction:

Substituting the values:

Answer:
a.8m/s is my ans it may help you
Answer:

Explanation:
We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

Similarly,

Now, the angle between velocity and acceleration vectors can be found.
The angle between any two vectors can be found by scalar product of them:

So,

At time t = 0, this equation becomes

Answer:
a) F = 3.2 10⁻¹⁰ N
, b) v = 9.9 10⁷ m / s
Explanation:
a) The electric force is
F = q E
The electric field is related to the potential reference
V = E d
E = V / d
Let's replace
F = e V / d
Let's calculate
F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²
F = 3.2 10⁻¹⁰ N
b) For this part we can use kinematics
v² = v₀ + 2 a d
v = √ 2 ad
Acceleration can be found with Newton's second law
e V / d = m a
a = e / m V / d
a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²
a = 3,516 10⁻¹⁷ m / s²
Let's calculate the speed
v = √ (2 3,516 10¹⁷ 1.4 10⁻²)
v = √ (98,448 10¹⁴)
v = 9.9 10⁷ m / s