Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the
1 answer:
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1) The projectile's motion follows
,
In order to determine the velocity, we must compute the derivative of h(t): Next, we will compute the speed at t=2 s and t=4 s: The negative value of the second speed suggests that the projectile has already attained its highest point and is now descending. 2) The maximum height of the projectile occurs when its speed equals zero: Thus, we have And solving yields
3) To determine the maximum height, we substitute the time at which the projectile reaches this peak into h(t), specifically t=2.30 s: 4) The time at which the projectile lands is when the height reaches zero; h(t)=0, which leads to This results in a second-degree equation, producing two answers: the negative root can be disregarded as it lacks physical significance; the second root is
, which indicates the landing time of the projectile. 5) The moment the projectile impacts the ground corresponds to the velocity at time t=4.68 s:
, carrying a negative sign to denote a downward direction.
,
In order to determine the velocity, we must compute the derivative of h(t): Next, we will compute the speed at t=2 s and t=4 s: The negative value of the second speed suggests that the projectile has already attained its highest point and is now descending. 2) The maximum height of the projectile occurs when its speed equals zero: Thus, we have And solving yields
3) To determine the maximum height, we substitute the time at which the projectile reaches this peak into h(t), specifically t=2.30 s: 4) The time at which the projectile lands is when the height reaches zero; h(t)=0, which leads to This results in a second-degree equation, producing two answers: the negative root can be disregarded as it lacks physical significance; the second root is
, which indicates the landing time of the projectile. 5) The moment the projectile impacts the ground corresponds to the velocity at time t=4.68 s:
, carrying a negative sign to denote a downward direction.
Answer:
A = 4.76 x 10⁻⁴ m²
Explanation:
Given data:
Person's weight = 625 N
Bike's weight = 98 N
Pressure per tire = 7.60 x 10⁵ Pa
Find: Contact area per tire
Total system weight = 625 + 98 = 723 N
Let F represent the force supported by each tire
2F = 723 N
Therefore, F = 361.5 N
Using the formula F = P × A
Contact area, A = 4.76 x 10⁻⁴ m²