Answer:
0.00712 m
Explanation:
Given:
Charge on first particle (q₁) = 75 nC = 
Charge on second particle (q₂) = 75 nC =
Force (F) = 1.00 N
Separation (d) = ?
The magnitude of force is given by Coulomb's law which states that, the magnitude of force acting between two charged particles separated by a distance is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.
Therefore, the magnitude of force is given as:
Where, 
is the coulomb's constant.
Plug in the given values and solve for 'd'. This gives,
Therefore, the distance between the charges is 0.00712 m.
Answer: D. 500 J
=========================================================
Explanation:
To find the amount of work done, we multiply the force by displacement
work = force*displacement
work = (100 N)*(5 m)
work = (100*5) Nm
work = 500 J
In this case, "Nm" refers to "Newton meters" and not "nanometers"
1 newton meter is equal to 1 joule
Answer:
The mass of the sand that will fall on the disk to decrease the is 0.3375 kg
Explanation:
Moment before = Moment after
where;
I is moment of inertia = Mr² = 0.3 x (0.3)² = 0.027 kg.m²
substitute this in the above equation;
![m = \frac{ 0.027[3(2 \pi) - 2(2 \pi)]} {0.2^2 * 6\pi } = \frac{ 0.027[6 \pi - 4\pi]} {0.2^2 * 4\pi }\\\\m = 0.3375kg](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B%200.027%5B3%282%20%5Cpi%29%20%20-%202%282%20%5Cpi%29%5D%7D%20%7B0.2%5E2%20%2A%206%5Cpi%20%7D%20%3D%20%5Cfrac%7B%200.027%5B6%20%5Cpi%20%20-%204%5Cpi%5D%7D%20%7B0.2%5E2%20%2A%204%5Cpi%20%7D%5C%5C%5C%5Cm%20%3D%200.3375kg)
Therefore, the mass of the sand that will fall on the disk to decrease the is 0.3375 kg
(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.
(b) The work done in firing the projectile is 2,500 J.
<h3>
Kinetic energy of the projectile at maximum height</h3>
The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;
K.E = ¹/₂mv₀ₓ²
where;
- m is mass of the projectile
- v₀ₓ is the initial horizontal component of the velocity at maximum height
<u>Note:</u> At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.
K.E = (0.5)(2)(30²)
K.E = 900 J
<h3>Work done in firing the projectile</h3>
Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.
W = K.E(i) = ¹/₂mv²
where;
- v is the resultant velocity
v = √(30² + 40²)
v = 50 m/s
W = (0.5)(2)(50²)
W = 2,500 J
Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.
The work done in firing the projectile is 2,500 J.
Learn more about kinetic energy here: brainly.com/question/25959744
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