Answer:
The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.
Explanation:
The application of coulonb's law is used to approach the question as shown in the attached file.
Answer:
f3 = 102 Hz
Explanation:
To find the frequency of the sound produced by the pipe you use the following formula:
n: number of the harmonic = 3
vs: speed of sound = 340 m/s
L: length of the pipe = 2.5 m
You replace the values of n, L and vs in order to calculate the frequency:
hence, the frequency of the third overtone is 102 Hz
Answer:
The other angle is 30 degrees.
Explanation:
The range of projectile is given by :
Here,
u is the speed of launch of projectile
Here,
We need to find the other launch angle when the projectile have the same range, such that,
So, the other angle is 30 degrees. Hence, this is the required solution.
Answer:
The fraction of mass that was thrown out is calculated by the following Formula:
M - m = (3a/2)/(g²- (a²/2) - (ag/2))
Explanation:
We know that Force on a moving object is equal to the product of its mass and acceleration given as:
F = ma
And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²
Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.
Case 1:
Hot balloon of mass = M
acceleration = a
Upward force due to hot air = F = constant
Gravitational force downwards = Mg
Net force on balloon is given as:
Ma = Gravitational force - Upward Force
Ma = Mg - F (balloon is moving downwards so Mg > F)
F = Mg - Ma
F = M (g-a)
M = F/(g-a)
Case 2:
After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:
Net Force is given as:
-m(a/2) = mg - F (Balloon is moving upwards so F > mg)
F = mg + m(a/2)
F = m(g + (a/2))
m = F/(g + (a/2))
Calculating the fraction of the initial mass dropped: