I can think of two possible and logical questions for the problem given. First, you can calculate for the maximum height reached by the blue ball. Second, you can compute the length of time for the two balls to be at the same height. If so, the solution are as follows:
When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.
1. For the maximum height, the equation is
H = v₀²/2g
where
v₀ is the initial speed
g is the acceleration due to gravity equal to 9.81 m/s²
For the blue ball, v₀ = 21.8 m/s. Substituting the values:
H = (21.8 m/s)²/2(9.81m/s²)
H = 24.22 m
The maximum height reached by the blue ball is 24.22 m + 0.9 = 25.12 m.
2. For this, you equate the y values of both balls:
y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(10.4 m/s)t + 0.5(9.81 m/s²)(t²) + 26.6 m = (21.8 m/s)t + 0.5(9.81 m/s²)(t²) + 0.9 m
Solving for t,
t = 2.25 seconds
Thus, the two balls would be at the same height after 2.25 seconds.
Answer:
The force exerted on the 
is 
Explanation:
From the question we are told that
The area is 
The magnitude of charge placed on them is 
The charge placed between the plate is 
The electric field generated around the plate is mathematically represented as
Substituting values
The force exerted the charge is mathematically represented as
Substituting values
Answer:
The time it takes the ball to stop is 0.021 s.
Explanation:
Given;
mass of the softball, m = 720 g = 0.72 kg
velocity of the ball, v = 15.0 m/s
applied force, F = 520 N
Apply Newton's second law of motion, to determine the time it takes the ball to stop;
Therefore, the time it takes the ball to stop is 0.021 s.
<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
Since the Earth is almost spherical in shape, we are actually to find first the volume of the spherical segment at a depth of 1,000 m. The radius of the Earth is 6,371,000 meters. The volume of a spherical segment is:
V = 1/3*πh²(3r - h)
Substituting the values and making sure the units is in mm,
V = 1/3*π(1000 m * 1000 mm/1 m)²[3(6,371,000 m * 1000 mm/1 m) - (1000 m * 1000 mm/1 m)]
V = 2×10²² mm³
Thus, the total amount of bacteria is:
2×10²² mm³ * 100 bacteria/1 mm³ = 2×10²⁴ bacteria
