After one day, the rate of increase in Delta Cephei's brightness is;0.46
We are informed that the function has been used to model the brightness of the star known as Delta Cephei at time t, where t is expressed in days;
B(t)=4.0+3.5 sin(2πt/5.4)
Simply said, in order to determine the rate of increase, we must determine the derivative of the function that provides
B'(t)=(2π/5.4)×0.35 cos(2πt/5.4)
Currently, at t = 1, we have;
B'(1)=(2π/5.4)×0.35 cos(2π*1/5.4)
Now that the angle in the bracket is expressed in radians, we can use a radians calculator to determine its cosine, giving us the following results:
B'(1)=(2π/5.4)×0.3961
B'(1)≈0.46
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Option B is the correct answer.
In a basketball game, a player shoots a jump shot then the floor pushes up on the player.
<h3>Newton's Third Law</h3>
Newton's third law states that when two bodies interact with each other, they apply forces to one another that are equal in magnitude and opposite in direction. The third law is also known as the law of action and reaction.
In the given situation, a player shoots a jump shot. It means that the player pushes the floor downward direction.
Newton's third law is applicable in this situation where the player pushes the floor downward direction, at the same time the floor pushes the player upward. The amount of force applied to the floor by the player is equal in magnitude and opposite in direction as compared to the force applied to the player by the floor.
Hence the option B is the correct answer.
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Answer:
60 km/h
Explanation:
In the first part of the trip, the speed is
v = 80 km/h
while the time interval is
t = 3 h
So, the distance covered is:
d = vt = (80)(3)= 240 km
The problem states that this distance is half distance between home and the destination - so, the total distance between home and the destination is

The time taken to cover the second part of the trip is 5 h, so the total time taken is
T = 3 h + 5 h = 8 h
Therefore, the average velocity for the entire trip is

Refer to the diagram shown below.
The initial KE (kinetic energy) of the system is
KE₁ = (1/2)mu²
After an inelastic collision, the two masses stick together.
Conservation of momentum requires that
m*u = 2m*v
Therefore
v = u/2
The final KE is
KE₂ = (1/2)(2m)v²
= m(u/2)²
= (1/4)mu²
= (1/2) KE₁
The loss in KE is
KE₁ - KE₂ = (1/2) KE₁.
Conservation of energy requires that the loss in KE be accounted for as thermal energy.
Answer: 1/2
A magnetic field does not have any influence on wind. The Coriolis effect, pressure gradient, and friction do affect wind.
The solution is B.