I'm pretty sure that there should be an options to choose. Anyway, I've seen this question before and I know that this is an example of <span>the phi phenomenon.</span>
Answer:
The stitches and dimples around a baseball and a golf ball respectively, disturbs the air drag on the balls once they are in motion, allowing the them to travel more easily.
Explanation:
The stitches on a baseball disturbs the air drag on the ball when the ball is in motion, allowing the ball to travel more easily. Depending on the orientation of the ball in flight, the drag changes as the flow is disturbed by the stitches.
A smooth ball with no stitches or dimples has more air drag that opposes the motion.
A golf ball is smooth ball with dimples to create a thin turbulent boundary layer of air that clings to the ball's surface. This allows the smoothly flowing air to follow the ball's surface a little farther around the back side of the ball, thereby decreasing the size of the wake, and allowing the ball to travel more easily.
Answer:
In a time-position graph (s-t graph):
slope = velocity
In a time-velocity graph (v-t graph):
slope = acceleration
area under graph = change in displacement (distance travelled)
In a time-acceleration graph (a-t graph):
area under graph = change in velocity
Answer:
301.48 J/s
Explanation:
We are given;
Temperature of the sky dropping to −40∘C: T_o = -40°C = -40 + 273 = 233 K
Temperature of your skin and clothes: T = 30°C = 30 + 273 = 303 K
Body surface area of human body is around 2 m². But here only half of the body is facing the sky, Thus Area is: A = 2/2 = 1 m²
To solve this, we will use the equation for thermal heat transfer known as the Stefan bolt Mann equation.
ΔQ/Δt = εσA(T⁴ - (T_o)⁴)
Where;
ΔQ/Δt is the rate at which you body loses energy by radiation
ε is the emissivity of the human body with a value of 0.97
σ is Stefan boltzmann constant with a value of 5.67 X 10^(-8) W/m².K⁴
Thus;
ΔQ/Δt = 0.97 × 5.67 X 10^(-8) × 1(303⁴ - 233⁴)
ΔQ/Δt = 301.48 J/s
To calculate the acceleration of the truck, we use the equation of motion
Here, v is final velocity of the truck, u is initial velocity of truck, a is acceleration of the truck and t is time taken.
Given, u = 6 m/s, v = 14 m/s and t = 4 s.
Substituting these values in above equation, we get
.
Thus, the truck acceleration is 
