It is overhead at the equator, it is because the sun ray’s
will be moving vertically as this will be directed at the equator. It is
because if it moves vertically, it will hit or overhead the equator and this
usually happens in spring and fall.
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:
460 g
Explanation:
Heat lost by the warm water = heat gained by the cold water
-mCΔT = mCΔT
-m (4.184 J/g/K) (37°C − 85°C) = (1000 g) (4.184 J/g/K) (37°C − 15°C)
-m (37°C − 85°C) = (1000 g) (37°C − 15°C)
-m (-48°C) = (1000 g) (22°C)
m = 458 g
Rounded to two significant figures, you need a mass of 460 g of water.
Answer:
.10/KWh
Explanation:
divide 606 by 61.37 and you get .1012...
ANSWER
T₂ = 10.19N
EXPLANATION
Given:
• The mass of the ball, m = 1.8kg
First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,
In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,
The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,
We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,
Solve for T₁,
Now, we use the second equation to find the tension in the horizontal string,
Solve for T₂,
Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.
