A cross country skier moves from location A to location B to location C to location D. Each leg of the back and forth motion tak
es 1 minute to complete; the total time is 3 minutes.
What is the distance travelled by the skier in the first minute?
What is the displacement of the skier in the first minute?
What is the distance travelled by the skier in the first two minutes?
Please help it’ll be greatly appreciated!!
What is the distance travelled by the skier in the first minute?
What is the displacement of the skier in the first minute?
What is the distance travelled by the skier in the first two minutes?
Please help it’ll be greatly appreciated!!
1 answer:
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Answer:
maximum height is y = b²/18g √ (12L/b)³
Explanation:
Let's analyze the situation first we have a projectile subjected to an acceleration depends on time, so we must use the definition of acceleration to find the speed when it is at distance L, then we will use the projectile launch equations
Acceleration dependent on t
a = dv / dt
dv = adt
∫dv =∫ (b t²) dt
v = b t³ / 3
The initial speed is zero for zero time
we use the definition of speed
v = dy / dt
dy = v dt
∫dy = ∫b t³ / 3 dt
y = b/3 t⁴ / 4
y = b/12 t⁴
we evaluate from the initial point where the height is zero for the zero time
Let's calculate the time to travel the length (y = L) of the canyon
t = (12 y / b) ¼
t = (12 L / b) ¼
Taking the time, we can calculate the projectile's output speed
v = b/3
This is the speed of the body, which is the initial speed for the projectile launch movement. Let's calculate the highest point where the zero speed
Vy² = v₀² - 2 g y
0 = Vo² - 2 g y
2 g y = v₀²
y = v₀²/ 2g
y = 1/2g [b/3 ) ] 2
y = 1 / 2g [b²/9 ]
y = b²/18g √ (12L/b)³
Answer:
Time period of the planet is
Explanation:
As we know by Kepler's law of time period
square of time period of planet is proportional to the cube of its orbital radius
here we know for Earth
now we have
Answer:
Value of angle between vector a and b is .
Explanation:
Vectors a and b have scalar product 6.00
Let be the angle between a and b.
ab cos = 6 ...(1)
Vectors a and b have magnitude of vector product 9.00
ab sin = 9 ...(2)
Dividing equation (2) by (1) we get
tan = 1.5
=
Thus, value of angle between vector a and b is .