Answer:
a) 447.21m
b) -62.99 m/s
c)94.17 m/s
Explanation:
This situation we can divide in 2 parts:
⇒ Vertical : y =-200 m
y =1/2 at²
-200 = 1/2 *(-9.81)*t²
t= 6.388766 s
⇒Horizontal: Vx = Δx/Δt
Δx = 70 * 6.388766 = 447.21 m
b) ⇒ Horizontal
Vx = Δx/Δt ⇒ 70 = 400 /Δt
Δt= 5.7142857 s
⇒ Vertical:
y = v0t + 1/2 at²
-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²
v0= -7 m/s ⇒ it's negative because it goes down.
v= v0 +at
v= -7 + (-9.81) * 5.7142857
v= -62.99 m/s
c) √(70² + 62.99²) = 94.17 m/s
Answer:
B. Trial 2
Explanation:
Trial 2, because the student’s finger applied the largest force to the sensor.
Because the trial 2 student finger applied to largest force.
Answer:
the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time
Explanation:
Since airplane is moving horizontally with constant speed v
so when object is dropped from the plane then the speed of the object will be same as that of the speed of the airplane
so we can say that two object when dropped after some interval of time then they always lie in same vertical line
now we know that they both have same acceleration in vertical line so the motion of two objects relative to each other in vertical direction is always uniform motion because they have no acceleration with respect to each other
So the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time
Answer:
A. 261.6 hz.
B. 0.656 m.
Explanation:
A.
When yhe tube is open at one end and closed at the other,
F1 = V/4*L
Where,
F1 = fundamental frequency
V = velocity
L = length of the tube
When the tube is open at both ends,
F'1 = V/2*L
Where
F'1 = the new fundamental frequency
Therefore,
V/2*L x V/4*L
F'1 = 2 * F1
= 2 * 130.8
= 261.6 hz.
B.
F1 = V/4*L
Or
F'1 = V/2*L
Given:
V = 343 m/s
F1 = 130.8
L = 343/(4 * 130.8)
= 0.656 m.
