Answer:
The ball reaches Barney head in 
Explanation:
From the question we are told that
The rise velocity is 
The height considered is 
The horizontal velocity of the large object is 
Generally from kinematic equation
Here s is the distance of the object from Barney head ,
u is the velocity of the object along the vertical axis which is equal but opposite to the velocity of the helicopter
So
So
= 
Solving the above equation using quadratic formula
The value of t obtained is
Conservation of linear momentum:
m*v inital = m*v final
0.06*0.7 + 0.03*0 = 0.06*(-0.2) + 0.03*v
(my algebra, or use ur calculator: 0.06*.07=0.042, etc ... or ur teacher may think you got some help)
0.06*(0.7+0.2)=0.03*v, v = 0.06*0.9/0.03=1.8 m/s
Answer 1.8 m/s (positive, to the right).
Answer:
a) (95.4 i^ + 282.6 j^) N
, b) 298.27 N 71.3º and c) F' = 298.27 N θ = 251.4º
Explanation:
a) Let's use trigonometry to break down Jennifer's strength
sin θ = Fjy / Fj
cos θ = Fjx / Fj
Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be
T = 90 -32 = 58º
Fjy = Fj sin 58
Fjx = FJ cos 58
Fjx = 180 cos 58 = 95.4 N
Fjy = 180 sin 58 = 152.6 N
Andrea's force is
Fa = 130.0 j ^
We perform the summary of force on each axis
X axis
Fx = Fjx
Fx = 95.4 N
Axis y
Fy = Fjy + Fa
Fy = 152.6 + 130
Fy = 282.6 N
F = (95.4 i ^ + 282.6 j ^) N
b) Let's use the Pythagorean theorem and trigonometry
F² = Fx² + Fy²
F = √ (95.4² + 282.6²)
F = √ (88963)
F = 298.27 N
tan θ = Fy / Fx
θ = tan-1 (282.6 / 95.4)
θ = tan-1 (2,962)
θ = 71.3º
c) To avoid the movement they must apply a force of equal magnitude, but opposite direction
F' = 298.27 N
θ' = 180 + 71.3
θ = 251.4º
Answer:
160 Hz , 240 Hz , 400 Hz
Explanation:
Given that
Frequency of forth harmonic is 320 Hz.
Lets take fundamental frequency = f₁
f₁=80 Hz
Frequency of first harmonic = f₂
f₂=2 f₁
f₂ =2 x 80 = 160 Hz
Frequency of second harmonic = f₃
f₃= 3 f₁=3 x 80 = 240 Hz
Frequency of fifth harmonic = f₅
f₅= 5 f₁= 5 x 80 = 400 Hz
Three frequencies are as follows
160 Hz , 240 Hz , 400 Hz
Change in velocity = d(v)
d(v) = v2 - v1 where v1 = initial speed, v2 = final speed
v1 = 28.0 m/s to the right
v2 = 0.00 m/s
d(v) = (0 - 28)m/s = -28 m/s to the right
Change in time = d(t)
d(t) = t2 - t1 where t1 = initial elapsed time, t2 = final elapsed time
t1 = 0.00 s
t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s
Average acceleration = d(v) / d(t)
(-28.0 m/s) / (5.00 s)
(-28.0 m)/s * 1 / (5.00 s) = -5.60 m/s² to the right
