Answer:
a) v_{p} = 2.83 m / s
, b) 50.5º north east
Explanation:
This is a vector problem.
The speed of the ball with respect to the ground is the speed of the ball with respect to Mia plus the speed of Mia with respect to the ground
To make the sum we decompose the speed of the ball in its components
The angle of 30 east of the south, measured from the positive side of the x axis is
θ = 30 + 270 = 300
= 
cos 300
= v_{b} sin. 300
v_{bx} = 3.60 cos 300 = 1.8 m / s
v_{by} = 3.60 sin 300 = -3,118 m / s
Let's add speeds on each axis
X axis
vₓ = v_{bx}
vₓ = 1.8 m / s
Y Axis
= v1 - vpy
v_{y} = 5.30 - 3.118
v_{y} = 2.182 m / s
The magnitude of the velocity can be found using the Pythagorean theorem
= √ (vₓ² + v_{y}²)
v_{p} = √ (1.8² + 2.182²)
v_{p} = 2,829 m / s
v_{p} = 2.83 m / s
b) for direction use trigonometry
tan θ = / vₓ
θ = tan ⁺¹ v_{y} / vₓ
θ = tan⁻¹ 2.182 / 1.8
Tea = 50.48º
This address is 50.5º north east
Answer: relative motion between observer and the sound source.
Explanation: The Doppler effect states that when there is a relative motion between an observer and a sound source the frequency of sound perceived by the observer is different in frequency from the original from the source.
The mathematical back up for this claim is given below.
f' = (v+v') /(v-vs) × f
Where f' = observed frequency
v = speed of sound in air
v' = velocity of observer
vs = velocity of source
f = frequency of sound source.
From the formulae, it can be seen that a change in the value of the velocity of observer (v') and source (vs) produces different value of observed frequency (f').
Note, frequency of sound (f) is a constant.
Answer:
Acceleration of object=
Explanation:
Given:-
Velocity (v)=24 cm/s
Time (t)=3 s
To calculate acceleration (a) of object=?
Now,
Therefore object acceleration =
Answer:
Explanation:
When beam is balanced and not rotating with Suki standing on it , let reaction force on the supports be R₁ and R₂. Then
R₁ +R₂ = 336 + 590
= 929
Now the moment beam begins to tip , reaction on distant support R₁ = 0
only R₂ will exists on the support near to Suki.
Taking torque about this support of weight of beam acting from the middle point and weight of suki of 590N ,who is x distance from the support towards the other end.
336 x 1.5 = 590 x
x = .85 m
ie , from second support , Suki can not go beyond a distance of .85 m towards the second end.
