The velocity of the air relative to the runner is 5 m/s.
<h3>What is the relative velocity?</h3>
We must recall that velocity is a vector quantity and the relative velocity must be obtained vectorially. Thus we know that;
Velocity of the runner = 4m/s. due west
Velocity of the wind = 3m/s due south
The relative velocity is;
Vr = √(4)^2 + (3)^2
Vr = 5 m/s
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Birth rate 10/1000 = +population
Death rate 9/1000 = -population
Immigration rate 3/1000 = +population
Emigration rate 7/1000 = -population
10/1000 - 9/1000 + 3/1000 - 7/1000= -3/1000
Growth rate = (-3/1000)(100%)= -0.3%
Answer:
The work done by the child to reach the cat is, W = 3684.8 J
Explanation:
Given data,
The mass of the child, m = 40 kg
The height of the cat from the ground, h = 9.4 m
The force acting on the child due to gravitational force is,
F = mg
The minimum force required by the child to climb the tree against the gravitational force is,
F = mg
Hence, the work done by the child is,
W = F · h
= 40 x 9.8 x 9.4
= 3684.8 joules
Hence, the work done by the child to reach the cat is, W = 3684.8 J
Answer:
47347.1936 milligrams per centimeter
Explanation:
1 pound is equal 453592 milligrams.
one meter is 100cm
10.43*453952/100= 47347.1936milligrams per cm
Answer: a) 0.78 s; b) 899.83 m; c) Δx=1150*Δt
Explanation: In order to solve this problem we have to use the kinematic equations for the independent two axis (x-y). The following expressions are:
y=yo+voy-(g*t^2)/2 g is 9.8 the aceletarion of the gravity on the y axis
vfy=voy-g*t
x=xo+vox+(a*t^2)/2 a=o there is not acceleration in the x axis.
vfx=vox+a*t
voy and vox ara the initial velocities for each axis.
yo and xo are the initial positions of teh bullet.
From these equations we can obtain the time
y=0=3-(g*t^2)/2
t^2=3*2/g
t=
t= 0.78 s
To calculate xf= voy*t= 1150 * 0.78= 899.83 m
Finally considering with the last equation we ontain the error in Δx
in the form:
Δx=1150*Δt since is linear the dependance with time.
