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yulyashka [42]
3 years ago
14

On a safari, a team of naturalists sets out toward a research station located 4.63 km away in a direction 38.7 ° north of east.

After traveling in a straight line for 2.13 km, they stop and discover that they have been traveling 25.9 ° north of east, because their guide misread his compass.
What are (a) the magnitude and (b) the direction (as a positive angle relative to due east) of the displacement vector now required to bring the team to the research station?
Physics
1 answer:
aleksandrvk [35]3 years ago
5 0

Exactly, it is a vector subtraction problem.  Let t=theta,

t1=38.7, d1=4.63;  t2=25.9, d2=2.13

v1=d1 <cos(t1), sin(t1)>=<3,6134, 2.8949>

v2=d2 <cos(t2), sin(t2)>=<1.9161, 0.9304>

Final vector

v3 = v1-v2

=<v3x, v3y>

=<1.6973, 1.9645>

where

v3x=d1*cos(t1)-d2*cos(t2)

v3y=d1*sin(t1)-d2*sin(t2)

The final vector v3 has therefore a magnitude of

||v3||=sqrt(1.6973^2+1.9645^2)=2.5962, and a direction of

theta=atan(1.9645/1.6973)=49.17 degrees  north of east

Note: all three vectors are in the direction of the first quadrant.

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avanturin [10]

Answer:

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\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow \\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{3.3\times 10^{-2}\times 10.8^2}{125}}\\\Rightarrow A=0.17547\ m

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3 years ago
Consider a 100 g object dropped from a height of 1 m. Assuming no air friction (drag), when will the object hit the ground and a
Katyanochek1 [597]

Answer:

speed and time are Vf = 4.43 m/s and  t = 0.45 s

Explanation:

This is a problem of free fall, we have the equations of kinematics

      Vf² = Vo² + 2g x

As the object is released the initial velocity is zero, let's look at the final velocity with the equation

      Vf = √( 2 g X)

      Vf = √(2 9.8  1)

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This is the speed with which it reaches the ground

 

Having the final speed we can find the time

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X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}

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Kelly,

\Delta t=\frac{1}{3600}hr.

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\Delta X=X_{k}-X_{C}=0.021miles

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