An archer shoots an arrow at an 85.0 m distant target; the bulls eye of the target is at same height as the release height of th
1 answer:
Answer:
a) 14.1°
b) over
Explanation:
The usual model of ballistic motion assumes that the only force on the flying object is that due to gravity. When an object is launched with initial velocity v0 at some angle θ with respect to the horizontal, the distance it travels is ...
d = (v0)²sin(2θ)/g
Using this relation, we can find the launch angle to make the object travel a given distance:
θ = 1/2arcsin(dg/v0²) . . . . where g is the acceleration due to gravity
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<h3>a)</h3>For the arrow to hit a target 85 m away at the same height it was launched with speed 42.0 m/s, the launch angle must be ...
θ = 1/2arcsin(dg/v0²) = 1/2(arcsin(85·9.8/42²)) ≈ 14.0893°
The arrow must be released at an angle of about 14.1°.
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<h3>b)</h3>The flight time to the tree at a distance of 42.5 m will be that distance divided by the horizontal speed:
t = 42.5/(42cos(14.0893°)) ≈ 1.0433 . . . . seconds
The height at that time is ...
h(t) = -4.9t² +42sin(14.0893°)t ≈ 5.33 . . . meters
The arrow will go <em>over</em> the branch.
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<em>Additional comment</em>
Since gravity provides the only force on the arrow, its horizontal speed is constant at vh = v0·cos(θ), when the arrow is launched with speed v0 at angle θ above the horizontal. Its vertical speed will be reduced by the acceleration of gravity, so will be vv = v0·sin(θ) -gt. The height is the integral of the vertical speed, so is ...
h(t) = (1/2)gt² +v0·sin(θ)t
The height will be 0 at t=0 and at t=2v0sin(θ)/g, so the horizontal distance traveled will be ...
d = vh·t
= (v0·cos(θ))(2v0·sin(θ)/g) = (v0²/g)(2·sin(θ)cos(θ))
= v0²sin(2θ)/g
Note that this is all simplified by the fact that the target and launch point are at the same level (h=0).
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So if you need to measure how much flux goes through a loop, you need the flux density times the area of the loop Φ=BA. The units work out like
Φ=[Wb/m²][m²]=[Wb], which is really just the amount of flux. The H field alone can't tell you this because without μ, we don't know the "number of field" lines that were caused in the material (even in vacuum) by that H field. And the flux cares about the number of lines, not the field intensity.
I'm way into magnetic fields, my PhD research is in this area so I could go on forever. I have included a picture that also shows M, the magnetization of a material along with H and B. M is like the polarization vector, P, of dielectric materials. If you need more info let me know but I'll leave you alone for now!
So if you need to measure how much flux goes through a loop, you need the flux density times the area of the loop Φ=BA. The units work out like
Φ=[Wb/m²][m²]=[Wb], which is really just the amount of flux. The H field alone can't tell you this because without μ, we don't know the "number of field" lines that were caused in the material (even in vacuum) by that H field. And the flux cares about the number of lines, not the field intensity.
I'm way into magnetic fields, my PhD research is in this area so I could go on forever. I have included a picture that also shows M, the magnetization of a material along with H and B. M is like the polarization vector, P, of dielectric materials. If you need more info let me know but I'll leave you alone for now!
A) 0.189 N
The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:
where
G is the gravitational constant
8.7×10^13 kg is the mass of the asteroid
m = 130 kg is the mass of the man
R = 2.0 km = 2000 m is the radius of the asteroid
Substituting into the equation, we find
B) 2.41 m/s
In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:
where
v is the speed of the astronaut
Solving the formula for v, we find the minimum speed at which the astronaut should launch himself and then orbit the asteroid just above the surface: