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mars1129 [50]
3 years ago
14

a model rocket flies horizontally off the edge of a cliff at a velocity of 40.0m/s. if the canyon below is 110.0m deep, how far

from the edge of the cliff does the model rocket land?
Physics
1 answer:
Over [174]3 years ago
5 0

Answer:

189.6 m

Explanation:

First of all, we study the vertical motion of the rocket in order to find the time it takes for it to land. The suvat equation for the vertical motion is

s=ut+\frac{1}{2}gt^2

where, taking downward as positive direction:

s = 110.0 m is the vertical displacement

u = 0 is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time of flight

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(110)}{9.8}}=4.74 s

Now we can just analyze the horizontal motion, which is a uniform motion with constant velocity, which is equal to

v_x = 40.0 m/s

So the distance travelled horizontall after a time t is

d=v_x t

So, when the rocket lands (t = 4.74 s), the horizontal distance travelled is

d=(40.0)(4.74)=189.6 m

Therefore, the rocket lands 189.6 m far from the edge of the cliff.

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Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surfac
alexgriva [62]

Answer:

a. A = 0.0859 m^2

b. A = 0.0178 m^2

Explanation:

Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surface is tilted at an angle of from the horizontal, and a net magnetic flux of 8.4 103 Wb passes through it. The same net magnetic flux passes through the second surface. (a) Determine the area of the first surface. (b) Find the smallest possible value for the area of the second surface.

take note that the question has not specified th angle which the surface is tilted so i assume the angle is at 12^{0} to the horizontal

flux = BAcos(\alpha)

B=magnetic flux in Weber

A=area of the flat surface in m^2

\alpha=the angle to the horizontal

a) 8.4 x10^-3= (.47)Acos(78)

alpha has to be the angle from the normal and not the horizontal so 90-12=78,

 8.4 x10^-3

/(.47)cos(78)

A = 0.0859 m^2

b) If flux remains the same then for it to be the smallest possible area it needs to be perpendicular to the magnetic field so alpha would be 0.

8.4 x10^-3 = (.47)Acos(0)

A = 0.0178 m^2

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4 years ago
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Alona [7]

Answer:

Explanation:

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A small block of mass M = 0.10 kg is released from rest at point 1 at a height H = 1.8 m above the bottom of a track, as shown i
Ede4ka [16]

Answer:

D

Explanation:

A) is not correct, because the gravitation potential energy will depend on the height the block is located at. It will be calculated with the formula:

U=mgh.

If we take the ground as a zero height reference, then on point 2 the potential energy will be:

U_{2} = 0.10kg(9.81 m/s^{2})(0.6m)

U_{2}=0.59 J

While on point 3, the potential energy will be greater.

U_{3}=0.10kg(9.81 m/s^{2})(1.2m)

U_{3}=1.18 J

B) is not the right answer because the kinetic energy will vary with the height the block is located at in the fact that the energy is conserved (this is if we don't take friction into account or air resistance) so in this case:

U_{2}+K_{2}= U_{3}+K_{3}

We already know the potential energy at point 2. We can calculate the kinetic energy at point 3 like this:

K_{3} =\frac{1}{2}mv_{3}^{2}

K_{3} =\frac{1}{2}(0.10kg)(2.5 m/s)^{2}

K_{3} =0.31 J

So the kinetic energy at point 2 is given by the equation:

K_{2}  =U_{3}-U_{2}+K_{3}

so:

K_{2} = (1.18J)-(0.59J)+0.31J

K_{2} =0.9J

As you may see the kinetic energy at point 2 is greater than the kinetic energy at point 3.

C) Is not correct because according to the first law of thermodinamics, energy is not lost, only transformed. So, since we are not taking into account friction or any other kind of loss, then we can say that the amount of mechanical energy at point 1 is exactly the same as the mechanical energy at point 3.

D) Because of what we talked about on part C, this will be the true situation, because the mechanical energy of the block will be the same no matter theh point you measure it at.

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Answer:

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