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Fudgin [204]
3 years ago
12

A plane flies from base camp to lake a, 200 km away in the direction 20.0° north of east. after dropping off supplies it flies t

o lake b, which is 230 km at 30.0° west of north from lake
a. graphically determine the distance and direction from lake b to the base camp.
Physics
1 answer:
Liono4ka [1.6K]3 years ago
8 0

Distance of lake a is 200 km at 20 degree north of east

distance between lake a and b is 230 km at 30 degree west of north

now the distance between base and lake b is given as

d = d_1 + d_2

given that

d_1 = 200 cos20 i + 200 sin20 j

d_1 = 187.94 i + 68.4 j

d_2 = -230 sin30 i + 230 cos30 j

d_2 = -115 i + 199.2 j

now the total distance is

d = (187.94 - 115)i + (199.2 + 68.4)j

d = 72.94 i + 267.6 j

now the magnitude of the distance is given as

d = \sqrt{72.94^2 + 267.6^2}

d = 277.4

also the direction is given as

\theta = tan^{-1}\frac{267.6}{72.94}

\theta = 74.7 degree

<em>so it is 277.4 km at 74.7 degree North of East</em>

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To solve this problem we will use the concepts related to the Impulse-Momentum Theorem for which it is specified as the product between force and change in time

\Delta p = F\Delta t

And

\Delta p = m\Delta v

Where,

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\Delta t = \text{Change in Time}

\Delta v = \text{Change in velocity}

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Rearranging to find the Force we have that

F = \frac{\Delta p}{\Delta t}

Using the expression between mass and velocity

F = \frac{m(v_f-v_i)}{\Delta t}

Our values are given as,

m = 50.2kg\\v_i = 0m/s \\v_f = 2.8m/s \\\Delta t = 20.1s

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7 0
3 years ago
A spring with a mass of 5 Kg has natural length 0.5m. A force of 35.6 N is required to maintain it stretched to a length of 0.5m
aleksandrvk [35]

Answer:

Explanation:

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= 35.6 / 0.5

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angular frequency ω of oscillation by spring mass system

\omega = \sqrt{\frac{k}{m} }

where m is mass of the body attached with spring

Putting the values

\omega = \sqrt{\frac{71.2}{5} }

ω = 3.77 radian / s

The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2

so the equation for displacement from equilibrium position that is middle point can be given as follows

x = .5 sin ( ω t + π / 2 )

= 0.5 cos ω t

= 0.5 cos 3.77 t .

x = 0.5 cos 3.77 t .

8 0
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Answer:

10m/s^2

Explanation:

Force = mass x acceleration

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The thrower's movement imparts kinetic energy to a ball thrown vertically. The maximum height that can be achieved after leaving the hand will depend on the actual velocity. Air resistance causes some of this energy to be lost to the air as frictional dissipation, which warms the air in the area as well as the ball's surface.

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The gravitational potential energy begins to rise as the ball moves vertically upward at precisely the same pace as it loses kinetic energy. The ball experiences a steady downward acceleration of 9.81 m/s2, which causes it to initially decline until it briefly comes to a stop at its highest point.

Due to its current position in the Earth's gravitational field relative to its initial position, all of the energy at this point is gravitational potential energy. As the ball experiences constant downward acceleration, its motion immediately becomes apparent in that direction because the acceleration easily transforms gravitational potential energy back into kinetic energy.

As a result, at every point along the trajectory, the total of these interchangeable forms of energy remains constant.

To learn more about what happens when a ball is thrown vertically upward:

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3 0
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Stels [109]

Answer:

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Explanation:

^

6 0
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