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Alinara [238K]
2 years ago
7

. Emergency rations are to be dropped from a plane to some stranded hikers. The search and rescue plane is flying at an altitude

of 1500 m at 70 m/s. a. Determine where the ideal drop point would be (measured horizontally from the hikers). (Answer is 1.22 km I need help solving b) b. The pilot notices (too late!) that they have passed the ideal drop point; the supplies now need to be launched vertically in order to land near the hikers. Calculate the vertical launch velocity required, given that the plane is now 1 km away (horizontally) from the hikers.
Physics
2 answers:
iragen [17]2 years ago
6 0

Answer: 35 m/s will be the initial vertical launch velocity.

Explanation:

So this is where learning about quadratic equations becomes useful. First we have have to calculate the amount of time it takes the basket to reach the hikers which can be solved by:

1000 m / (70 m/s) = 14.28 s [We know the horizontal displacement and velocity, so calculating time is simple. Convert km to m to make life easier, and remember to handle vertical and horizontal velocity individually.]

The hard part is to find vertical velocity. To do so, set up your equation:

d=v•t+ 1/2 at^2

Vertical displacement we know is 1500 m, time is 14.28s and half acceleration is 4.9m/s^2. Your equation should look like this when the numbers are plugged in:

1500(m)=v•×14.28(s)+4.9(m/s^2)×14.28(s^2)

To convert this into a solvable equation, we will set the left side to 0 and arrange the equation, to look like this:

0=v•14.28(s)+4.9(m/s^2)×14.28(s^2)-1500(m)

Isolate v• and solve.

(Note: v• stands for initial vertical velocity if confused)

almond37 [142]2 years ago
3 0

Answer:

35 m/s down

Explanation:

The horizontal speed of the package is 70 m/s.  So the time needed to reach the hikers is:

1000 m / (70 m/s) = 14.28 s

Taking down to be positive, the initial velocity needed is:

Δy = v₀ t + ½ at²

1500 m = v₀ (14.28 s) + ½ (9.8 m/s²) (14.28 s)²

v₀ = 35 m/s

The package must be launched down with an initial velocity of 35 m/s.

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Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.
KIM [24]

Answer:

The velocity is v = 4.76 \ m/s

Explanation:

From the question we are told that

   The first distance is   d_1  =  4.0 \ km  =  4000 \ m

   The  first speed  is  v_1 =  5.0 \ m/s

    The  second distance is  d_2  =  1.0 \ km  =  1000 \ m

    The  second speed  is  v_2  =  4.0 \ m/s

Generally the time taken for first distance is  

      t_1 =  \frac{d_1 }{v_1 }

        t_1 =  \frac{4000}{5}

       t_1 =  800 \ s

The time taken for second  distance is

           t_1 =  \frac{d_2 }{v_2 }

        t_1 =  \frac{1000}{4}

       t_1 =  250 \ s

The total time is mathematically represented as

     t =  t_1 + t_2

=>   t =  800 + 250

=>    t =  1050 \ s

Generally the constant velocity that would let her finish at the same time is mathematically represented as

      v =  \frac{d_1 + d_2}{t }

=>    v =  \frac{4000 + 1000}{1050 }

=>    v = 4.76 \ m/s

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2 years ago
Which of the following characteristics of Earth's relationship to the Sun explains the existence of Earth's seasons? Choose all
Aneli [31]

Answer:

Earth's axis is tilted relative to its orbital plane.

Earth orbits around the Sun, completing one orbit each year

Explanation:

The earth tilt at an angle causes the sun rays to hit the earth surface around the globe differently. Due to the oblique angle that the rays hit the subtropics and poles, there is less heat intensity compared to the equator where the sun rays hit the earth's surface at a more or less right angle.

The earth rotation around the sun also causes seasons coupled with the earth’s tilts. As the earth rotates, in one point in the orbit, the northern or southern hemispheres will be tilted towards the sun. The phenomenon varies the local temperatures of particular regions of the earth hence driving seasonal climatic changes.

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for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha
VLD [36.1K]

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

R=\dfrac{u^2\ \sin2\theta}{g}

Here,

u is the speed of launch of projectile

Here, \theta=30^{\circ}

We need to find the other launch angle when the projectile have the same range, such that,

\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}

\sin(60)=\sin2\alpha

\dfrac{\sqrt3}{2}=\sin2\alpha

\alpha =30^{\circ}

So, the other angle is 30 degrees. Hence, this is the required solution.

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A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At
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Answer:

E/4

Explanation:

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E = σ/(2ε₀)

Where;

E is the electric field

σ is the surface charge density

ε₀ is the electric constant.

Formula to calculate σ is;

σ = Q/A

Where;

Q is the total charge of the sheet

A is the sheet's area.

We are told the elastic sheet is a square with a side length as d, thus ;

A = d²

So;

σ = Q/d²

Putting Q/d² for σ in the electric field equation to obtain;

E = Q/(2ε₀d²)

Now, we can see that E is inversely proportional to the square of d i.e.

E ∝ 1/d²

The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.

From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;

E_new = E/4

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The fundamental force responsible for the cohesion of the water molecules leaving the faucet is the electromagnetic force.
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2 years ago
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