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jek_recluse [69]
3 years ago
12

How does the range of the pumpkin change if its initial velocity is tripled (keeping the angle fixed and less than 90∘)? How doe

s the range of the pumpkin change if its initial velocity is tripled (keeping the angle fixed and less than )? The pumpkin's range is nine times as far. The pumpkin's range is eighteen times as far. The pumpkin's range is three times as far.
Physics
1 answer:
Deffense [45]3 years ago
5 0

Answer:

The pumpkin's range is nine times as far.

Explanation:

Given,

In the first case,

Let a pumpkin is thrown with initial velocity u with an angel theta above the horizontal axis.

Therefore the range of the pumpkin is,

R\ =\ \dfrac{u^2sin2\theta}{g}\,\,\,\,\,\,\,\,\,\,\,\,eqn(1)

Now in the second case

Initial velocity of the pumpkin is three times the first case,

\therefore U\ =\ 3u

Let R' be the new range of the pumpkin.

New range of the pumpkin is,

\therefore R'\ =\ \dfrac{U^2sin2\theta}{g}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn(2)

From eqn (1) and (2), we get,

\therefore R'\ =\ \dfrac{(3u)^2sin2\theta}{g}\\\Rightarrow R'\ =\ \dfrac{9u^2sin2\theta}{g}\\\Rightarrow R'\ =\ 9 R

Hence the pumpkin's range is nine times of the initial case.

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the answer is 2.

Explanation:

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2 years ago
Structures on a bird feather act like a diffraction grating having 7000 7000 lines per centimeter. What is the angle of the firs
Anestetic [448]

Answer:

θ = 28.9°

Explanation:

We are given;

Wavelength; λ = 602nm = 602 x 10^(-9) m

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Thus, the distance between 2 adjacent lines is;

d = 1/700000 = 1.43 x 10^(-6) m

The angle at which diffracted light is formed is given by the formula

mλ = d sinθ

Where;

m is the mth order of the diffraction

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d is the distance separating the centres of 2 adjacent slits

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From the question, m is 1 because it says first order.

Thus, plugging in the relevant values into mλ = d sinθ, we have;

1 x 602 x 10^(-9) = 1.43 x 10^(-6) sinθ

sinθ = 602 x 10^(-9)/(1.43 x 10^(-6))

sinθ = 0.42098

θ = sin^(-1) 0.42098

θ = 28.9°

8 0
2 years ago
The graph below shows the displacement of an object as a constant
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In deep space, there is very little friction. Once they launch a probe into deep space, where there are no external forces actin
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Answer:

move at constant velocity.

Explanation:

Newton's first law (also known as law of inertia) states that:

"when the net force acting on an object is zero, the object will keep its state of rest or if it is moving, it will continue moving at constant velocity".

In the case of the probe, friction in deep space is negligible, therefore when the engine is shut down, there are no more forces acting on the probe: the net force therefore will be zero, so the probe will move at constant velocity.

5 0
3 years ago
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A Boeing 747 ""Jumbo Jet"" has a length of 59.7 m. The runway on which the plane lands intersects another runway. The width of t
Reika [66]

Answer:

1.7 seconds

Explanation:

To clear the intersection, the total distance to be covered = 59.7 + 25 =84.7m

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u^2 = 2990.58

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