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nalin [4]
3 years ago
6

An object is thrown with an initial velocity v0 forming an angle θ with an inclined plane, which a In turn it forms an α-angle α

with a horizontal plane.
A) Find the distance X from the launch point to the point where the object falls. X = F (v0, θ, α, g)

B) Find the distance for v0 = 20m / s, θ = 53o , α= 36 ° , g = 9.8m / s2

Physics
1 answer:
xxMikexx [17]3 years ago
7 0
Refer to the figure shown below, which is based on the given figure.

d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.

Part A
From the geometry, obtain
d = X cos(α)                     (1a)
h = X sin(α)                      (1b)

The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α)               (2a)
u = v₀ cos(θ - α)             (2b)

If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0             (3a)
ut = d                                (3b)

Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain
0.5(9.8)( \frac{d}{u})^{2} -v_{0} sin(\theta -  \alpha ) \frac{d}{u} - h = 0
4.9[ \frac{X cos \alpha }{v_{0} cos(\theta -  \alpha }  ]^{2} - v_{0} sin(\theta -  \alpha ) [ \frac{X cos \alpha }{v_{0} cos(\theta -  \alpha } ] - X sin \alpha  = 0
Hence obtain
aX^{2}-bX=0 \\ where \\ a=4.9[ \frac{cos \alpha }{v_{0} cos(\theta -  \alpha )}]^{2} \\  b = cos \alpha \,  tan(\theta -  \alpha ) + sin \alpha
The non-triial solution for X is
X= \frac{b}{a}

Answer:
X= \frac{sin \alpha  + cos \alpha  \, tan(\theta -  \alpha )}{4.9 [ \frac{cos \alpha }{v_{0} \, cos(\theta -  \alpha )}  ]^{2}}

Part B
v₀ = 20 m/s
θ = 53°
α = 36°

sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423

X = 0.8351/(4.9*0.0423²) = 101.46 m

Answer:  X = 101.5 m

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Answer:

20 m

Explanation:

Given:

v₀ = 15 m/s

v = -25 m/s

a = -10 m/s²

Find: Δy

v² = v₀² + 2aΔy

(-25 m/s)² = (15 m/s)² + 2 (-10 m/s²) Δy

Δy = 20 m

5 0
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A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len
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Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}

\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}

\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}

q' =66.41 cm

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A planned high-speed train between Houston and Dallas will travel a distance of 386 kilometers in 5.40 × 10^3 seconds. What is t
Mazyrski [523]

¡Hellow!

For this problem, first, lets convert the seconds in hours:

5,4x10³\rightarrow 5400

h = sec / 3600

h = 5400 s / 3600

h = 1,5

Let's recabe information:

d (Distance) = 386 km

t (Time) = 1,5 h

v (Velocity) = ?

For calculate velocity, let's applicate formula:

                                                    \boxed{\boxed{\textbf{d = v * t} } }

Reeplace according we information:

386 km = v * 1,5 h

v = 386 km / 1,5 h

v = 257,33 km/h

The velocity of the train is of <u>257,33 kilometers for hour.</u>

<u></u>

Extra:

For convert km/h to m/s, we divide the velocity of km/h for 3,6:

m/s = km/h / 3,6

Let's reeplace:

m/s = 257,33 km/h / 3,6

m/s = 71,48

¿Good Luck?

7 0
3 years ago
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