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devlian [24]
3 years ago
10

Which of the following is a pure substance?

Physics
2 answers:
astraxan [27]3 years ago
7 0
Yes I agree that it is D as well
11Alexandr11 [23.1K]3 years ago
6 0
The answer to your question is D
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If a wave has a frequency of 50 Hz and a wavelength of 3 meters, what is its Velocity? (Use
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A projectile is launched with speed v0 at an angle of θ0 above the horizontal. Find an expression for the maximum height it reac
Step2247 [10]

Answer:

h= \frac{(v_{o})^{2} sin^{2} \theta o }{2g}

Explanation:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

vfy= v₀y -gt Equation (3)

vfy²= v₀y²-2gH Equation (4)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

H= hight that reaches the projectile above its starting point (m)

Data

v₀  : total initial speed

θ₀ : angle of v₀ above the horizontal.

g acceleration due to gravity

Calculation of the componentes x-y of the v₀

v₀x = v₀*cos θ₀

v₀y = v₀*sin θ₀

Calculation of the maximum hight that reaches the projectile above its starting point

When the projectile reaches its maximum height (h), vy = 0:

in the Equation (4)

vfy²= v₀y²-2gh

0= v₀y²-2gh

2gh= v₀y²h= \frac{(v_{o})^{2}sin^{2} \theta o }{2g}

2gh=  (v₀*sin θ₀)²

h= \frac{v_{o}^{2} sin^{2} \theta o }{2g}

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