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3 years ago

A snowboarder travels 150 m down a mountain slope that is 65 degrees above horizontal. What is his vertical displacement?

1 answer:

3 0

This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:

sin 65 = h / 150

Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters

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A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains air

Ganezh [65]

Answer:

0.435atm

Explanation:

cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains air with a volume of 0.185 m3 at a pressure of 0.740 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.315 m3. If the temperature remains constant, what is the final value of the pressure?

Given

Initial pressure P1= 0.740atm

Initial volume V1= 0.185 m3

Final pressure P2= ?

Final volume V2= 0.315 m3

At constant temperature, the pressure of a syste is inversely proportional to volume, by Boyles law then

P1V1=P2V2

P2=P1V1/V2

=(0.185*0.740)/0.315

0.1369/0.315

= 0.435atm

Therefore, final pressure is 0.435atm

7 0

3 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

nignag [31]

The position of the particle is given by:

x(t) = t³ - 12t² + 21t - 9

Differentiate x(t) with respect to t to find the velocity x'(t):

x'(t) = 3t² - 24t + 21

Differentiate x'(t) with respect to t to find the acceleration x''(t):

x''(t) = 6t - 24

5 0

3 years ago

A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon

Anettt [7]

First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed $v_x=30 m/s$ , and an accelerated motion on the y-axis, with initial speed $v_y=20 m/s$ and acceleration $g=9.81 m/s^2$ :
$S_x(t)=v_xt$
$S_y(t)=v_y t- \frac{1}{2} gt^2$
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
$S_y(t)=0$
Therefore:
$v_y t - \frac{1}{2}gt^2=0$
which has two solutions:
$t=0$ is the time of the beginning of the motion,
$t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s$ is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
$S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m$

4 0

3 years ago

Visible light with a wavelength of 480 nm appears

DedPeter [7]

Green would be best
hope it help

7 0

3 years ago

1) Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as

telo118 [61]

Explanation:

The tangential speed of Andrea is given by :

$v=r\omega$

Where

r is radius of the circular path

ω is angular speed

The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'

r' = 2r

New angular speed,

$v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v$

New angular speed is twice that of the Chuck's speed.

8 0

3 years ago