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Ghella [55]
3 years ago
12

How might the construction of a new recreational path between the river and the railroad track affect the stability of the railr

oad?
Physics
1 answer:
Alex_Xolod [135]3 years ago
3 0
Had to look for the missing details and here is my answer.

This question is based on Hazard City: Landslide Hazard Assessment. And based on the given illustration, how the construction of a new recreational path between the <span>railroad track and the river would influence the railroad stability is that it would make the railroad more vulnerable to landslide due to increasing slope. Hope this helps.</span>
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A baseball is hit that just goes over a wall that is 45.4m high. If the baseball is traveling at 46.2 m/s at an angle of 32.7° b
mario62 [17]

Answer:

54.9 m/s at 44.9 degrees

Explanation:

If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.

Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s

Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s

SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.

Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use

Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2

We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m

If we take the first derivative of the equation of position, we get the equation for speed

V(t) = Vy0 + a * t

We know that being t2 the moment the ball goes over the wall

V(t2) = -25 m/s

Y(t2) = 45.4 m

So:

45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2

-25 = Vy0 + a * t2

Then:

Vy0 = -25 - a * t2

So:

45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2

0 = -44.4 - 25 * t2 - 1/2 * a * t2^2

a = -9.81 m/s^2

0 = -44.4 - 25 * t2 + 4.9 * t2^2

Solving this quadratic equation we get:

t1 = -1.39 s

t2 = 6.5 s

Since we are looking for a positive value we disregard t1.

Now we can obtain Vy0:

Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s

Since horizontal speed is constant Vx0 = 38.9 m/s

By Pythagoras theorem we obtain the value of the initial speed:

V0 = \sqrt{Vx0^2 + Vy0^2} = \sqrt{38.9^2 + 38.76^2} = 54.9 m/s

The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90

a = atan(Vy0/Vx0) = 44.9 degrees

5 0
3 years ago
Samanthawalks along a horizontal path in the direction shown the curved path is a semi circle with a radius of 2 m while the hor
Anna11 [10]

Answer:

Explanation:

Samantha walks along a horizontal path in the direction shown the curved path is a semi circle with a radius of 2 m while the horizontal part is for me what is the magnitude of displacement

Displacement is given by the straight line distance between P and Q. Displacement will be length of straight line joining P and Q

a semi circle with a radius of 2 m

Length of this straight line=4+diameter

=4+(2*2)

=8 m

7 0
3 years ago
The diagrams show objects’ gravitational pull toward each other. Which statement describes the relationship between diagram X an
creativ13 [48]

' C ' is the only correct statement on the list.  We don't know anything about diagram-x or diagram-y because we can't see them.

8 0
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The two main states of mechanical energy are ___________ and potential energy.
shutvik [7]
<span>Potential energy and Kinetic energy</span>
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3 years ago
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What speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0.30?
lorasvet [3.4K]

The momentum of the x-ray photon is p = h/lambda . Lambda is the wavelength (0.30nm=3x10^(-9)m) and h is Planck's constant,(h=6.62607004 × 10-34<span> m2 kg / s).The momentum is: 2.2 x 10^(-25).</span>

The momentum can be calculated also as: p=mv, where m is the mass of the electron and v is the speed.

So v=p/m,p is known,and also the mass of the electron (m=9.10938356 × 10-31<span> kilograms).</span>

v=2.2 x 10^(-25)/9.10938356 × 10-31<span> kilograms=0.24 x 10^6 m/s</span>

8 0
3 years ago
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