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crimeas [40]
3 years ago
5

A long jumper leaves the ground at an angle of 25 degrees with respect to the horizontal with a resultant velocity of 30 ft/s. W

hat was the horizontal velocity of the jumper at takeoff? What was the vertical velocity at takeoff? How high did the COM rise above the point of takeoff?
Physics
1 answer:
astra-53 [7]3 years ago
5 0

Answer:

The horizontal component of the velocity is 27.19 ft/s

The vertical component of the velocity at takeoff is 12.68 ft/s

The vertical distance traveled by the jumper is 8.2 ft

Explanation:

Given;

angle of projection or takeoff, θ = 25°

the resultant velocity, v = 30 ft/s

The horizontal component of the velocity is given by;

v_x = vcos \theta\\\\v_x = 30 * cos(25)\\\\v_x = 27.19 \ ft/s

The vertical component of the velocity at takeoff is given by;

v_y = vsin \theta\\\\v_y = 30 * sin(25)\\\\v_y = 12.68 \ ft/s

The vertical distance traveled by the jumper is given by;

v_y_f^2 = v_y_o^2 +2(-g)h\\\\v_y_f^2 = v_y_o^2 -2gh\\\\0 = 12.68^2 - (2*9.8)h\\\\0 = 160.78 - 19.6h\\\\19.6h = 160.78\\\\h = \frac{160.78}{19.6}\\\\ h = 8.2 \ ft

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Answer:

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Explanation:

The question above is highly related to the topic about "Impulse" in Physics.

"Impulse"<em> refers to an object's change in momentum (the amount of motion in an object) when a force acts upon it for an interval time.</em> When it comes to providing safety to people when it comes to vehicular crashes, impulse plays a vital role.

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Another example are the arrestor beds for trucks. Arrestor beds have been designed in order for trucks to stop, since it's hard to maneuver them. <u>With the help of arrestor beds, trucks are able to come to a stop with a longer time interval, but decreased force.</u>

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3 years ago
An object that is falling has the following type(s) of energy. Ignore air resistance.
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Potential and kinetic
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A pen rolls off a 0.55–meter high table with an initial horizontal velocity of 1.2 meters/second. At what horizontal distance fr
NARA [144]
To find the horizontal distance multiple the horizontal velocity by the time. Since there is no given time it must be calculated using kinematic equation.

Y=Yo+Voyt+1/2at^2
0=.55+0+1/2(-9.8)t^2
-.55=-4.9t^2
sqrt(.55/4.9)=t
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8 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

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Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

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3 years ago
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Answer:

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t = Time taken

u = Initial velocity

v = Final velocity

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v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-6.95^2}{2\times -9.81}\\\Rightarrow s=2.4619\ m

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v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 16.7619+0^2}\\\Rightarrow v=18.1347\ m/s

The speed at which the stone reaches the ground is 18.1347 m/s

8 0
3 years ago
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