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1 year ago

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In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de

grees? A. 15 degrees. B. 65 degrees. C. 70 degrees. D. 80 degrees. E. 90 degrees.

1 answer:

snow_tiger [21]1 year ago

5 0

As we know that range of the projectile motion is given by

$R = \frac{v^2 sin(2\theta)}{g}$

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

so the two angles must be

$\theta, 90 - \theta$

so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

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19.99 kg m²/s

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2 years ago

At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig

Elena-2011 [213]

Complete Question

The complete Question is shown on the first uploaded image

Answer:

a

The torque acting on the particle is $\tau = 48t \r k$

b

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

The position of the particle is $\= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j$

The mass of the particle is $m = 3.0 kg$

The time is t

The torque acting on the particle is mathematically represented as

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where $\r l$ is change in angular momentum which is mathematically represented as

$\r l = m (\r r \ \ X \ \ \r v)$

Where X mean cross- product

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Substituting for $\r r$

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Now the cross product of $\r r \ and \ \r v$ is mathematically evaluated as

$\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]$

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$\r r \ \ X \ \ \r v = 8t^2 \r k$

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Substituting for m

$\r l = 3 * (8t^2 \r k)$

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Substituting into equation for torque

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The magnitude of the angular momentum can be evaluated mathematically as

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Explanation:

<em><u>The knowable variables are </u></em>

$d_{t}=25m$

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$t_{2}=3.9 s$

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Since the three traffic signs are <u>equally spaced</u>, the <u>distance between each sign is $\frac{25}{3} m$ </u>

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

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