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4 years ago

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In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vi

gorous 21 m/s, catching the ball at the highest point in his jump. Right after catching the ball, how fast is the receiver moving?

1 answer:

irinina [24]4 years ago

8 0

To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where,

$m_{1,2}$ = Mass of each object

$v_{1,2}$ = Initial velocity of each object

$v_f$ = Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

$m_2v_2 = (m_1+m_2)v_f$

$(0.42)(21) = (90+0.42)v_f$

$v_f = 0.0975m/s$

Therefore the velocity right after catching the ball is 0.0975m/s

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ioda

Answer:

The magnitude of the net force is 5430N

Explanation:

I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

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acting in the direction of the Drag, and the plane-perpendicular component is:

$W_L=-3600N\cdot\cos 27^\circ=-3208N$

with negative sign since it counteracts the Lift.

So the components of the netforce F are:

$F_h=T-D-W_D=(8000-1000-1634)N=5366N\\F_v=L+W_L=(4100-3208)N=829N$

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3 years ago

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3 years ago

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star.

Vsevolod [243]

Answer:

The angular speed of the neutron star is 3130.5 rad/s.

Explanation:

Given that,

Initial radius $r_{1}=7\times10^{5}\ km$

Final radius $r_{2}=18 km$

Density of a neutron $\rho= 10^{14}$

Equal masses of two stars $m_{1}=m_{2}$

Suppose, If the original star rotated once in 35 days, find the angular speed of the neutron star

Time period of original star T = 35 days = 3024000 s

We need to calculate the initial angular speed of original star

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$\omega=\dfrac{2\pi}{T}$

Put the value into the formula

$\omega_{1}=\dfrac{2\pi}{3024000}$

$\omega_{1}=0.00207\times10^{-3}\ rad/s$

Let the initial moment of inertia of the star is

$I_{1}=m_{1}r_{1}^2$

Final moment of inertia of the star is

$I_{2}=m_{2}r_{2}^2$

From the conservation of angular momentum

$I_{1}\omega_{1}=I_{2}\omega_{2}$

$\omega_{2}=\dfrac{I_{1}\omega_{1}}{I_{2}}$

$\omega_{2}=\dfrac{m_{1}r_{1}^2\omega_{1}}{m_{2}r_{2}^2}$

Put the value into the formula

$\omega_{2}=\dfrac{(7.0\times10^{5})^2\times0.00207\times10^{-3}}{18^2}$

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