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Oksi-84 [34.3K]
3 years ago
7

A football is kicked from a tee at 12 m/s at 72° above the horizontal. What is the flight time of the football? __s

Physics
1 answer:
Ierofanga [76]3 years ago
8 0
2.3 seconds    

Ignoring air resistance, the flight time is merely a function of gravity and vertical velocity. The vertical velocity will be the initial velocity multiplied by the sine of the angle above the horizon. So: 

 V = sin(72)*12 m/s 

 V = 0.951056516 * 12 m/s 

 V = 11.4126782 m/s   

 Gravitational acceleration is 9.8 m/s, so divide the vertical velocity by gravitational acceleration to get how long it takes for the ball to reach its apex. 

 11.4126782 m/s / 9.8 m/s^2 = 1.164559 s   

 And the old saying "What goes up, must come down" really applies here. And conveniently, it's also symmetric, in that the time it takes to fall will match the time it takes to reach its apex. So multiply the time by 2.  



1.164559 s * 2 = 2.329117999 s   

 Rounding the result to 2 significant figures gives 2.3 seconds.
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maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
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Light waves corresponds to the visible part of the electromagnetic spectrum, which includes all the different types of electromagnetic waves (which consist of oscillations of electric and magnetic fields that are perpendicular to the direction of propagation of the wave): therefore, they are transverse waves.

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Explanation:

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