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levacccp [35]
3 years ago
15

A grasshopper leaps into the air at a 62° angle above the horizontal, and follows a parabolic arc in free fall after it leaves t

he ground. Mark all of the following that are true at the top of that parabolic arc?
a. The grasshopper is in equilibrium at the top of the arc.
b. The grasshopper's total velocity vector is zero.
c. The grasshopper's acceleration vector is zero.
d. None of the above.
Physics
1 answer:
harina [27]3 years ago
5 0

Answer:

d. None of the above.

Explanation:

In a parabolic motion, you have that in the complete trajectory the component velocity is constant and the vertical component changes in time. Then, the total velocity vector is not zero.

In the complete trajectory the gravitational acceleration is always present. Then, the grasshopper's acceleration vector is not zero.

At the top of the arc the grasshopper is not at equilibrium because the gravitational force is constantly acting on the grasshopper.

Then, the correct answer is:

d. None of the above.

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Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an
8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 m/s^2 where as radial the component of acceleration is 8.77 m/s^2

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

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r=e^u

So the transverse component of acceleration are given as

a_{\theta}=(ru''+2r'u')\\

Here

r=e^u\\r=e^{\pi/4}\\r=2.1932 m

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So

a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

The transverse component of acceleration is 26.32 m/s^2

The radial component is given as

a_r=r''-r\theta'^2

Here

r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m

So

a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2

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6 0
3 years ago
A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved​ along, a marble was fir
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Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.  

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v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now,  the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

speed of the train i.e. 2.716 m/s.

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A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg initially at rest. The trains collide. After the collisi
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Explanation:

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To maintain your body temperature your body converts chemical potential energy into thermal energy true or false
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