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2 years ago

A flea jumps straight up to a maximum height of 0.540 m . what is its initial velocity v0 as it leaves the ground?

1 answer:

3 0

For an uniformly accelerated motion, we can write
$2ah=v_f^2-v_0^2$
where $a=g=-9.81 m/s^2$ is the acceleration of this motion, which in this problem is the gravitational acceleration, with a negative sign because it points downward, against the direction of the motion; h=0.540 m is the distance covered by the flea, and $v_0$ is the initial velocity.

At the maximum height, the velocity is zero, so $v_f =0$ . Therefore we can solve to find $v_0$ :
$v_0 = \sqrt{2ah}= \sqrt{2(9.81 m/s^2)(0.540 m)} =3.25 m/s$

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3 years ago

Three persons wants to push a wheel cart in the direction marked x in Fig. The two person push with horizontal forces F1 and F2

Svetllana [295]

Answer:

I had to search the Figure on Google to solve this question.

a) The magnitude of the force F₃ is:

$F_{3} = 87.47 N$

And the direction of F₃:

$\alpha = 79.04 ^{\circ}$ (with respect to the y-direction, in the third quadrant)

b) P = 4.22 N

Explanation:

I had to search the Figure on Google to solve this question.

a) We can find the force of the third person as follows:

$\Sigma F_{x} = F_{1x} + F_{2x} + F_{3x} = 0$

$\Sigma F_{y} = F_{1y} - F_{2y} + F_{3y} = 0$

So, in x-direction we have:

$\Sigma F_{x} = 45 N*cos(70) + 75 N*cos(20) + F_{3x} = 0$

$F_{3x} = -85.87 N$

In y-direction we have:

$\Sigma F_{y} = 45 N*sin(70) - 75 N*sin(20) + F_{3y} = 0$

$F_{3y} = -16.63 N$

The magnitude of the force F₃ is:

$F_{3} = \sqrt{F_{3x}^{2} + F_{3y}^{2}} = \sqrt{(-85.87 N)^{2} + (-16.63 N)^{2}} = 87.47 N$

To find the direction of F₃ we need to calculate its angle with respect to the y-direction (in the third quadrant):

$tan(\alpha) = \frac{|F_{3x}|}{|F_{3y}|} = \frac{85.87 N}{16.63 N}$

$\alpha = 79.04 ^{\circ}$

b) If the third person exerts the force found in part (a) the car will stop, so the only way for the cart to accelerate at 200 m/s² is that the third person does not exert the force found in a.

To find the weight of the cart when it accelerates at 200 m/s², we need to consider: F₃ = 0.

First, we need to find the cart's mass. Since the car is moving in the x-direction we have:

$\Sigma F_{x} = F_{1x} + F_{2x} = ma$

$45 N*cos(70) + 75 N*cos(20) = m*200 m/s^{2}$

$m = \frac{45 N*cos(70) + 75 N*cos(20)}{200 m/s^{2}} = 0.43 kg$

Now, the weight of the cart is:

$P = mg = 0.43 kg*9.81 m/s^{2} = 4.22 N$

I hope it helps you!

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2 years ago