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3 years ago

How long Tina, a ballerina, in the air when she leaps straight up with a speed of 1.8 m/s?

1 answer:

3 0

The acceleration of gravity on or near the surface of the Earth is 9.8 m/s².
Anything acted on only by gravity loses 9.8 m/s of upward speed, or gains
9.8 m/s of downward speed, every second.

Leaping straight upward at 1.8 m/s, Tina keeps rising until she runs out of
upward speed. That happens in (1.8/9.8) = 0.1837 second after the leap.

After that, Finkel's First Law of Motion takes over:
"What goes up must come down."

The dropping part of the leap is symmetrical with the first. Please don't
make me go through proving it. Tina hits the floor at the same speed of
1.8 m/s with which she left it, and it takes the same amount of time to drop
from the peak to the floor as it took to rise from the floor to the peak.

So her total time out of contact with the floor is

2 x (0.1837 sec) = 0.367 second (rounded)

You might be interested in

The Pacific Plate is moving 29 mm/year toward the north and 20 mm/year toward the west relative to the North American Plate. Sho

adell [148]

Answer:

7 mm per year

Explanation:

It is given that :

The Pacific plate is moving towards north at = 29 mm per year

The Pacific plate is moving towards west at = 20 mm per year

We have to calculate the total relative motion towards the northwest.

So we have to find the resultant of the two motions.

Since the two movements are perpendicular, therefore the angle between the two motions is 90 degree.

Therefore, finding their resultant,

$R^2=(29)^2+(20)^2$

$R^2=(49)^2$

R = 7

Therefore, total relative motion towards the northwest is 7 mm per year.

4 0

2 years ago

A power lifter performs a dead lift, raising a barbell with a mass of 305 kg to a height of 0.42 m above the ground, giving the

Ann [662]

Answer:

Explanation:

Before it hits the ground:

The initial potential energy = the final potential energy + the kinetic energy

mgH = mgh + 1/2 mv²

gH = gh + 1/2 v²

v = √(2g (H - h))

v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))

v ≈ 2.0 m/s

When it hits the ground:

Initial potential energy = final kinetic energy

mgH = 1/2 mv²

v = √(2gH)

v = √(2 * 9.81 m/s² * 0.42 m)

v ≈ 2.9 m/s

Using a kinematic equation to check our answer:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)

v ≈ 2.9 m/s

3 0

2 years ago

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24

Oksi-84 [34.3K]

Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}$
Now At xp we have:
$\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0$
$\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0$
Which is a second order equation, using the quadratic formula to solve for xp would give us:
$xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
or
$xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.