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2 years ago

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A mom pushes her 19.3 kg daughter on the swing. If she gives her an initial velocity of 7.5 m/s at the bottom of the swing and t

he swing sits 0.6 m above the ground at it's lowest point, what height does she reach above the ground?

1 answer:

kiruha [24]2 years ago

3 0

Answer:

3.17333333333? I hope I get it right

Explanation:

..................hello

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A 210 kg meteoroid is heading toward the earth accelerates by 2.4 × 105 m/s2.

lidiya [134]

According to Newton's second law of motion, Force is the product of mass and acceleration of the object.
So, F = m * a

Here, m = 210 Kg
a = 2.4 * 10⁵ m/s²

Substitute their values,
F = 210 * 2.4 * 10⁵ N
F = 504 * 10⁵ N
F = 5.04 * 10⁷ N

In short, Your Answer would be Option B

Hope this helps!

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Which definition best explains coping strategy? the way a person handles an abusive relationship the thoughts and actions used t

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Answer:

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Explanation:

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The electric field strength E₀ is measured at a perpendicular distance R from an infinitely large, thin sheet that contains a un

Troyanec [42]

Answer:

Explanation:

E=(σ/ε0)

As noted by Dirac the field is the same no matter how far you are from the sheet. When your charge covers a conducting plane, as in your case, the field is, D/eo ,(D is charge density). Because the field inside the conductor (no matter how thin) is zero. The only time the field is, D/2eo, is when you have just a sheet of charge, by itself, not on a conducting plane."

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The geocentric theory states that the center of the universe is

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Suppose you lived in a pre-industrial society and needed to lift a heavy (20 kg) block a height of 5 m and had two choices for h

igomit [66]

Let's break the question into two parts:

1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.

1. Ramp Scenario:
In an incline, the only component of cart's weight(mg) that is in the direction of motion is $mgsin \alpha$ . Therefore the effort force in this case must be equal or greater than $mgsin \alpha$ .

Now we need to find $\alpha$ . $\alpha$ is the angle between the incline of the ramp and the ground.

Since the height is 5m and the length of the ramp is 8m, $sin \alpha$ would be 5/8 or 0.625. Now that you have $sin \alpha$ , mutiple it with mg.

=> m*g* $sin \alpha$ = 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is 125N.

2. Lever Scenario:
Just apply "moment action" in this case, which is:
$F_{e} d_{e} = F_{r} d_{r}$

$F_{e}$ = ?

$F_{r}$ = mg = 20 * 10 = 200N
$d_{e}$ = 10m
$d_{r}$ = 1m

Plug-in the values in the above equation:
$F_{e}$ = 200/10= 20N

As 20N << 125N, the best choice is to use lever.

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