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Tanya [424]
3 years ago
7

a ball is dropped and falls with an acceleration of 9.8m/s^2 downward. it hits the ground with a velocity of 49m/s downward. how

long did it take the ball to fall to the ground

Physics
1 answer:
yaroslaw [1]3 years ago
8 0
The answer below...........

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12000 inches to yards
Nutka1998 [239]

ANSWER

\begin{equation*} 333.33\text{ yds} \end{equation*}

EXPLANATION

We want to convert 12000 inches to yards.

To do this, divide the value in inches by 36:

\begin{gathered} 1\text{ in }=\frac{1}{36}\text{ yd} \\  \\ 12000\text{ in }=\frac{12000}{36}\text{ yds }=333.33\text{ yds} \end{gathered}

That is the answer.

4 0
11 months ago
Remy wonders if the height of the mountain has anything to do with the eventual size of the tsunami wave. How should Remy test t
Anna71 [15]

Answer:

<em>The answer is B</em>

Explanation:

<em>I got this from study island</em>

5 0
3 years ago
A long point-object, mass = 1.0 kg, moves in a circular path at a radial distance = 0.5 m from the axis of rotation. What is the
Vinvika [58]

Answer:0.25\ kg-m^2

Explanation:

Given

mass of Point object m=1 kg

Distance r=0.5 m

Since mass is moving in circular path therefore every time mass is at distance of r from center .

Also Moment of Inertia tells about  the distribution of mass over the given region with respect to center of mass.

Therefore I=mr^2

I=1\times 0.5^2

I=0.25\ kg-m^2

5 0
3 years ago
Which most describes this table?
kati45 [8]

Answer:

In the table there we always eat.

It just based on my own...

3 0
2 years ago
child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s
Strike441 [17]

Answer:

θ = 13.7º

Explanation:

  • According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
  • The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
  • As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

       \Delta K + \Delta U = W_{nc} (1)

  • ΔK, is the change in kinetic energy, as follows:

       \Delta K = \frac{1}{2}* m* (v_{f} ^{2}  - v_{0} ^{2}) (2)

  • ΔU, is the change in the gravitational potential energy.
  • If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

        \Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)

  • Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

        W_{nc} = F_{f} * d * cos 180\º \\\\  = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)

  • Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

      \frac{1}{2}* (v_{f} ^{2}  - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d

  • Replacing by the givens and the knowns, we can solve for sin θ, as follows:              \frac{1}{2}*( (4.30 m/s) ^{2}  - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2}  = 0.236⇒ θ = sin⁻¹ (0.236) = 13.7º
8 0
2 years ago
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