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Marina86 [1]
2 years ago
12

Energy is the ability to do work. True or false

Physics
1 answer:
Blababa [14]2 years ago
4 0
Work can be defined as the energy transferred from a body to its sorroundings, the energy spent to move a body, or the energy you need to alter a charged particle, so no energy, no work; thus, the statement is true.
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Diego is trying to lift a piano to the second floor of his house. Diego uses a pulley system and gives a big lift to the piano.T
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Answer:he to

Explanation:

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2 years ago
During free fusion, the eyes _______ in order to view a stereogram without a stereoscope. A. use the pictorial depth cue B. are
Andru [333]

Answer:

D

Explanation:

During free fusion, the eyes converge or diverge in order to view a stereogram without a stereoscope.

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A scientist is conducting research to determine how much pressure a dam can withstand. Why are good observations important to hi
Natalija [7]

The things that a scientist should consider while observing the force is the environmental conditions,the force that is expected to act on the dam, the means to contain that force,  and compare different types of designs in accordance with the location of the dam

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2 years ago
You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r
Brut [27]

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

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v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

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Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

8 0
3 years ago
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit
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As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

\Delta KE = \Delta PE

\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1

Where,

m = mass

v_{f,i} = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2

\frac{1}{2} (v_f^2-v_0^2) = gh_2

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v_f = \sqrt{2gh_2+v_0^2}

v_f = \sqrt{2(9.8)(23.5)+13.6^2}

v_f = 25.4m/s

7 0
3 years ago
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