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Mazyrski [523]
3 years ago
12

The weight of an object is the product of its mass m, and acceleration of gravity , g . If an objects mass is m=10 . Kg what is

its weight?
Physics
1 answer:
zzz [600]3 years ago
8 0

Its weight is 10g .  The 'g' is the acceleration of gravity
wherever the object happens to be at the time.

On Mars, g=3.71 m/s², the object weighs 37.1 newtons.

On the Moon, g=1.62 m/s², the object weights16.2 newtons.

On Earth, g=9.81 m/s², the object weighs 98.1 newtons.
.
.
etc.

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Energy cannot be created nor destroyed. A. Ella said "a lot of energy in the hot water has disappeared." Explain why her stateme
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Answer:

See Explanation

Explanation:

The principle of conservation of energy states that; energy can neither be created nor destroyed but is converted from one form to another.

In view of this principle, Ella can not be correct when she says that a lot of energy has disappeared. The use of the term "disappeared" connotes the idea that the energy no longer exists which does not happen.

Hence, energy can not "disappear" from hot water rather the energy in the water may be transferred to the surroundings.

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Which statement best describes the benefits that would make the investment worthwhile?
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<span>Fusion produces large amounts of energy, and the fuel is found on Earth.</span>
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Answer:

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Explanation:

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How do i solve the ones in red? A student fires a cannonball horizontally with a speed of 24.0m/s from a height of 51.0m. Neglec
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Horizontal speed = 24.0 m/s

height of the cliff = 51.0 m

For the initial vertical speed will are considering the vertical component. Therefore,

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let's find how long the ball remained in the air.

\begin{gathered} 0=51-\frac{1}{2}(9.8)t^2 \\ 4.9t^2=51 \\ t^2=\frac{51}{4.9} \\ t^2=10.4081632653 \\ t=\sqrt[]{10.4081632653} \\ t=3.22 \\ t=3.22\text{ s} \end{gathered}

Finally, let's find the how far from the base of the building the ball landed(horizontal distance)

\begin{gathered} S_x=u_xt+\frac{1}{2}a_xt^2 \\ S_x=24\times3.23 \\ S_x=\text{horizontal distance=77.28 m} \\ \text{note} \\ a_x=0m/s^2 \end{gathered}

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1 year ago
To what potential should you charge a 0.900 μf capacitor to store 1.50 j of energy
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The energy stored in a capacitor is given by
U= \frac{1}{2} CV^2
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In our problem, C=0.900 \mu F=0.9 \cdot 10^{-6} F, while the energy is U=1.5 j. We can then solve the formula to find V:
V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2\cdot 1.5 J}{0.9\cdot 10^{-6} F} }=1826 V
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