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raketka [301]
3 years ago
5

If a coin has a mass of 29.34g and drops from a height of 14.7 meters, what is its loss in gravitational potential energy? Show

work.
Physics
1 answer:
Serga [27]3 years ago
7 0
U= mgh
U=((29.34/1000)kg)(9.81 \frac{m}{s^{2} } )(14.7m)=4.321J

do not forget to change grams to kilograms and if it were in centimeters convert them to meters.
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Dos cargas puntuales están fijas en el eje x: q1 = 6.0µC está en el origen, O, con x1 = 0.0 cm, y q2 = –3.0 µC está situada en e
erik [133]

Answer:

E_total = 1.30 10¹⁰ C / m²

Explanation:

The intensity of the electric field is

     E = k q / r²

on a positive charge proof

The total electric field at the midpoint is

as q₁= 6 10⁻⁶ C the field is outgoing to the right

for charge q₂ = -3 10⁻⁶ C, the field is directed to the right, therefore

E_total = E₁ + E₂

E_total = k q₁ / r₁² + k q₂ / r₂²

r₁ = r₂ = r = 4 10⁻² m

E_total = k/r² (q₁ + q₂)

 we calculate

E_total = 9 10⁹ / (4 10⁻²)²   (6.0 10⁻⁶ +3.0 10⁻⁶)

E_total = 1.30 10¹⁰ C / m²

8 0
3 years ago
Strong evidence for the existence of dark matter comes from observations of
adoni [48]

Answer:

NASA's Chandra X-ray Observatory and other telescopes such as the Hubble telescope.

Explanation:

Dark matter and normal matter have been wrenched apart by the tremendous collision of two large clusters of galaxies.

The above observations have provided the strongest evidence yet that most of the matter in the universe is dark.

5 0
3 years ago
When you push a 1.90-kg book resting on a tabletop, you have to exert a force of 2.10 N to start the book sliding. Once it is sl
zimovet [89]
Static frictional force = ƒs = (Cs) • (Fɴ)
              2.26 = (Cs) • m • g
              2.26 = (Cs) • (1.85) • (9.8)
           Cs = 0.125 
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
              1.49 = (Cκ) • m • g
              1.49 = (Cκ) • (1.85) • (9.8)
           Cκ = 0.0822 
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3 years ago
"a wind shift from the south or southwest to the northwest is commonly associated with the passage of which type of front
cluponka [151]
If i wouldve know i would tell you sorry
4 0
3 years ago
If 36 grams of water is to be heated from 24.0°C to 48°C to make a cup of tea, how much heat must be added? The specific heat of
Vinvika [58]

We will have the following:

\begin{gathered} Q=mc\Delta T\Rightarrow Q=(36)(4.18)(48-24) \\  \\ \Rightarrow Q=3611.52 \end{gathered}

So, the heat to add is 3611.52 Joules.

3 0
1 year ago
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