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alukav5142 [94]
3 years ago
9

A 5kg object is moving downward at a speed of 12m/s. If it is currently 2.6m above the ground, what is its potential energy? Use

g=10m/s^2
Physics
1 answer:
motikmotik3 years ago
4 0
Don’t trust the people that give you these file things to click on it’s a scam!
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Les propriétés de l’air?
crimeas [40]

Answer:

Lorsque l'on détend l'air son volume augmente et sa pression diminue. L'air qui est un mélange de gaz est compressible et expansible. – Lorsque l'on comprime l'air, son volume diminue et sa pression augmente. – Lorsque l'on détend l'air, son volume augmente et sa pression diminue.

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2 years ago
A wave passes along the surface of the water in a ripple tank. Describe the motion of a molecule on the surface as the water pas
laila [671]

Answer:

Explanation:

Water waves are generally a transverse wave which do not cause permanent displacement of molecules of the medium. Transverse waves are waves in which the direction of propagation of the wave is perpendicular to the direction of vibration of the particles of the medium.

As the wave propagates from one point to another on the surface of water transferring energy, a molecule of water on its surface vibrates upwards and downwards. Its motion is perpendicular to the direction of propagation of the wave. After the vibration, it comes back to its initial position.

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3 years ago
A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.
densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

KE = \frac{1}{2} mv^2

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (5*10^{-3})(300)^2

KE =  225J

On the other hand the required Energy to heat up t melting point is

Q_1 = mC_p \Delta T

Q_2 = L_f m

Where,

m = Mass

C_p =Specific Heat

\Delta T =Change at temperature

L_f = Latent heat of fussion

Heat required to heat up to melting point,

Q = Q_1+Q_2

Q = mC_p \Delta T+L_f m

Q = 5*0.128*(327-20) + 5*24.7

Q = 310J

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

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3 years ago
Which of the following is a key assumption of the scientific method
lesantik [10]

Answer:

Though you have not gave the choices, I do believe it is “testing”

Explanation:

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What type of radioactive decay is shown in this equation?
svp [43]
There is no <span>radioactive decay</span>
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