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ASHA 777 [7]
3 years ago
10

Where is the deepest cave in the world? How far down is it located?

Physics
1 answer:
Sergeeva-Olga [200]3 years ago
6 0

Answer:

Geology Notes

1 Veryovkina Cave 2212[1] 13.5 km (8.4 mi)[1] Abkhazia / Georgia 43°23′52″N 40°21′37″E.

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Write the energy conversion when climbing a ladder​
Volgvan
A child climbing a ladder is transforming kinetic energy into potential energy.
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2 years ago
150c passes through cell in 30 seconds. cell has a potential difference of 12v. what is current in the circuit
zzz [600]

The current in the circuit is 5 A

Explanation:

The intensity of current is given by the equation:

I=\frac{q}{t}

where

I is the current

q is the amount of charge passing through a given point of the circuit in a time interval of t

For the cell in this problem, we have

q = 150 C is the charge

t = 30 s is the time interval

Substituting into the equation, we f ind

I=\frac{150}{30}=5 A

Learn more about current:

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brainly.com/question/10597501

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#LearnwithBrainly

6 0
3 years ago
What would the person’s mass be on the earth? Part B pls
Dafna1 [17]

Answer:

I hopes it helps

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4 0
2 years ago
These nerve cells are needed in muscle and tendons to detect degree of stress and stretching as a sensory activity. A. Ganglia.
SOVA2 [1]

Answer:

D.Proprioceptors

Explanation:

"Proprioceptors is Sensory receptors found in muscle an tendons that detect their degree of stretch"

4 0
3 years ago
The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re
lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

\Delta x = \Delta y

Where:

\Delta x - Range of the bullet, measured in meters.

\Delta y - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

K_{1} = U_{2}

Where:

K_{1} - Kinetic energy at point 1, measured in joules.

U_{1} - Gravitational potential energy at point 2, measured in joules, and:

U_{2} = m\cdot g \cdot \Delta y

Where:

m - Mass of the bullet, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

K_{1} = m\cdot g \cdot \Delta y

\Delta y = \frac{K_{1}}{m\cdot g}

If K_{1} = 1066\,J, m = 0.25\,kg and g = 9.81\,\frac{m}{s^{2}}, the maximum height is now computed:

\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta y = 434.791\,m

\Delta y = 0.435\,km

Lastly, the range of the bullet is 0.435 kilometers.

3 0
3 years ago
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