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Anna11 [10]
3 years ago
14

How much power does it take to lift 30.0 N 10.0 m high in 5.00 s?

Physics
2 answers:
Rudik [331]3 years ago
7 0

Power = work/time = (Force times distance)/time

= (30N *10.0m)/5.00s = 300/5 = 60 Watts

Anna007 [38]3 years ago
5 0

Answer: P = 60 W

Explanation: Power is the ratio of energy per unit time. Energy can also be substituted by work.

P = W / t

= 30 N ( 10 m) / 5.00 s

= 60 W

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ioda

Answer:

= 54,000 Joules or 54 kJ

Explanation:

Electrical energy is given by the formula;

E = VIt; where V is the potential difference in volts, I is the current and t is the time in seconds.

Therefore;

Electrical energy = 120 V × 0.50 A × 15 ×60 seconds

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Thus; the electrical energy is 54,000 joules or 54 kJ

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2 years ago
If a rock climber accidentally drops a 52.5-g piton from a height of 325 meters, what would its speed be just before striking th
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Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.  
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6 0
3 years ago
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larisa [96]
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8 0
3 years ago
An example of an atom that has no change is one that has A. 1 proton, 2 electrons, and 3 neutrons B. 3 protons,2 electrons, and
Vikki [24]

Your question asks which is an example of an atom that has no change, or charge.

Your answer would be C). 2 protons, 2 electrons, and 1 neutron

These atoms would be known as a neutral atom.

In order for an atom to be neutral, it must have the same amount of protons and electrons.

We know that Protons are positive

We also know that Electrons are negative

In order for the atom to be neutral, they must have an equal amount of protons and neutrons to "cancel out"

In answer choice C, you would see that there are 2 protons and 2 electrons.

2- 0 = 0, in which allows this atom to be neutral.

Therefore, answer choice C. 2 protons, 2 electrons, and 1 neutron would be your answer.

4 0
3 years ago
Read 2 more answers
A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part
mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t

v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}

At <em>t</em> = 5.0 s, the particle has velocity

v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}

v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}

The speed at this time is the magnitude of the velocity :

\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}

The direction of motion at this time is the angle \theta that the velocity vector makes with the positive <em>x</em>-axis, such that

\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}

4 0
2 years ago
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