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andre [41]
3 years ago
14

What are the metric units of length?​

Physics
1 answer:
Anastaziya [24]3 years ago
5 0
Inches, feet, meters
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How much work is done when a force of 15.0 newtons is used to lift a box 3.0 meters into the air?
denis-greek [22]

Added potential energy = (mass) x (gravity) x (height)

or

Added potential energy = (weight) x (added height)

If you need to lift a 15N box 3m straight up, you have to increase its potential energy by (15 N) x (3 m) = 45 Joules .

Where is that added potential energy supposed to come from ?  You could use an electric winch, a steam engine, a gasoline-powered motor, thousands of hamsters running on little treadmills that are are connected to the main pulley somehow, or your own arm muscles.  But howEVER you do it, you have to provide <em>45 Joules</em> of WORK in order to increase the potential energy of the box by just that much.  

8 0
3 years ago
Read 2 more answers
1. Describe how the periodic table differentiates between metals and nonmetals.
IgorC [24]
There would be a symbol on names of the elements that would say which state it would usually be in.


there are more metals in the periodic table of elements (currently) if that's what the second question is asking about
7 0
3 years ago
Which of the following is a category of mechanical wave?
givi [52]

Answer:

a

because the mechanical wave is when it goes over and over again

8 0
3 years ago
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A stationary skateboarder I with a mass of 50 kg pushes a stationary skateboarder II with a mass of 75 kg. After the push the sk
Mice21 [21]

The final velocity of the skateboarder I after the collision is 3 m/s to the left.

<h3>Velocity of skateboarder I</h3>

The velocity of skateboarder I is determined from the principle of conservation of linear momentum.

m1u1 + m2u2 = m1v1 + m2v2

50(0) + 75(0) = 50v1 + 75(2)

0 = 50v1 + 150

v1 = (-150)/50

v1 = - 3

v1 = 3 m/s to the left

Thus, the final velocity of the skateboarder I after the collision is 3 m/s to the left.

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

8 0
2 years ago
A 70-kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.3 m/s. Neglect air resistan
yKpoI14uk [10]

Answer:

h_{B} = 6.083\,m

Explanation:

Let assume that pole vaulter begins running at a height of zero. The pole vaulter is modelled after the Principle of Energy Conservation:

K_{A} = K_{B} + U_{g,B}

\frac{1}{2}\cdot m \cdot v_{A}^{2} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + m\cdot g \cdot h_{B}

The expression is simplified and final height is cleared within the equation:

\frac{1}{2}\cdot (v_{A}^{2} - v_{B}^{2}) = g\cdot h_{B}

h_{B} = \frac{(v_{A}^{2}-v_{B}^{2})}{2\cdot g}

h_{B} = \frac{[(11\,\frac{m}{s} )^{2}-(1.3\,\frac{m}{s} )^{2}]}{2\cdot (9.807\,\frac{m}{s^{2}} )}

h_{B} = 6.083\,m

5 0
3 years ago
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