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gladu [14]
3 years ago
5

Which statements are part of the safety protocol for this lab experiment? Check all that apply.

Physics
2 answers:
tekilochka [14]3 years ago
8 0

Answer:

Always wear safety goggles when performing an experiment.

Use caution when dropping objects and/or launching them into the air.

Use only the materials that your teacher provides or approves.

Check the soda bottle and its cap for cracks and chips prior to use.

Report all accidents—no matter how big or small—to your teacher.

Mazyrski [523]3 years ago
5 0

Answer:

1. Always wear a lab coat and safety goggles when performing an experiment

2. Use the right gear, such as tongs and thermal mitts, to handle hot objects

3. Check glassware, such as beakers and flasks, for cracks and chips prior to use.

Explanation:

It is in the safety section of the student guide for edge.

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I have a question, it concerns hydrostatic buoyancy and Archimedes' law.
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How Does Earth's gravitational force field act on objects that aren't touching Earth's surface?
natali 33 [55]

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8 0
2 years ago
how much gravitational potential energy do you give a 70 kg person when you lift him up 3 m in the air?
SCORPION-xisa [38]

Given gravitational potential energy when he's lifted is 2058 J.

Kinetic energy is transferred to the person.

Amount of kinetic energy the person has is -2058 J

velocity of person = 7.67 m/s².

<h3>Explanation:</h3>

Given:

Weight of person = 70 kg

Lifted height = 3 m

1. Gravitational potential energy of a lifted person is equal to the work done.

PE_g=W=m\times g\times h\\Acceleration due to gravity = g = 9.8 \ m/s^2 \\PE_g= m = m\times g\times h= 70\times 9.8 \times 3 = 2058\ kg.m/s^2 = 2058\ J

Gravitational potential energy is equal to 2058 Joules.

2. The Gravitational potential energy is converted into kinetic energy. Kinetic energy is being transferred to the person.

3. Kinetic energy gained = Potential energy lost = -PE_g = -2058\ kg.m/s^2

Kinetic energy gained by the person = (-2058 kg.m/s²)

4. Velocity = ?

Kinetic energy magnitude= \frac{1}{2} m\times v^2 = m\times g \times h

Solving for v, we get

v=\sqrt{2gh} =\sqrt{2\times 9.8 \times 3} = \sqrt{58.8} = 7.67 m/s^2

The person will be going at a speed of 7.67 m/s².

4 0
3 years ago
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