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Sloan [31]
4 years ago
14

Name two things that are important to remember when handling electric equipment

Physics
2 answers:
miv72 [106K]4 years ago
6 0

Answer:

Electric equipment should only be used as instructed by a manual or teacher. It should not be handled with wet hands, and should be turned off or unplugged once the investigation is complete.

Explanation:

..

vfiekz [6]4 years ago
5 0

always wear electric resistant material / clothing

always be cautious when handling wires and rods

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A family is skating at an ice rink. The 58.2 kg mother is holding the
MariettaO [177]

Answer:

When I got this question I had to draw it out so if you have to do that, draw 3 stick figures holding hands, one representing the mother, father, and daughter. Then you write their weights on top of them and then draw an arrow pointing from the father to the mother.

Explanation:

use this formula :

a_{y} = \frac{Fdadshandy}{msys}

then you fill it in :

a_{y} = \frac{100N}{35.5kg+58.2kg}

a_{y} = \frac{100N}{93.7kg}

a_{y} = 1.0672 m/s^{2}

then you multiply that with the daughters weight :

T_{2} x= m_{2} a_{y}

T_{2} x = 35.5kg (1.0672 m/s^{2})

T_{2} x = 37.89N

and that's the answer :) : 37.89N

5 0
3 years ago
MEASUREMENT
lubasha [3.4K]

Answer:

You would need to type the numbers in the question or you could have added a picture but you didn't so there is no way to answer this question. Have a nice day!

5 0
3 years ago
What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?
adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 10973731.6 m⁻¹

n_f = Higher energy level = 3

n_i= Lower energy level = 2

Putting the values, in above equation, we get:

\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )

\lambda=6.56\times 10^{-7}m

Now we have to calculate the energy.

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 6.56\times 10^{-7}m

Putting the values, in this formula, we get:

E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}

E=3.03\times 10^{-19}J

Therefore, the energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

3 0
3 years ago
Can anyone explain how tides work
Mila [183]
The position of the sun and the moon affect how high the tide is 
6 0
4 years ago
Read 2 more answers
A pendulum oscillates 12 times in 4 seconds. what is the length of the pendulum?
seraphim [82]

Answer:

L = 2.8 cm

Explanation:

Period T = 4 / 12 = 1/3 s

T = 2π√(L/g)

L = (T/2π)²g

L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm

6 0
3 years ago
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