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MariettaO [177]
2 years ago
15

Two friends of different masses are on the playground. They are playing on the seesaw and are able to balance it even though the

ir masses are different. show answer No Attempt How are they able to balance on the seesaw? They are able to balance their individual energies. All of these. Their densities are the same. They are able to balance forces due to gravity. They are able to balance their individual momenta. They are able to balance torques due to gravity.
Physics
1 answer:
Westkost [7]2 years ago
5 0

Answer:

They are able to balance torques due to gravity.

F_1 L_1 = F_2L_2

Explanation:

When two friends of different masses will balance themselves on see saw then at equilibrium position the see saw will remain horizontal

This condition will be torque equilibrium position where the see saw will not rotate

Here we can say

F_1 L_1 = F_2L_2

here we know that force is due to weight of two friends

and their positions are different with respect to the lever about which see saw is rotating

since both friends are of different weight so they will balance themselves are different positions as per above equation

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Sveta_85 [38]

Answer:

wipe the area that came into contact with the electrodes with an alcohol pad

Explanation:

According to my research on the procedure for iontophoresis, I can say that based on the information provided within the question the OTR should wipe the area that came into contact with the electrodes with an alcohol pad. This is because the alcohol pad kills any bacteria that is lingering on the skin and therefore prevents infections from occurring. Especially since the iontophoresis procedure can increase skin permeability which makes it easier for infections to arise.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

4 0
2 years ago
You hold a piece of wood in one hand and a piece of iron in the other. both pieces have the same volume, and you hold them fully
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You hold a piece of wood in one hand and a piece of iron in the other. Both pieces have the same volume, and you hold them fully under water at the same depth. At the moment you let go of them, which one experiences the greater buoyancy force?<span>
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5 0
2 years ago
"The predictions of Einstein’s Theory of General Relativity were tested on a double pulsar system in January of 2004. His equati
Rasek [7]

Answer:

99.95%

Explanation:

A double pulsar system named PSR J0737-3039A/B  in Puppis constellation was discovered in the year 2003. Pulsars are second most densest object in the universe after black holes and they emit radio waves at regular intervals. This pair presented a great and natural setup to test the Theory of General Relativity presented by Einstein in 1915. In this theory Einstein had presented a set of equations on how the space-time fabric will be curved because of the very dense objects such as Neutron stars. It also predicted how the gravitational waves are created because of stars orbiting each other.

A team of astrophysicists led by Michael Kramer, conducted a study on how these gravitational waves will impact the time in which the radio waves emitted by pulsars will reach Earth. The result of the study proved the theory of General Relativity to be accurate up to 99.95%.

8 0
2 years ago
A square block of steel with volume 10 cm3 and mass of 75 g is cut precisely in half. The density of the two smaller pieces is n
dem82 [27]
A. Density only depends on the substance. It doesn't matter whether you have a little chip of it or a supertanker full of it ... the density doesn't change.
4 0
2 years ago
Read 2 more answers
A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel
dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

3 0
2 years ago
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