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3 years ago

What is the experimental group(s)?

1 answer:

5 0

<h3>I think it B The group(s) that gets the special treatment.</h3><h3 /><h3>I hope this is correct.</h3><h3 /><h3 /><h3 />

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A tennis ball is tossed up off a building with a velocity of 22m/s. It takes 6.4s to reach the ground. How high is the building

Amanda [17]

The height of the tennis ball,relative to the ground is H=h max+h-->h max-the maximum height that the tennis ball reaches relative to the roof of the building; h-the height of the building;h max =v0^2/2g=24,2m(g=10m/s^2).H=gt^2/2=>24,2+h=gt^2/2=>h=gt^2/2-24,2=180,6m

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3 years ago

Why type of energy is present after a solar panel changes forms?

NNADVOKAT [17]

Explanation:

Solar technologies convert sunlight into electrical energy either through photovoltaic (PV) panels or through mirrors that concentrate solar radiation. This energy can be used to generate electricity or be stored in batteries or thermal storage.

8 0

2 years ago

Read 2 more answers

Which evidence supports the big band theory select 3 options

ruslelena [56]

Answer:

Explanation:

There are different theories and evidence about the big bang, in this case, we're going to see three evidence.

The galaxies are moving from us, this means space is expanded, this in consequence Big Bang's explosion.

The cosmic microwave background radiation is related to the early warmth of the universe.

The observed abundance of hydrogen, helium, deuterium, lithium, these are checked from the spectra of the oldest stars.

5 0

2 years ago

What force must the deltoid muscle provide to keep the arm in this position?

ruslelena [56]

Answer:

Deltoid Force, $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, $T_{Deltoid}$

2. The torque of the arm, $T_{arm}$

Assuming the arm is just being stretched and there is no rotation going on,

$T_{Deltoid}$ = 0

$T_{arm}$ = 0

⇒ $T_{Deltoid} = T_{arm}$

$r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}$

Where,

$r_{d}$ is radius of the deltoid

$F_{d}$ is the force of the deltiod

$\alpha_{d}$ is the angle of the deltiod

$r_{a}$ is the radius of the arm

$F_{a}$ is the force of the arm , $F_{a} = mg$ which is the mass of the arm and acceleration due to gravity

$\alpha_{a}$ is the angle of the arm

The force of the deltoid muscle is,

$F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}$

but $F_{a} = mg$ ,

∴ $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

7 0

2 years ago

Having just enough weight to achieve all three states of buoyancy with only minor adjustments in the water is the definition of:

eduard

Answer:

Proper weighting

Explanation:

Proper weighing involves the condition of a scuba diver that is fully geared having a near empty tank and the BCD emptied with a held breadth is expected to float at eye level

The fundamental of adequate or good buoyancy of a scuba diver is to ensure proper weighting when diving, With proper weighting, there is more control for the diver when a safety stop is required. There is less need to carry excess weight that increases drag and gas consumption.

6 0

2 years ago