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3 years ago

Which of these statements best supports the conclusion that a substance is in the plasma phase?

Physics

1 answer:

emmasim [6.3K]3 years ago

3 0

The colorless watery fluid of the blood and lymph that containing

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A person weighs 60 kg. The area under the foot of the person is 150 cm2. Find the pressure exerted on the ground by the person.

9966 [12]

Answer:

40000 N/m²

Explanation:

Applying,

P = F/A................... Equation 1

Where P = Pressure, F = Force, A = Area.

From the question,

The force(F) exerted by the person's foot is thesame as it's weight.

F = W = mg............ Equation 2

Where m = mass of the person, g = acceleration due to gravity.

Substitute equation 2 into equation 1

P = mg/A................ Equation 3

Given: m = 60 kg, g = 10 m/s², A = 150 cm² = (150/10000) m² = 0.015 m²

Substitute these values into equation 3

P = (60×10)/0.015

P = 600/0.015

P = 40000 N/m²

3 0

2 years ago

To determine a waves' frequency, you must know the

givi [52]

The correct answer is<span> number of oscillations in a given period of time

This is measured in what is called the Hertz measurement and the period of time is usually taken to be per second.</span>

6 0

3 years ago

How is force proportional to mass?

sukhopar [10]

Answer:

a = F / m

Explanation:

force same -> mass variable

more mass -> less force

8 0

3 years ago

A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing a

hodyreva [135]

Answer:

5 m/s2

Explanation:

The total acceleration of the circular motion is made of 2 components: centripetal acceleration and linear acceleration of 4 m/s2. They are perpendicular to each other.

The centripetal acceleration is the ratio of instant velocity squared and the radius of the circle

$a_c = \frac{v^2}{r} = \frac{30^2}{300} = \frac{900}{300} = 3 m/s^2$

So the magnitude of the total acceleration is

$a = \sqrt{a_c^2 + a_l^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 m/s^2$

4 0

3 years ago

The international Space Station (ISS) orbits the Earth once every 90 mins at an altitude of 409 km. How high would it have to be

Oksi-84 [34.3K]

It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.

To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>

It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
Mathematically, T²∝a³

<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>

The time period of geosynchronous orbit is 24 hours or 1440 minutes.
As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
a1= (T1/T2)⅔×a2

= (1440/90)⅔×6780

= 43,090 km

Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km

Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.

Learn more about the Kepler's third law here:

brainly.com/question/16705471

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6 0

1 year ago