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jolli1 [7]
3 years ago
12

The average density of living matter on earthâs land areas is 0.10 g/cm2 . what mass of living matter in kilograms would occupy

an area of 0.125 ha?
Physics
1 answer:
Margaret [11]3 years ago
4 0
<span>Density is a value for mass, such as kg, divided by a value for volume or area. Density is a physical property of a substance that represents the mass of a certain substance per unit volume or unit area. From the problem statement, we are given the density of the living matter in a particular area and the area. To calculate mass, we simply multiply the area to the density. However, we should remember that the units should be homogeneous or the same and if not, we convert.

Mass = density x area
Mass = 0.10 g / cm^2 (0.125 ha) ( 1x10^8 cm^2 / 1 ha )
Mass = 1250000 g or 1250 kg</span>
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a scooter has a mass of 25 kg. a constant force is exerted on it for 5 s during the time the force is exerted the scooter increa
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Answer:

Answer was deleted first time, the answer is 917 N

Explanation:

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5 0
3 years ago
A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend
Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}.

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ



According to the question, we have -

Entering Velocity (v) = 20 i  m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j  m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $\frac{2\pi r}{4} = $\frac{\pi R}{2}

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $\frac{\pi R}{2v} = \frac{\pi R}{40}

Hence, the time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}

To solve more questions on Force on charged particle, visit the link below-

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6 0
2 years ago
What must happen to an atom of magnesium in order to become a magnesium ion Mg+2?
igomit [66]

Answer:

Answer is: c. It must lose two electrons and become an ion.

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.

Explanation:

4 0
3 years ago
Read 2 more answers
The drag force that resists the motion of a car traveling at 80 km h^- 1 is 300 N.
kobusy [5.1K]

The power require to keep the car traveling is 6,666 W.

The power of the engine at the given efficiency is 3,999.6 W.

<h3>What is Instantaneous power?</h3>

This the product of force and velocity of the given object.

The power require to keep the car traveling is calculated as follows;

P = Fv

P = 300\ N \ \times  \ \frac{80 \ kmh^{-1}}{3.6 \ km h^{-1}/m/s} \\\\&#10;P = 300 \ N \times 22.22 \ m/s\\\\&#10;P = 6,666 \ W

The power of the engine at the given efficiency is calculated as follows;

E = \frac{P_{out}}{P _{in}} \times 100\%\\\\&#10;60\% = \frac{P_{out}}{6,666} \times 100\%\\\\&#10;0.6 = \frac{P_{out}}{6,666} \\\\&#10;P_{out} = 3,999.6 \ W

Learn more about efficiency here: brainly.com/question/15418098

8 0
2 years ago
PLEASE SHOW STEPS!
SOVA2 [1]

Answer:

660 J/kg/°C

Explanation:

Heat lost by metal = heat gained by water

-m₁C₁ΔT₁ = m₂C₂ΔT₂

-(0.45 kg) C₁ (21°C − 80°C) = (0.70 kg) (4200 J/kg/°C) (21°C − 15°C)

C₁ = 660 J/kg/°C

8 0
3 years ago
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