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faltersainse [42]
3 years ago
8

HELP ASAP PLEASE!!!

Physics
1 answer:
lana [24]3 years ago
7 0
A .Earthquakes
Explaination
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My Exams are coming.so, please tell me some ways to score good marks?​
Citrus2011 [14]

Instead of asking this question go and study

5 0
2 years ago
Read 2 more answers
What happens when a colder drier air mass pushes against a warmer moister air mass
andrew-mc [135]
A cold<span> front </span>occurs when a cold air mass advances into a region occupied by a warm air mass<span>. If the boundary between the </span>cold<span> and </span>warm air masses<span> doesn't move, it is called a stationary front. This is due to the (usually) higher humidity in the </span>air<span> of </span>warm<span> fronts compared to that of </span>cold<span> fronts.

Hope this helps</span>
6 0
4 years ago
A spherical balloon has a radius of 7.15 m and is filled with helium. The density of helium is 0.179 kg/m^3, and the density of
tekilochka [14]

The largest mass of cargo the balloon can lift is 791.06 kg

First, we need to calculate the mass of helium.

Since the radius of the spherical balloon is r = 7.15 m, its volume is V = 4πr³/3.

The volume of the balloon also equals the volume of helium present.

Now, the mass of helium m = density of helium, ρ × volume of helium, V

m = ρV

Since ρ = 0.179 kg/m³

m = ρV

m = ρ4πr³/3.

m = 0.179 kg/m³ × 4π(7.15 m)³/3

m = 0.179 kg/m³ × 4π(365.525875 m³)/3

m = 0.179 kg/m³ × 1462.1035π m³/3

m = 261.7165265π/3 kg

m = 822.207/3 kg

m = 274.07 kg

Since the mass of the skin and structure of the balloon is 910 kg, the total mass, M of the balloon = mass of skin and structure + mass of helium gas is 910 kg + 274.07 kg = 1184.07 kg.

The weight of this mass W = Mg where g = acceleration due to gravity.

The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.

W' = m'g

Now, the mass of air m' = density of air, ρ' × volume of air displaced, V'

We know that the volume of air displaced, V' = volume of balloon, V

So, V' = V = 4πr³/3.

Since the density of air, ρ' = 1.29 kg/m³,

m' = ρ'V

m = 1.29 kg/m³ × 4π(7.15 m)³/3

m = 1.29 kg/m³ × 4π(365.525875 m³)/3

m = 1.29 kg/m³ × 1462.1035π m³/3

m = 1886.113515π/3 kg

m = 5925.4/3 kg

m = 1975.13 kg

So, the net weight W" that the balloon can lift is W" = W' - W = m'g - Mg = (m' - M )g = (1975.13 kg - 1184.07 kg)g = 791.06g.

So, the net mass m" = W"/g = 791.06g/g = 791.06 kg

This net mass is the largest mass of cargo that the balloon can lift.

Thus, the largest mass of cargo the balloon can lift is 791.06 kg

Learn more about balloons here:

brainly.com/question/21890581

8 0
3 years ago
A cube has sides of length 10 cm and weighs 10 N.
Lorico [155]

Answer:

P = r g h   where r here is the density m / V and then P is the pressre.

W (weight) = m g

m = 10 N / 9.8 m / s^2 = 1.02 kg

r (density) = m / V = 1.02 kg / 1000 cm^3 = .00102 kg/cm^3

P = .00102 kg / cm3 * 9.8 m/s^2 * 10 cm

P = .1 kg m / (s^2 cm^2)

Since 1  N = kg m / s^2    then P = .1 N / cm^2

More simply P = F / A = 10 N / 100 cm^2 = .1 N / cm^2

8 0
3 years ago
Uses of hydrometer in points​
Arisa [49]

Answer:

A hydrometer is an instrument used to determine specific gravity. It operates based on the Archimedes principle that a solid body displaces its own weight within a liquid in which it floats. Hydrometers can be divided into two general classes: liquids heavier than water and liquids lighter than water

7 0
3 years ago
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