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Snowcat [4.5K]
3 years ago
5

When we go up From the earth surface.what happens about the atmospheric pressure​

Physics
1 answer:
topjm [15]3 years ago
7 0

Answer:

There are two reasons why air pressure decreases as altitude increases: density and depth of the atmosphere. Most gas molecules in the atmosphere are pulled close to Earth's surface by gravity, so gas particles are denser near the surface.

Explanation:

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A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft
elena-s [515]

Answer:

The water level rises more when the cube is located above the raft before submerging.

Explanation:

These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.

Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.

When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.

Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]

Density is given by:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyancy force can be calculated using the following equation:

F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]

Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.

7 0
4 years ago
On Mars gravity is one-third that on Earth. What would be the mass on Mars of a person who has a mass of 90 kilograms (kg) on Ea
snow_tiger [21]

Answer: The person will still have a mass of 90kg on Mars

Explanation: The Truth is, the mass of a body remains constant from place to place. It is the weight which is equal to {mass of body * acceleration due to gravity{g}} that varies from place to place since it is dependent on {g}.

In this case the person will have a Weight of 90*9.8 = 882N on Earth.

{ "g" on Earth is 9.8m/s²}

And a Weight of 90*3.3 = 297N on Mars.

{ From the question "g" on Mars is {9.8m/s²}/3 which is 3.3m/s²}

From this analysis you notice that the WEIGHT of the person Varies but the MASS remained Constant at 90kg.

4 0
3 years ago
How much work is done on an object that is moved to acquire a displacement of 5 meters when 500 Newtons of force was exerted?​
grandymaker [24]

Answer:

2,500 Joules (J) or Newton Meter (N M)

Explanation:

Work = Force x Distance

The force in this equation is 500 Newtons. The distance (displacement) is 5 meters. Plug it into the equation above.

Work = 5m x 500n

Work = 2,500 Joules or Newton-Meters.

Therefore 2,500 Joules or Newton Meters of work is done on an object.

3 0
3 years ago
How does the radius of a string affect centripetal force.
KiRa [710]

Answer:

because a raduis is half of 25% of a cicrle.

Explanation:

5 0
3 years ago
Particles q1, 92, and q3 are in a straight line.
NNADVOKAT [17]

The net force on q₃ will be 17.51 N. The net force is the algebraic sum of the two forces on the pleading q₃

<h3 /><h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The force,by the charge q₁ on the q₃;

\rm F_{31}} = \frac{Kq_1q_3}{r^2} \\\\ \rm F_{31}}  = \frac{9 \times 10^9 \times -28.1 \times 10^{-6}\times \times -47.9 \times 10^{-6}}{(0.600)^2} \\\\ F_{31}} =33.64 \ N

The force,by the charge q₂ on the q₃;

\rm F_{32}} = \frac{Kq_2q_3}{r^2} \\\\ \rm F_{32}}  = \frac{9 \times 10^9 \times 25.5 \times 10^{-6}\times \times -74.9\times 10^{-6}}{(0.300)^2} \\\\ F_{32}} =-19.09  \ N

The net force is the sum of the two forces;

\rm F_{net}=F_{32}+F_{31}\\\\\ \rm F_{net}=36.6-19.9 \\\\ \rm F_{net}=17.51 \ N

Hence, the net force on q₃ will be 17.51 N.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

#SPJ1

8 0
2 years ago
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