Mountain climbers have climbed to the top of mt. everest at an altitude of 8.85 km above sea level. the air pressure at this level is approximately _________ mb, and there is _______% of the atmosphere above the mountain climbers.

The greatest biodiversity on earth is found in the tropical rainforest biome.

Biome is actually another name for ecosystem. Rainforests are basically the wettest ecosystems and very diverse due to some reasons such as very high annual rainfall, high average temperatures, nutrient-poor soil, and high levels of biodiversity. Biomes or ecosystems are characterized by their climate and on that basis we can find which type of animals and plants can be found there. The greatest biodiversity on earth is found in the tropical rainforest biome.

The greatest biodiversity on earth is found in the tropical rainforest biome. Hence, option D is correct.

Biodiversity or biological diversity is the measure of variation at the species, genetic, and ecosystem levels. It comprises all the different kinds of life and supports life on Earth.

Biodiversity is important because it supports the entire life on the Earth including plants, animals, microorganisms, etc. Without biodiversity, a healthy ecosystem is not possible. Biome is called an ecosystem. Rainforest has the wettest ecosystems and has a higher annual rainfall and nutrient-rich soil.

The greatest biodiversity on earth is found in the tropical rainforests biome and hence, the ideal solution is option D.

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i just did it and the answer us A

The work done due to friction is [tex]\boxed{59.58{\text{ N}}}[/tex].

Further explanation:

Here, we have to calculate the total work done due to frictional force.

Frictional force is a non-conservative force and the work done by a non-conservative force is path dependent.

Given:

The mass of the physics book is [tex]1.9{\text{ kg}}[/tex].

Coefficient of friction between the book and floor is [tex]0.2[/tex].

The distance between observer that is (you) and Beth is [tex]8{\text{ m}}[/tex]

The distance between Beth and Carlos is [tex]8{\text{ m}}[/tex].

Formula and concept used:

The work done by the conservative force is equal to the product of force and the displacement.

Now, we will know about displacement,

Displacement: The shortest distance between the initial and final position of the object is known as displacement.

But, in case of the non-conservative forces, the work done is equal to the product of the force and the total distance travel by the object.

So, friction force can be calculated as,

[tex]{F_r} = \mu mg[/tex]

Here, [tex]m[/tex] is the mass of the book, [tex]\mu[/tex] is the coefficient of friction.

So, work done due to friction when observer slides the book to the Beth and Beth slides the book to Carlos.

[tex]\boxed{W = {F_r} \cdot d}[/tex]

Here, [tex]{F_r}[/tex] is the friction force, [tex]d[/tex] is the total distance travel against friction force.

[tex]\boxed{\begin{aligned}d&=8+8\\&=16\text{ m}\end{aligned}}[/tex]

Substitute the value of [tex]{F_r}[/tex] in above equation.

[tex]W = \mu mgd[/tex]                                                                       …… (1)

Calculation:

Substitute [tex]0.2[/tex] for [tex]\mu[/tex], [tex]1.9{\text{ kg}}[/tex] for [tex]m[/tex] and [tex]16{\text{ m}}[/tex] for [tex]d[/tex] in equation (1).

[tex]\begin{aligned}W&=\left( {0.2} \right)\left( {1.9} \right)\left( {9.8} \right)\left({16}\right)\\&=59.58{\text{ N}}\\\end{gathered}[/tex]

The work done due to friction is [tex]\boxed{59.58{\text{ N}}}[/tex].

Learn more:

1. Acceleration against friction: https://brainly.com/question/7031524

2. Water is a compound because: https://brainly.com/question/4636675

3. Conservation of momentum during collision https://brainly.com/question/9484203

Answer detail:

Grade: Senior School

Subject: Physics

Chapter: Work and Energy

Keywords:

Book slide, Beth, Carlos, work done, friction, four corners, square, 8m length, 59.58 N, connecting these people, displacement.

The distance of every object in a round trip is always positive

Distance is a scalar quantity, the total distance of an object in any trip is the total path covered by the object from the starting point to the finish point.

On the other hand, the displacement of the object will be zero. Displacement is the change in the position of an object.

Thus, we can conclude that the distance of every object in a round trip is always positive while the displacement is zero.

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Final answer:

The number 6.3 is expressed in scientific notation as 6.3×10^0.

Explanation:

To express the number 6.3 in scientific notation, we need to follow a standard format where the number is between 1 and 10 followed by an exponent of 10.

For 6.3, this is already the case, so it can be expressed as 6.3×10^0 because any number raised to the power of 0 is equal to 1, so this does not change the value of 6.3.

Therefore, when writing in scientific notation, we are just acknowledging that we could move the decimal zero places and maintain the same value.

The mass (kg) of benzene is about 4470 kg

This problem is about Density.

Density is the ratio of mass to the volume of the object.

[tex]\large {\boxed {\rho = \frac{ m }{ V } } }[/tex]

ρ = density of object ( kg / m³ )

m = mass of object ( kg )

V = volume of object ( m³ )

Given:

Diameter of Cylinder = d = 2.00 m

Radius of Cylinder = R = d/2 = 2.00/2 = 1.00 m

Length of Cylinder = L = 4.00 m

Liquid Depth = H = 0.85 m

Density of Benzene = ρ = 0.879 g/cm³ = 879 kg/m³

Unknown:

mass of benzene  = m = ?

Solution:

This problem is about Liquid Volume in Partially Filled Horizontal Tanks

Firstly we will calculate the volume of Benzene by using following formula:

[tex]V = A \times L[/tex]

[tex]V = ( \texttt{Area of Sector - Area of Triangle} ) \times L[/tex]

[tex]V = [ R^2 \cos^{-1}(\frac{R - H}{R}) - (R - H)\sqrt {(2RH - H^2)} ] L[/tex]

[tex]V = [ 1^2 \cos^{-1}(\frac{1 - 0.85}{1}) - (1 - 0.85)\sqrt {(2(1)(0.85) - 0.85^2)} ] 4[/tex]

[tex]V = [ \cos^{-1} (0.15) - 0.15 \sqrt{ 0.9775} ] 4[/tex]

[tex]V \approx \boxed {5.0877 ~ m^3}[/tex]

[tex]m = \rho \times V[/tex]

[tex]m = 879 \times 5.0877[/tex]

[tex]m \approx \boxed {4470 ~ kg}[/tex]

Grade: High School

Subject: Mathematics

Chapter: Density

Keywords: Temperature , Density , Iron , Sphere , Volume , Mass

The current carried by the wire and the power rating of the heater are 20 A and 2400 W respectively.

Given data:

The potential difference across the electric heater is, V' = 120 V.

The total resistance of the nichrome wire is, [tex]R= 6.00 \;\rm \Omega[/tex].

First we need to apply the Ohm's law to find the current through the wire. The expression for the Ohm's law is given as,

[tex]V'= I \times R\\\\I =\dfrac{V'}{R}\\\\I =\dfrac{120}{6}\\\\I=20 \;\rm A[/tex]

Now, the expression for the electric power through the heater is given as,

[tex]P= V \times I\\\\P= 120 \times 20\\\\P =2400 \;\rm W[/tex]

Thus, we can conclude that the current carried by the wire and the power rating of the heater are 20 A and 2400 W respectively.

Learn more about the Ohm's law here:

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It is the second one or the one with the letter F

or an object with mass  moving at a velocity , the angular momentum  with respect to a reference point is defined using the cross product as:

where  is the position vector of the object that describes the object’s position with respect to the reference point. The units for measuring angular momentum is kg m2 s-1. Since angular momentum is defined in terms of a cross product, the direction of the angular momentum vector is taken to be in a direction perpendicular to both the particle’s position vector  and its velocity vector .

Imma answer so other dude can have brainliest.  

<3 :p

We can use the temperature coefficient of resistance to determine the resistance of the nichrome wire at 0.0 °C. The temperature coefficient of resistance (α) is the amount by which the resistance of a material changes per degree Celsius of temperature change.

Given information:

Resistance of the nichrome wire at 11.5°C = 200.00 Ω

Temperature at which resistance is to be found = 0.0°C

We can use the following formula to find the resistance of the nichrome wire at 0.0°C:

R₂ = R₁ [1 + α (T₂ - T₁)]

Where,

R₁ = Resistance of the wire at temperature T₁

R₂ = Resistance of the wire at temperature T₂

α = Temperature coefficient of resistance

T₁ = Temperature at which R₁ is given

T₂ = Temperature at which R₂ is to be found

Since we are given the resistance of the nichrome wire at 11.5°C, we can take this as R₁ and T₁ as 11.5°C. We also know that the temperature coefficient of resistance for nichrome wire is 0.0004 per °C.

Substituting the given values into the formula, we get:

R₂ = 200.00 Ω [1 + (0.0004/°C) (0.0°C - 11.5°C)]

R₂ = 200.00 Ω [1 - 0.0046]

R₂ = 200.00 Ω (0.9954)

R₂ = 199.08 Ω

Therefore, the resistance of the nichrome wire at 0.0°C is 199.08 Ω.

Nichrome wire resistance at 0.0°C is 199.08 Ω.

Calculate nichrome wire resistance at 0.0°C using its temperature coefficient and resistance at 11.5°C.

The resistance R of a conductor at a temperature T is given by the formula:

[tex]\[ R_T = R_0 \left(1 + \alpha \Delta T\right) \][/tex]

Rearrange the formula to find [tex]\( R_0 \)[/tex] (resistance at 0.0°C):

[tex]\[ R_0 = \frac{R_{11.5}}{1 + \alpha \Delta T} \][/tex]

[tex]\( \Delta T \)[/tex]= 11.5 °C - 0.0°C = 11.5°C

Substitute the values:

[tex]\[ R_0[/tex]=  200.00Ω/(1 + (0.0004°C* 11.5°C))

[tex]\[ R_0[/tex] = 200.00Ω/(1 + 0.0046)

[tex]\[ R_0[/tex] = 200.00Ω/(1.0046)

[tex]\[ R_0[/tex] = 199.08 Ω

Answer:

It takes 2.04s to get to the maximum height

Explanation:

This is a vertical throw problem so it can be  treated as a uniform accelerated rectilinear motion. For computing time we are going to use the formula:

[tex]v_{f}=v_{o}+g*t[/tex]

where[tex]v_{f}[/tex] is the final velocity, [tex]v_{o}[/tex] is the initial velocity, [tex]t[/tex] is the time and, [tex]g[/tex] is the gravity.

For solving this kind of problems we need at least three values. The values we have are:

Now:

[tex]v_{f}=v_{o}+g*t[/tex]

[tex]v_{f}-v_{o}=g*t[/tex]

[tex]\dfrac{v_{f}-v_{o}}{g}=t[/tex]

[tex]\dfrac{0-20}{-9.8}=t[/tex]

[tex]t=2.04s[/tex]

Answer:

I do not think the FCC should allow in-flight calls, because they would make the flight noisy and could make it difficult for passengers to hear the pilot or flight attendants.

Explanation:

Answer:

(2) 2s

Explanation:

Remember that the time that it takes an object to fall from a certain distance is only determined by the force with which the object is pulled towards the center of the earth which is gravity, so any object with 0 velocity will drop at the same rate to the ground when dismissing resistance from the air, in to calculate this you just have to use the next formula:

[tex]H=\frac{g*t^2}{2}\\ t=\sqrt{\frac{2H}{g} }[/tex]

So we just insert the data that we have into the formula:

[tex]t=\sqrt{\frac{2H}{g} }\\t=\sqrt{\frac{2*19,6}{9,81} }\\t=\sqrt{4}\\ t=2 seconds[/tex]

Final answer:

To determine the wreckage's velocity after a perfectly inelastic collision, calculate the vector sum of the car's and truck's momenta, then divide by the total mass. Use Pythagorean theorem for the magnitude and arctan for the direction.

Explanation:

To find the velocity of the wreckage immediately after a perfectly inelastic collision, we apply the principle of conservation of momentum. Since the collision is inelastic, both vehicles stick together and move with a common velocity after the impact.

To calculate this, we will:

Determine the momentum of each vehicle before the collision.

Sum the momenta vectorially.

Divide the resultant momentum by the total mass of the system to find the velocity of the wreckage.

For the 1040 kg car and 3360 kg truck:

The momentum of the car is given by its mass times its velocity (1040 kg * 1.80 m/s south).

The momentum of the truck is given by its mass times its velocity (3360 kg * 8.25 m/s west).

Now we add these momenta vectorially, using the components in each direction:

Southward momentum (car's): 1040 kg * 1.80 m/s

Westward momentum (truck's): 3360 kg * 8.25 m/s

With the sum of momenta:

Total southward momentum = 1040 kg * 1.80 m/s = 1872 kg·m/s

Total westward momentum = 3360 kg * 8.25 m/s = 27720 kg·m/s

The magnitude of the wreckage's velocity can be found using Pythagoras' theorem:

√(1872^2 + 27720^2)

Finally, the direction of the velocity is given by the arctangent of the ratio between the southward and westward momentum components.

After calculating the values:

Magnitude of velocity: √(1872^2 + 27720^2) kg·m/s {÷} (Total mass 1040 kg + 3360 kg)

Direction of velocity: arctan(1872 kg·m/s {÷} 27720 kg·m/s)

This resultant velocity is the combined speed and direction of the wreckage after the collision.

Answer:

Four conditions need to be met for an object to be seen-an object, an eye, a source of light, and a direct, unblocked path between the object and the eye

hope this helps

The rate of heat loss from a hot-water pipe by natural convection is calculated using the formula Q = h * A * ΔT. After plugging in the given values and conducting the appropriate calculations, the rate of heat loss turns out to be 2512 W.

Let's determine the rate of heat loss from a hot-water pipe by natural convection. The formula to calculate the rate of heat loss though natural convection is: Q = h * A * ΔT, where:

Let's plug the numbers in:

First calculate the surface area, A = π * 0.05 m * 10 m = 1.57 m2. Then, to find Q, we use the formula Q = 25 w/m2·k * 1.57 m2 * 64 K = 2512 W. Therefore, the rate of heat loss from the pipe by natural convection is 2512 W.

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The rate of heat loss from the pipe by natural convection is 2512 W.

To determine the rate of heat loss from a hot-water pipe, we can use the formula :

Q = h ×A×ΔT

Where:

First, we calculate the surface area of the pipe:

The external diameter of the pipe is given as 0.05 m, and the length is 10 m. The surface area of a pipe is :

A = π ×D×L

Substituting the values:

A = π × 0.05 m × 10 m = 1.57 m²

Next, we calculating the temperature difference:

ΔT = T(pipe) - T(air) = 80°C - 16°C = 64°C

Finally, using heat transfer formula:

Q = h×A×ΔT = 25 W/m²·K × 1.57 m² × 64 = 2512 W

The rate of heat loss from the pipe by natural convection is  2512 W.

Answer:

D.  1.4 × 10^2 kilogram meters/second

We know that in an elastic collision the moments of the both the objects is conserved i.e

Intial momentum of object1 + Intial momentum of object2 =Final momentum of object1 + Final momentum of object2

Intial momentum of object1 =12 kg.m/s

Intial momentum of object2 = P (assume that P is the momentum)

Final momentum of object1= 14 kg.m/s

Final momentum of object2 =16 kg.m/s

On substituting all values we get

12 + P =14 kg.m/s + 16 kg.m/s

P = -18 kg.m/s

(-ve sign indicates the second object was moving in the opposite direction to the object1 before collision )

Therefore the initial momentum of the second object was 18 kg.m/s.

Answer:

Interference pattern- alternate bright and dark fringes

Explanation:

When a monochromatic light is incident on a thin barrier having two slits, an interference pattern can be observed.

Dark and bright fringes would appear on the screen placed in front of it. It would not be random pattern. It would be a smooth pattern with distinctive dark fringe and bright fringe alternatively.

When the light from two source would superimpose, a constructive interference would lead to bright fringe and destructive interference would lead to dark fringe.