Approximating Venus's atmosphere as a layer of gas 50 km thick, with uniform density 21 kg/m3, calculate the total mass of the atmosphere.Express your answer using two significant figures.m venus atmosphere = ____ kg

Answer:

Explanation:

Part A:

Using the formula for electric potential due to a charged spherical shell:

V1 = kq1/R1 and V2 = kq2/R2

where k is the Coulomb constant.

For r < R1, the electric potential due to both shells is:

V = V1 + V2 = kq1/R1 + kq2/R2

For R1 < r < R2, the electric potential due to the outer shell is:

V = V2 = kq2/R2

For r > R2, the electric potential due to both shells is:

V = V1 + V2 = kq1/r + kq2/R2

Substituting the given values, we get:

V = (9.0x10^9 Nm^2/C^2)(5.00x10^-6 C)/0.025 m + (9.0x10^9 Nm^2/C^2)(-6.00x10^-6 C)/0.15 m

V = -4.32x10^5 V

Therefore, the electric potential due to the two shells at a distance of 2.50 cm from their common center is -4.32x10^5 V.

Part B:

For r < R1, the electric potential due to both shells is:

V = V1 + V2 = kq1/R1 + kq2/R2

For R1 < r < R2, the electric potential due to both shells is:

V = V1 + V2 = kq1/r + kq2/R2

For r > R2, the electric potential due to both shells is:

V = V1 + V2 = kq1/r + kq2/r

Substituting the given values, we get:

V = (9.0x10^9 Nm^2/C^2)(5.00x10^-6 C)/0.1 m + (9.0x10^9 Nm^2/C^2)(-6.00x10^-6 C)/0.1 m

V = 4.80x10^5 V

Therefore, the electric potential due to the two shells at a distance of 10.0 cm from their common center is 4.80x10^5 V.

The best way of presenting the information on density graphically would be to use a D, bar graph.

A bar graph is a type of chart that uses rectangular bars to represent data. The bars are typically arranged in columns, with the independent variable (in this case, the type of wood) on the x-axis and the dependent variable (in this case, the density) on the y-axis.

A bar graph is the best choice for this data because it allows for easy comparison of density of different types of wood. We can see at a glance which type of wood is the densest and which type of wood is the least dense.

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Answer:

The greater weight increases the terminal velocity by acting as an extra force against gravity and air resistance.

Answer:

you need 7 points and the other team just needs to stop you from scoring

Explanation:

The big bang is how astronomers explain the way the universe began. It is the idea that the universe began as just a single point, then expanded and stretched to grow as large as it is right now—and it is still stretching!

What's This Big Bang All About?

In 1927, an astronomer named Georges Lemaître had a big idea. He said that a very long time ago, the universe started as just a single point. He said the universe stretched and expanded to get as big as it is now, and that it could keep on stretching.

What an Idea!

The universe is a very big place, and it’s been around for a very long time. Thinking about how it all started is hard to imagine.

Some More Information

Just two years later, an astronomer named Edwin Hubble noticed that other galaxies were moving away from us. And that’s not all. The farthest galaxies were moving faster than the ones close to us.

This meant that the universe was still expanding, just like Lemaître thought. If things were moving apart, it meant that long ago, everything had been close together.

Everything we can see in our universe today—stars, planets, comets, asteroids—they weren't there at the beginning. Where did they come from?

A Tiny, Hot Beginning

When the universe began, it was just hot, tiny particles mixed with light and energy. It was nothing like what we see now. As everything expanded and took up more space, it cooled down.

The tiny particles grouped together. They formed atoms. Then those atoms grouped together. Over lots of time, atoms came together to form stars and galaxies.

The first stars created bigger atoms and groups of atoms. That led to more stars being born. At the same time, galaxies were crashing and grouping together. As new stars were being born and dying, then things like asteroids, comets, planets, and black holes formed!

The formula for calculating attraction is F=GMmr².

Gravitational attraction is the force of attraction that occurs on all bodies

which have a mass as a result of the force of gravitational force.

The Force is directly proportional to the product of their masses  and

inversely proportional to the square of the distance between them.

Gravitational attraction =  F=GMmr².

where F is the Force, G is gravitational constant, M and m are masses and r²

is square of the distance between the masses.

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The net force on q₂ will be  1.07 x 10⁻² N, pointing to the left.

To find the net force on particle q₂, we need to calculate the force due to q₁  and q₃ individually and then add them up vectorially. We can use Coulomb's law to calculate the force between two point charges:

F = k × (q₁ × q₂) / r²

where F is the magnitude of the force, k is Coulomb's constant (k = 8.99 x 10⁹ Nm²/C²), q1 and q2 are the charges of the two particles, and r is the distance between them.

The force due to q₁ on q₂ can be calculated as:

F₁ = k × (q₁ × q₂) / r₁²

where r1 is the distance between q₁ and q₂ (r₁ = 0.180 m).

Similarly, the force due to q₃ on q₂ can be calculated as:

F₂ = k × (q₃ × q₂) / r₃²

where r₃ is the distance between q₂ and q₃ (r₃= 0.230 m).

The direction of each force can be determined by the direction of the electric field due to each charge. Since q₁ and q₃ have opposite signs, their electric fields point in opposite directions. Therefore, the force due to q₁ points to the left and the force due to q₃ points to the right.

To find the net force, we need to add up the forces vectorially. Since the forces due to q₁ and q₃ are in opposite directions, we can subtract the magnitude of the force due to q₃ from the magnitude of the force due to q₁ to get the net force on q₂:

Fnet = F₁ - F₃

Substituting the values we get:

Fnet = k × (q₁ × q₂) / r₁² - k × (q₃ × q₂) / r₃²

Plugging in the values we get:

Fnet = (8.99 x 10⁹ Nm²/C²) × [(9.33 x 10⁻⁶ C) × (4.22 x 10⁻⁶ C) / (0.180 m)² - (-8.42 x 10⁻⁶ C) × (4.22 x 10⁻⁶ C) / (0.230 m)²]

Fnet = 1.07 x 10⁻² N

Therefore, the net force on q₂ is 1.07 x 10⁻² N, pointing to the left.

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Hooke's Law can be given as follows sometimes:

The restoring force of a spring is equal to the spring constant multiplied by the displacement from its normal position:

F = -kx

Where, F = Restoring force of a spring (Newtons, N)

k = Spring constant (N/m)

x = Displacement of the spring (m)

The negative sign relates to the direction of the applied force and by convention, the minus or negative sign is present in F = -kx. The restoring force F is directly proportional to the displacement (x), according to Hooke's law. When the spring is compressed, the displacement (x) is negative. It is zero when the spring is at its original length and positive when the spring is extended.

Practically, Hooke's Law is applicable only within a limited frame of reference, and through experimenting, this statement proves to be true. Because materials cannot be compressed beyond a certain size or expanded beyond a certain size without some permanent deformation or change of their original state.

The law only applies under some conditions such as a limited amount of force or deformation. Factually, many materials will noticeably deviate from Hooke's law even before those elastic limits are reached.

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1. The momentum of the drone before the crash is 7.5 kg•m/s

2. The momentum of drone after the crash is 0 kg•m/s

The correct answer to the question is Option A. 7.5 kg•m/s before and 0 kg•m/s after

Momentum is simply defined as the product of mass and velocity i.e

Momentum = mass × velocity

With the above formula, we can obtain the answers to the questions given above.

1. Determination of the momentum before the crash

Mass = 0.5 Kg

Initial Velocity = 15 m/s

Momentum = mass × velocity

Momentum = 0.5 × 15

Thus, the momentum of the drone before the crash is 7.5 Kg•m/s

2. Determination of the momentum after the crash

Mass = 0.5 Kg

Final Velocity = 0 m/s

Momentum = mass × velocity

Momentum = 0.5 × 0

Momentum = 0 Kg•m/s

Thus, the momentum of the drone after the crash is 0 Kg•m/s

The correct answer to the question is Option A. 7.5 kg•m/s before and 0 kg•m/s after

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Answer:

1. 7.5 kg ⋅ m/s before and 0 kg ⋅ m/s after

2. 7.2 kg ⋅ m/s

3. 40,000 kg⋅m/s

4. It must have decreased.

5. She will move in the same direction at the same speed forever.

6. two balls colliding in deep space

7. magnetism

8. Determine the momentums of the two particles before the collision and add them together. Determine the momentums of the two particles after the collision and add them together. Verify that both sums are the same.

9. 2.0 m/s

10. 1.1 m/s

11. 81 kJ

12. Increase the input force and decrease the output displacement.

13. 4 m

Answer:

2.3*10^8m/s

Explanation:

Using

V = c/n

Where c= speed of light

n = refractive index of water

By substituting

We have

V= 3*10^8m/s/1.33

= 2.3*10^8m/s

Answer:

400J/kg°C

Explanation:

The specific heat capacity will be 400 J/kg °C. Specific heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

The amount of heat required to increase a substance's temperature by one degree Celsius is known as specific heat capacity.

Similarly

Mathematically, specific heat capacity is given by;

\(\rm C = \frac{E}{m \times \theta }\)

The given data in the problem is;

The specific heat of the substance is,C in J/kg °C.

The energy is,E =1000 JIs

The mass is,m= 25 g

\(\rm \theta\) Is the temperature difference =  100 °C.

On substituting the given data;

\(\rm C = \frac{E }{m \times \theta } \\\\ C = \frac{1000}{25\times 10^{-3} \times 100 } \\\\ C = 400 \ J/kg^0C.\)

Hence, the specific heat capacity will be 400 J/kg °C.

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Answer:

Step 1: List the mass and velocity of the object. Step 2: Convert any values into SI units (kg, m, s). Step 3: Multiply the mass and velocity of the object together to get the momentum of the object.

a) The final pressure and temperature for the isothermal compression are \(4.5*10^5 Pa\) and 293 K, respectively, while  b) the final pressure and temperature for the adiabatic compression are\(5.58*10^5 Pa\) and 515 K, respectively.

a. Isothermal compression:

For an isothermal process, the temperature remains constant. Therefore, we can use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.

Since the process is isothermal, we can write:

\(P_1V_1 = P_2V_2\)

where P1 and V1 are the initial pressure and volume, and\(P_2\)and\(V_2\)are the final pressure and volume.

We are given that the volume is compressed to 1/3 of its original volume, so\(V_2 = (1/3)V_1\). Substituting this into the equation above gives:

\(P_2 = (V_1/V_2)P_1 = 3P_1\) = \(4.5*10^5 Pa\)

To find the final temperature, we can use the ideal gas law again:

PV = nRT

Rearranging, we get:

T = PV/(nR)

Substituting the values we know, we get:

T = (\(1.5*10^5\)Pa)(V1)/(nR)

Since the process is isothermal, the temperature remains constant, so the final temperature is the same as the initial temperature:

T2 = T1 = 293 K

b. Adiabatic compression:

For an adiabatic process, there is no heat transfer between the gas and its surroundings. Therefore, we can use the adiabatic equation:

PV^γ = constant

where γ = Cp/Cv is the ratio of specific heats.

Since the process is adiabatic and reversible, we can write:

\(P_1V_1\)^γ = \(P_2V_2\)^γ

We are given that the volume is compressed to 1/3 of its original volume, so V2 = (1/3)V1. Substituting this into the equation above gives:

\(P_2 = P_1(V_1/V_2)\)^γ = \(P_1\)\((3)^{(1.4)\) = \(5.58*10^5 Pa\)

To find the final temperature, we can use the adiabatic equation again:

\(T_2 = T_1(P_2/P_1)\)^((γ-1)/γ) = T1(5.58/1.5)^(0.4) = 515 K

Therefore, the final pressure and temperature for the isothermal compression are \(4.5*10^5 Pa\)and 293 K, respectively, while the final pressure and temperature for the adiabatic compression are \(5.58*10^5\) Pa and 515 K, respectively.

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Answer:

a

Explanation:

it corrodes after repeated exposure to excess moisture and dry air the rust itself can be turned back into iron after melting it and removing the waste

Given:

Magnitude of vector A = 10.0 m

Magnitude of vector B = 20.0 m

Magnitude of vector C = 7.0 m

Let's find the magnitude and direction of the net displacement using the component method of vector addition.

The vector of magnitude r which makes an angle θ with the positive x-axis in vector form is:

Vector A:

From the figure, vector A makes an angle of -90° with the x-axis, thus, we have:

Vector B:

Vector B makes an angle of 45° with a magnitude of 20.0 m. Thus, we have:

Vector C:

Vector C makes an angle of -30° with a magnitude of 7.0 m. Thus, we have:

To find the net displacement, apply the formula:

Net displacement = A + B + C

Hence, we have the net diplacement below:

To find the magnitude of the net displacement, we have:

To find the direction, we have:

Therefore, the magnitude of the net displacement is 20.21 m while the direction is at 1.8 degrees.

ANSWER:

Magnitude = 20.21 m

Direction = 1.8°

The answer is that the time the diver was in the air is 1.13 seconds.

To determine the time the diver was in the air, we can use the kinematic equation:

Δy = viΔt + 1/2at²,

where Δy is the displacement, vi is the initial velocity, a is the acceleration due to gravity (g), and t is the time.The initial velocity, vi, is given as 5.5 m/s, and since the diver jumps upwards, the displacement, Δy, is equal to the height of the board, which is 3.0 m. The acceleration due to gravity, a, is -9.8 m/s² (negative because it acts downwards).Substituting the known values into the equation:3.0

m = (5.5 m/s)t + 1/2(-9.8 m/s²)t²

Simplifying, we get:

4.9t² + 5.5t - 3.0 = 0

We can solve for t using the quadratic formula:

t = (-5.5 ± √(5.5² - 4(4.9)(-3.0))) / (2(4.9))= (-5.5 ± 1.59) / 9.8= -0.47 s or 1.13 s

Since time cannot be negative, the time the diver was in the air is 1.13 seconds.

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Answer:

C number is write i think

The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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Answer:

Explanation:

R + 4.17sin(40) = mg

R = mg - 4.17sin(40)

= 2(9.8) - 4.17sin(40)

= 16.91957567

= 16.92 N

A 2.00 kg block is pulled across a flat, friction less floor with a 4.17 N force directed 40.0° above horizontal.  The normal force acting on the block  16.92 N.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Given in the question,

R + 4.17 sin(40) = mg

R = mg - 4.17 sin(40)

  = 2(9.8) - 4.17 sin(40)

  = 16.91957567

  = 16.92 N

A 2.00 kg block is pulled across a flat, friction less floor with a 4.17 N force directed 40.0° above horizontal.  The normal force acting on the block  16.92 N.

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Answer:

Qc>0; Qh>0

Explanation:

In the context of energy transfers with hot and cold reservoirs, the sign convention is QC > 0; QH < 0.

The conversion of one form of energy into another, or the movement of energy from one place to another is known as energy transfer.

Given that there are two energy reservoirs. Heat and work sign convention is:

As, heat is going out that from the hot reservoir ,it is taken as negative- QH < 0.

As heat is coming inside the cold reservoir that is why it is taken as positive - QC > 0.

Therefore, QC > 0 and QH < 0 is the correct answer.

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In order to find the tension in the wire, let's first decompose it in its vertical and horizontal components:

Now, since the system is stable, the sum of vertical forces is equal to zero, so we have:

So the tension in the wire is equal to 679.94 N.

Answer:

If x-rays get in touch with our body tissues, they create a picture of a metal film. The high energizing rays are unable to penetrate soft tissue, such as skin and organs and travel through the laser. The places where the x-rays pass soft tissues are Black areas of an X-ray.

Answer:

Explanation:

Mode of Operation: In the X-ray tube a Low voltage high p.d is applied between the hot cathode and the anode. Electrons are emitted from the cathode and are accelerated to an extremely high speed. ... They are produced by increasing the p.d. between the cathode and anode.

Answer: It is a balance equation

Explanation:

In the diagram tha shows snapshots of an oscillator at different times, the frequency of the oscillation is 0.555 Hz.

The period of the oscillation is the time taken for the for the object to return to its original position. (ie. Displacement = 0). From the above snapshot,

Period of oscillation = 1.80s.

From here, finding the frequency is simple as, Frequency = 1/Period. Hence,

Frequency = 1/1.80

= 0.555 Hz (3 sf).

The frequency of the oscillation is indeed 0.555 Hz. The frequency represents the number of oscillations or cycles per second. In this case, the object completes approximately 0.555 oscillations per second.

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Given :

Two charges q1 and q2 exert a force of 20 N on each other.

To Find :

If the charge of q1 is doubled, what will the new force be?

1) 40 N 2) 80 N 3) 20 N 4) 10 N​

Solution :

We know, electrostatic force between two charges is given by :

\(F = \dfrac{kq_1q_2}{r^2}\)

Here, F = 20 N.

\(20 = \dfrac{kq_1q_2}{r^2}\)

Now, if \(q_1\) is doubled :

\(F'= \dfrac{k(2q_1)q_2}{r^2}\\\\F'= 2\dfrac{kq_1q_2}{r^2}\\\\F'= 2\times F\\\\ F'= 2\times 20 \ N\\\\F'= 40 \ N\)

Therefore, if the charge of q1 is doubled, new force will be 40 N.

The diameter of the smaller end of the pipe is approximately 0.78 meters.

To determine the diameter of the smaller end of the pipe, we can use the principle of conservation of mass. According to this principle, the mass flow rate of water should remain constant throughout the pipe.

The mass flow rate is given by the equation:
Mass flow rate = density of water * cross-sectional area * velocity

Since the density of the water remains constant, we can write:
Cross-sectional area1 * velocity1 = Cross-sectional area2 * velocity2

Given that the velocity1 is 13 m/s, the diameter1 is 1.2 m, and the velocity2 is 30 m/s, we can solve for the diameter2 using the equation:
(pi * (diameter1/2)^2) * velocity1 = (pi * (diameter2/2)^2) * velocity2

Simplifying the equation:
(1.2/2)^2 * 13 = (diameter2/2)^2 * 30

Calculating the equation:
(0.6)^2 * 13 = (diameter2/2)^2 * 30

0.36 * 13 = (diameter2/2)^2 * 30

4.68 = (diameter2/2)^2 * 30

Dividing both sides by 30:
0.156 = (diameter2/2)^2

Taking the square root of both sides:
0.39 = diameter2/2

Multiplying both sides by 2:
0.78 = diameter2

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Explanation:

Yes, there are differences in the shapes of position-time graphs for uniform acceleration and uniform deceleration in different directions. Let's consider each case separately:\(\hrulefill\)

(1) - Uniform acceleration towards the positive direction:

In this case, the object is moving in the positive direction with a constant acceleration. The displacement-time graph will typically be a curve that starts from an initial position and shows a steady increase in displacement over time. The shape of the graph will depend on the specific acceleration value.

(2) - Uniform acceleration towards the negative direction:

In this case, the object is moving in the negative direction with a constant acceleration. The displacement-time graph will also be a curve, but it will show a steady decrease in displacement over time.

(3) - Uniform deceleration towards the positive direction:

In this case, the object is initially moving in the positive direction but is slowing down with a constant deceleration. The displacement-time graph will be a curve that starts with a positive slope and gradually levels off.

(4) - Uniform deceleration towards the negative direction:

In this case, the object is initially moving in the negative direction but is slowing down with a constant deceleration. The displacement-time graph will be a curve that starts with a negative slope and gradually levels off.