The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×10^3 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10^-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×10^5 m around the planet Glob. Calculate the speed of the satellite.

Answer:

Approaches mathematical learning through inquiry

-Explore real contexts, problems, situations, and models

-Learning through doing shifts the focus on the students

-Problems have multiple entry and exit points

-Links to other disciplines

Explanation:

quizlet

Answer:

12 grams of the isotope carbon-12.

Explanation:

hope it helps you and give me a brainliest

Therefore, the thrust due to ejected fuel is 2310 N, the thrust due to accelerated air passing through the engine is 98340 N, and the power delivered is 3260 hp.

Acceleration is the rate of change of velocity of an object with respect to time. It is a vector quantity, meaning that it has both magnitude and direction. Acceleration occurs when an object changes its speed, its direction, or both. A positive acceleration means that the speed of an object is increasing, while a negative acceleration (also called deceleration) means that the speed of an object is decreasing. The standard unit of acceleration is meters per second squared.

Here,

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a system is conserved when there are no external forces acting on it. In this case, we can assume that the airplane and the burned gases form a closed system, so the total momentum of the system is conserved.

Part A: To find the thrust due to ejected fuel, we can use the equation:

Thrust = (mass flow rate of fuel) x (exit velocity of burned gases)

Thrust = (4.2 kg/s) x (550 m/s) = 2310 N

Part B: To find the thrust due to accelerated air passing through the engine, we can use the equation:

Thrust = (mass flow rate of air) x (exit velocity of air) + (mass flow rate of fuel) x (exit velocity of burned gases)

The mass flow rate of air is 120 kg/s, and the exit velocity of air is the sum of the speed of the airplane and the speed of the air relative to the airplane. We can use the formula for the velocity addition to find the exit velocity of air:

exit velocity of air = speed of airplane + speed of air relative to airplane

exit velocity of air = 270 m/s + 550 m/s = 820 m/s

Now we can substitute the values into the equation:

Thrust = (120 kg/s) x (820 m/s) + (4.2 kg/s) x (550 m/s)

Thrust = 98340 N

Part C: To find the power delivered by the engine, we can use the equation:

Power = Thrust x Velocity

We can use the speed of the airplane as the velocity, since this is the speed at which the engine is delivering thrust to the airplane. The speed of the airplane is 270 m/s, which is equivalent to 603 mi/h. To convert the thrust from Newtons to pounds-force (lbf), we can divide by the conversion factor 4.448 N/lbf. Then we can use the following formula to convert the power from watts to horsepower:

1 hp = 746 W

Substituting the values into the equation, we get:

Power = (98340 N / 4.448 N/lbf) x (603 mi/h) / (3600 s/h) x (1 hp / 746 W)

Power = 3260 hp

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Answer:

Explanation:

To verify the hypothesis that the Earth has a liquid outer core and a solid inner core, we can use seismographs to study seismic waves that pass through the Earth's interior. The experiment can be set up as follows:

1. Select multiple seismograph stations around the globe to record seismic waves.

2. Choose a location for an earthquake to occur. The earthquake should be large enough to generate seismic waves that travel through the Earth's interior and be located far away from the selected seismograph stations.

3. Record the seismic waves generated by the earthquake at the various seismograph stations.

4. Analyze the seismic waves to determine how they interact with the Earth's interior. Specifically, we will study how the seismic waves pass through the Earth's outer and inner core.

5. Repeat the experiment using earthquakes of different magnitudes and at different locations, and record the resulting seismic waves.

Equipment needed for the experiment include seismographs, computers for data analysis, and earthquake monitoring systems. Seismographs can be installed in various locations around the globe to record the seismic waves generated by the earthquake. Data from these seismographs can be collected and analyzed using computer software to determine how the seismic waves interact with the Earth's interior.

Evidence that proves the outer core is liquid includes the observation of seismic waves that cannot travel through the liquid outer core, resulting in a shadow zone on the opposite side of the Earth from the earthquake. This shadow zone indicates that the seismic waves are refracted or absorbed by the liquid outer core. In contrast, evidence that proves the inner core is not liquid includes the observation of seismic waves that are reflected and refracted by the inner core boundary. This is due to the fact that the inner core is solid and has a different density and composition than the outer core.

To use repeatability to show whether the hypothesis is valid or not, we can repeat the experiment using earthquakes of different magnitudes and at different locations, and record the resulting seismic waves. If the results from multiple experiments are consistent with the hypothesis, then we can have greater confidence that the hypothesis is valid. If the results from multiple experiments are inconsistent, then we would need to investigate further to determine the cause of the inconsistency and revise the hypothesis accordingly.

(a) The work done by the 150 N force is 877.5 Joules.

(b) The coefficient of kinetic friction between the crate and the surface is approximately 0.437.

To answer this problem, we must take into account the work done by the applied force as well as the work done by friction.

(a) The applied force's work may be estimated using the following formula:

Work = Force * Distance * cos(theta)

where the force is 150 N and the distance is 5.85 m. Since the force is applied horizontally and the displacement is also horizontal, the angle theta between them is 0 degrees, and the cosine of 0 degrees is 1.

As a result, the applied force's work is:

Work = 150 N * 5.85 m * cos(0) = 877.5 J

So, the work done by the 150 N force is 877.5 Joules.

(b) Frictional work is equal to the force of friction multiplied by the distance. The work done by friction is identical in amount but opposite in direction to the work done by the applied force since the crate travels at a constant speed.

The frictional work may be estimated using the following formula:

Work = Force of Friction * Distance * cos(theta)

The net force applied on the crate is zero since it is travelling at a constant pace. As a result, the friction force must be equal to the applied force, which is 150 N.

Thus, the work done by friction is:

Work = 150 N * 5.85 m * cos(180) = -877.5 J

Since the work done by friction is negative, it indicates that the direction of the frictional force is opposite to the direction of motion.

The coefficient of kinetic friction may be calculated using the following equation:

Friction Force = Kinetic Friction Coefficient * Normal Force

The normal force equals the crate's weight, which may be computed as:

Normal Force = mass * gravity

where the mass is 35.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2.

Normal Force = 35.0 kg * 9.8 m/s^2 = 343 N

Now, we can rearrange the equation for the force of friction to solve for the coefficient of kinetic friction:

Force of Friction = coefficient of kinetic friction * Normal Force

150 N = coefficient of kinetic friction * 343 N

coefficient of kinetic friction = 150 N / 343 N ≈ 0.437

As a result, the kinetic friction coefficient between the container and the surface is roughly 0.437.

In summary, the work done by the 150 N force is 877.5 Joules, and the coefficient of kinetic friction between the crate and the surface is approximately 0.437.

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The distance that the car travel during each time is:

During this time, the car will travel a total distance of 250 meters.

This can be calculated by using the equation for displacement: displacement = initial velocity x time + 0.5 x acceleration x time^2.

Therefore, the displacement of the car after 10 seconds is:

D= (0 x 10 + 0.5 x 5 x 10^2) = 250 meters.

t= 12 sec

D= (0 x 12 + 0.5 x 5 x 12^2) = 300 meters.

t= 15 sec

D= (0 x 15 + 0.5 x 5 x 15^2) = 375 meters.

t= 18 sec

D= (0 x 18 + 0.5 x 5 x 18^2) = 450 meters.

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A) The approximate magnitude of the electric field between points A and B is 1.3 V/cm; B) The ranking from strongest to weakest electric field is B>A>C>D>E>F>P; C) The arrows indicating the electric field at points A, B, C, and D should be drawn perpendicular to the equipotential lines in the direction of decreasing potential.

Equipotential lines are imaginary lines that connect points in a space that have the same electric potential or voltage. They indicate areas of uniform electric potential in an electric field.

(A) To find the approximate magnitude of the electric field between points A and B, we need to use the formula:

E = ΔV / Δd

where ΔV is the potential difference between points A and B, and Δd is the distance between them. We can estimate ΔV by counting the number of equipotential lines between points A and B, which appears to be around 4. The distance Δd is approximately 3.0 cm. Therefore, the approximate magnitude of the electric field between points A and B is:

E = 4 / 3.0 ≈ 1.3 V/cm

(B) To rank the strength of the electric field at points ABCDEFP, we need to look at the spacing between the equipotential lines. The closer the lines are, the stronger the electric field. Based on the diagram, the ranking from strongest to weakest electric field is:

B > A > C > D > E > F > P

This ranking is determined by observing the spacing between the equipotential lines. The electric field is strongest at point B because the equipotential lines are closest together there, indicating a steep potential gradient.

(C) To draw arrows indicating the electric field at points A, B, C, and D, we need to draw the arrows perpendicular to the equipotential lines, in the direction of decreasing potential. The arrows should be longer where the equipotential lines are closer together. Based on the diagram, the arrows should be drawn as follows:

At point A, the arrow should point to the left, because the potential is decreasing in that direction.

At point B, the arrow should point to the left and be longer than the arrow at point A, because the potential gradient is steeper.

At point C, the arrow should point upward, because the potential is decreasing in that direction.

At point D, the arrow should point downward, because the potential is increasing in that direction.

Hence, The electric field between points A and B has an approximate magnitude of 1.3 V/cm; the order of the electric fields from strongest to weakest is B>A>C>D>E>F>P; and the arrows indicating the electric field at each of the four points should be drawn perpendicular to the equipotential lines in the direction of decreasing potential.

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The voltage of the power source flowing through the given wire is determined as 2.67 x 10⁻⁵ V.

The voltage of the power source is determined from ohm's law as shown below;

V = IR

where;

But, I = Q/t

V = (Q/t)R

\(V = \frac{QR}{t} \\\\V = \frac{(1\times 10^{15} \times 1.602\times 10^{-19} \ C) \times \ 10 \ ohms}{1 \times 60\ s} \\\\V = 2.67 \times 10^{-5} \ V\)

Thus, the voltage of the power source flowing through the given wire is determined as 2.67 x 10⁻⁵ V.

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Answer:

1. I pick the role of Mateo.

2. The main problems from Mateo's perspective are: He knows that the collision was a complete accident and that Deshawn did not injure Andrew on purpose. Because Andrew told everyone that Deshawn injured him on purpose, Deshawn did something unethical by sharing private information about Andrew on social media.

3. The ethical issues are: Andrew took his hurt and frustration of being injured and missing play time out on Deshawn, so he slandered him by saying that Deshawn hurt him on purpose. Deshawn then retaliated by sharing personal information about Andrew on social media, which is also unethical.

4. These issues are relevant to Mateo because he is Andrew's best friend AND he saw the incident happen and knows it was an accident.

5. If I was Mateo, I would talk to Andrew and explain to him that I know he is mad that he is injured and can't play for several weeks, but telling everyone that Deshawn did it on purpose wasn't right or good sportsmanship. I would encourage Andrew to meet with Deshawn with me and have Andrew apologize to Deshawn for saying it was his fault, and then encourage Andrew to go and tell everyone the real story. I would also encourage Deshawn to apologize for posting private information about Andrew, have Andrew ask him to take it down, and then ask Deshawn to post another post saying why he did it and that he was sorry. I would also encourage them to work together, since they are both star players.

Explanation:

The statement that is NOT true is "the spaces between the particles increased.

option D.

If we consider the solids and liquids, when the temperature increases the molecules gain energy and start moving in all directions. This expands the substance and the volume of the substance increases.

Similar, when the ball is heated, the volume of the ball increases due to thermal expansion.

As the temperature increases, the average kinetic energy of the particles within the ball also increases, causing them to move faster.

However, the spaces between the particles do not necessarily increase. In fact, the expansion of the ball occurs due to the particles themselves moving farther apart, but the intermolecular spacing within the ball remains relatively constant.

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Answer:

\(a=100\ m/s^2\)

Explanation:

Given that,

Initial velocity, u = -2 m/s

Final velocity, v = =.5 m/s

Time, t = 0.025 s

We need to find the rate of acceleration of a hammer. Acceleration is equal to the change in velocity divided by the time taken.

\(a=\dfrac{v-u}{t}\\\\a=\dfrac{0.5-(-2)}{0.025}\\\\a=100\ m/s^2\)

So, the acceleration of the jet is \(100\ m/s^2\).

Answer:

  q₂ = ± 1.306 10⁻⁶ C

   q₂ =  ± 3.073 10⁻⁶ C

Explanation:

We must solve this exercise using Coulomb's law, let's write the equilibrium equation for each case

first case

     F₁ = k q₁ q₂ / r²         (1)

where F1 is the force on sphere 1, with a value of F₁ = 0.110 N; r is the distance between them r = 0.573 m

second case.

The spheres were joined with a conductive wire, so the charge of the two is equal, then the wire is removed and

       -F₂ = k q₃ q₃ / r²           (2)

In this case the force is expelling and is equal to F₂ = 0.0489 N

when the spheres are joined with the wire the charge of the spheres is distributed evenly between them two

         q₁ + q₂ = 2 q₃             (3)

we can see that there are three equations with three unknowns for which the system can be solved

let's substitute equation 3 in 2

         - F₂ = k (q₁ + q₂)² / r²

we join this equation with equation 1

          F₁ = k q₁ q₂ / r²

          -F₂ = k (q₁ + q₂)²/4r²

          q₁ = F₁ r² / k q₂

         - F₂ = k (F₁ r² / k q₂ + q₂)²/4r²

Let's replace the values ​​and work out

         - 0.0489 = 9 109 (0.110 0.573² / (9 10⁹ q₂) + q₂)² / 0.573²

          - 0.0489 4 0.573²/9 10⁹ = (0.110 0.573² / (9 10⁹ q₂) + q₂)²

          - 7.1356 10⁻¹² = (4.013 10⁻¹² / q₂ + q₂)²

          - 7.1356 10⁻¹² = 1 / q₂² (4.013 10⁻¹² + q₂² )²

          - 7.1356 10⁻¹² q₂² = (16.104 10⁻²⁴ + 4.013 10⁻¹² q₂² + q₂⁴

          q₂⁴ + q₂² (4,013 10⁻¹² + 7,1356 10⁻¹²) + 16,104 10⁻²⁴ = 0

we change variables

        q’ = q₂²

          q'² + q'  11.1486 10⁻¹² + 16.104 10⁻²⁴ = 0

we solve this quadratic equation

          q '= [- 11.1486 10⁻¹² ±√ ((11.1486 10⁻¹² )² - 4 16.104 10⁻²⁴)] / 2

          q '= [- 11.1486 10 12 ±√ (59.8753 10⁻²⁴)] / 2

          q '= [- 11.1486 ± 7.7379] / 2 10⁻¹²

          q'1 = -1.70535 10⁻¹²

          q'2 = -9.44325 10⁻¹²

           q' = q₂²

           q₂ = √ q'

           q₂ = ± 1.306 10⁻⁶ C

           q₂ =  ± 3.073 10⁻⁶ C

since q 'is squared of these charges they can be positive or negative giving the same result.

Answer: A Punnett square can be used to predict genotype and phenotypes of offspring from genetic crosses. ... In the P generation, one parent has a dominant yellow phenotype and the genotype YY, and the other parent has the recessive green phenotype and the genotype yy.

Explanation:

The magnitude of the force exerted by the left cube on the right cube is 17.64N.

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction.

Two identical 3.0-kg cubes are placed on a horizontal surface in contact with one another. The cubes are lined up from left to right and a force F₁ is applied to the left side of the left cube causing both cubes to move at a constant speed v. The coefficient of kinetic friction between the cubes and the surface is 0.3.

From the equilibrium of forces in vertical direction

Normal force N= 2m x g

friction force f = μN =μ(2m)g

From the equilibrium of forces in horizontal direction

F₁ =ma =0

using Newton's third law of motion, we get

F₁  - f =0

F₁  =f = μ(2m)g

Put the values, we get

F₁  = 17.64N

Thus, magnitude of the force exerted by the left cube on the right cube is 17.64N.

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Carl Rogers, along with Abraham Maslowsignificantly contributed to the humanistic psychology movement through his development of the person-centered approach and his emphasis on the individual's subjective experience and potential for personal growth.

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Answer:

C

Explanation:

Gravity is the main reason that make our planets to pull each other

Answer:

The woman is walking at a speed of 0.08 m/s relative to the man on the walkway.

Explanation:

Notice that for the observer, the woman is walking at 0.38 m/s, and the man (who stands still in the walkway) has a speed of 0.3 m/s.

those 0.38 m/s reported from a stationary observer are in fact the addition of two velocities: that of the walkway (0.3 m/s), and that of the woman relative to the walkway (our unknown velocity "v"):

0.38 m/s = 0.3 m/s + v

0.38 m/s - 0.3 m/s = v

v = 0.08 m/s

Net forces in all above cases will be

A) F net = 0

B) F net = 1 N

C) F net = 5 N

D) F net = 5N

E) F net  = 5 N

F) F net = 5 N

G) F net = 9.9 N

H) F net = 9.9 N

A) net force = upward force - downward force

                   = 4 - 4 = 0

B) net force = rightward force - leftward force = 4 - 3 = 1 N  

C) net force  

using Pythagoras theorem

F net =  \(\sqrt{4^{2} + 3^{2} }\) = \(\sqrt{25}\) = 5 N

D) F net = \(\sqrt{-4^{2} + (-3)^{2} }\) = 5 N

E) balancing horizontal forces

   rightward force - leftward force  = 5 - 2 = 3 N

   using  Pythagoras theorem

   F net =  \(\sqrt{4^{2} + 3^{2} }\) = 5 N

F) balancing vertical forces

   -5 + 2 = -3 N

    using  Pythagoras theorem

     \(\sqrt{(-4)^{2} +(-3) ^{2} }\) = 5 N

G) using  Pythagoras theorem

    F net = \(\sqrt{7^{2} + 7^{2} }\) = 9.9 N

H ) balancing horizontal forces

     10 - 3 = 7 N

     \(\sqrt{7^{2} + 7^{2} }\) = 9.9 N

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Answer:

6%

Explanation:

6% it id heat rate 6% fornhight

Answer:

the velocity of the kid is 5.6 m/s

Explanation:

r is the radius and w is the frequency.

so we should know that the diameter is 18m and the diameter is equal to two times the radius, so r = 18m/2 = 9m

we should also know that the circumference of a circle is equal to c = 2pi*r, so each revolution has this length. if the kid does 5.9 revolutions in one minute then the kid spins at v = 5.9*2pi*9m/min

so we want to write this in meters per second and this means that we need to divide it by 60!

v = (5.9*2pi*9/60)m/s = 5.56 m/s

so your answer will be  5.6 m/s glad i could help!

0.25 m/s

Explanation:

The radius r of the merry-go-round is half its diameter D:

\(r = \frac{1}{2}D = \frac{1}{2}(1.5\:\text{m}) = 0.75\:\text{m}\)

We also need to convert the angular speed from rev/min to rad/s. We know that there are 60 seconds to a minute and that there are \(2\pi\) radians per revolution. Therefore,

\(\omega = 3.2\:\dfrac{\text{rev}}{\text{min}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}}×\dfrac{1\:\text{min}}{60\:\text{s}}\)

\(\:\:\:\:=0.335\:\text{rad/s}\)

Now that we know the angular speed in rad/s, the child's linear speed can be calculated as

\(v = r\omega = (0.75\:\text{m})(0.335\:\text{rad/s}) = 0.25\:\text{m/s}\)

Answer:

The value is  \(\rho_s = 4.026 *10^{-6} \ C/m^2\)

Explanation:

From the question we are told that

   The radius of the inner conductor  is  \(r_1 = 1 \ mm = 0.001 \ m\)

    The radius of the outer conductor is  \(r_2 = 3 \ mm = 0.003 \ m\)

    The potential at the outer conductor is  \(V = 1.5 kV = 1.5 *10^{3} \ V\)

Generally the capacitance per length of the capacitor like set up of the two conductors is

      \(C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}\)

Here \(\epsilon_o\) is the permitivity of free space with value  \(\epsilon_o = 8.85*10^{-12} C/(V \cdot m)\)

=>   \(C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}\)

=>   \(C= 50.6 *10^{-12} \ F/m\)

Generally given that the potential  of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line  charge density of the outer  conductor is mathematically represented as

      \(\rho_l = C * V\)

=>   \(\rho_d = 50.6*10^{-12} * 1.5*10^{3}\)

=>   \(\rho_d = 7.59*10^{-8} \ C/m\)

Generally the surface charge density is mathematically represented as

        \(\rho_s = \frac{\rho_l }{2 \pi * r_2 }\)    here \(2 \pi r = (circumference \ of \ outer \ conductor )\)

=>    \(\rho_s = \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }\)

=>    \(\rho_s = 4.026 *10^{-6} \ C/m^2\)

The vector with a magnitude of 5 and in the direction of the vector 4i - 3j + k is approximately (20/√26)i + (-15/√26)j + (5/√26)k.

The given vector can be normalized before being multiplied by the desired magnitude. This is how to locate the vector:

The vector that has been provided should be normalized by dividing each of its components by its magnitude. The Pythagorean theorem can be used to determine the magnitude of the vector 4i - 3j + k:

Magnitude = √(4² + (-3)² + 1²) = √(16 + 9 + 1) = √26

Normalize the vector by dividing each component by the magnitude:

Normalized vector = (4/√26)i + (-3/√26)j + (1/√26)k

Multiply the normalized vector by the desired magnitude:

To obtain a vector with a magnitude of 5, multiply each component of the normalized vector by 5:

Desired vector = 5 * ((4/√26)i + (-3/√26)j + (1/√26)k)

Simplifying the expression gives:

Desired vector ≈ (20/√26)i + (-15/√26)j + (5/√26)k

So, the vector with a magnitude of 5 and in the direction of the vector 4i - 3j + k is approximately (20/√26)i + (-15/√26)j + (5/√26)k.

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This question deals with the concept of displacement, vector, magnitude, and direction.

The displacement of the mail carrier is "one block to the east".

Displacement is a vector quantity because it contains both the magnitude and the direction of the physical quantity. In this case, as well both the direction and magnitude shall be considered to find out the displacement of the mail carrier.

Displacement = 6 blocks north + 6 blocks south + 1 block east

Since north and south are the opposite directions. Therefore, they will cancel each other.

Displacement = 1 block east

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The attached picture shows the concept of displacement.

Answer:

B. lnelastic collision, because interest kinetic energy is not conserved .

I hope it helps ❤❤

The type of collision experienced by the two objects is inelastic collision, because internal kinetic energy is not conserved.

There are two types of collisions that two objects moving directly or opposite each may experience:

In elastic collision, both momentum and kinetic energy of the to objects are conserved.

In inelastic collision, only momentum is conserved while kinetic energy decreases after the collision.

Thus, we can conclude that the type of collision experienced by the two objects is inelastic collision, because internal kinetic energy is not conserved.

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If the cloud is transparent, milky, thin layers, rain within the next 2 hours, then the cloud you are seeing is most likely a type of altocumulus cloud.

Altocumulus clouds are generally characterized by their white or gray color, and can sometimes appear milky or translucent. They often form in layers, and can be thin or thick depending on the conditions.

Altocumulus clouds are typically found at medium altitudes, between 6,500 and 20,000 feet and are often associated with unsettled weather conditions.

While they don't necessarily indicate that rain is imminent, altocumulus clouds can be a sign that a change in the weather is on the way.

Thus, if it is likely to rain in the next 2 hours, then the cloud must be altocumulus clouds.

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A free body diagram is a visual representation of all the forces acting on an object.

A free body diagram is a visual representation of all the forces acting on an object. It is a tool commonly used in physics to help analyze the motion of an object.

In the case of an air cart, the forces acting on the cart might include the force of gravity pulling the cart downward, the normal force of the surface supporting the cart pushing upward, and the force of air resistance acting against the cart as it moves through the air. There may also be other forces at play, depending on the specific situation, such as an applied force from a fan or a motor driving the cart forward.

If there is an extra force causing a discrepancy between your numbers, you would also include this force on the free body diagram, with an arrow representing the magnitude and direction of the force

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In compressing the spring by 0.5 m, Cho stores

1/2 (25 N/m) (0.5 m)² = 3.125 J

of potential energy. As the spring relaxes, some of this potential energy is converted into kinetic energy K in the box and the rest is lost to heat H due to friction:

K - H = 3.125 J

By the work-energy theorem, the total work W performed on the box during this process is equal to the change in the box's kinetic energy. It starts at rest when the spring is compressed, so

W = K - 0 = 1/2 (1.5 kg) v²

By Newton's second law,

• the net vertical force on the box is

∑ F (ver) = n - mg = 0

where n is the magnitude of the normal force from contact with the table, and mg is the weight of the box. It follows that

n = mg = (1.5 kg) (9.8 m/s²) = 14.7 N

• the net horizontal force is

∑ F (hor) = r - f = ma

where r is the mag. of the restoring force, f = 0.2 n is the mag. of kinetic friction, and a is the resulting acceleration of the box. Friction exerts an opposing force of

f = 0.2 (14.7 N) = 2.94 N

The work done by the spring as it relaxes is 3.125 J. The work done by friction is (2.94 N) (0.5 m) = 1.47 J. Therefore the total work done on the box is

W = 3.125 J - 1.47 J = 1.655 J

Find the speed of the box v when it returns to the equilibrium position:

1/2 (1.5 kg) v² = 1.655 J

v² = (3.31 J) / (1.5 kg)

v ≈ 1.49 m/s

I assume the rest of the table is not scarred. (If it is, the box would come to a stop well before 2.30 seconds have passed, or before the box can slide for 2.30 m.) Then the box leaves the table with speed v at an angle of 75°, so that the horizontal distance x that it covers in the air after t seconds is

x = v cos(75°) t

and its tolerable height y is

y = 2.5 m + v sin(75°) t - g/2 t²

(if y turns out negative for some t, that means it has fallen more than 2.5 m and cannot land safely)

The pile is 3.5 m away from the table, so with its launch speed the box travels this distance in time t such that

v cos(75°) t = 3.5 m

t = (3.5 m) / ((1.49 m/s) cos(75°)) ≈ 9.1 s

After such time, the box would have height

2.5 m + v sin(75°) (9.1 s) - g/2 (9.1 s)² ≈ -390.51 m

i.e. it would be nearly 400 m underground! So no, the box will not land safely.

Hi there!

The formula for velocity given acceleration:

v = at

Plug in given values:

v = 6.4(7) = 44.8 m/s

Explanation:

F = kx

mg = kx

(20 kg) (10 m/s²) = (1800 N/m) x

x = 0.11 m

Answer:

Energy

Explanation:

A wave is a disturbance that transfers energy from one place to another without transferring matter. Waves transfer energy away from the source, or starting place, of the energy.