Chemical weathering is greatest under conditions of

Answers

Answer 1
Answer: higher mean annual rainfall and temperatures.

Explanation:

Chemical weathering is the set of destructive processes through which rocky materials go trhough. These processes cause changes in the color, texture, composition, firmness and shape of the material.

It should be noted that this happens when the rocks come into contact with atmospheric agents such as oxygen and carbon dioxide.

Another important aspect is that rocks are able to break up more easily thanks to this type of weathering, since the mineral grains within the rock lose adherence and dissolve better under the action of some physical agents, such as humidity (rainfall included) and temperature.

Therefore:

Chemical weathering is greatest under conditions of higher mean annual rainfall and temperatures.

Answer 2
Final answer:

Chemical weathering is greatest under circumstances of hot temperatures and abundant moisture, such as in tropical rainforests. These conditions accelerate weathering reactions such as hydrolysis, oxidation and dissolution, impacting soil characteristics and promoting biological activity.

Explanation:

Chemical weathering occurs most under conditions of high temperature and abundant moisture. These conditions accelerate the reactions that lead to weathering. For example, processes like hydrolysis, oxidation, and dissolution, which are facilitated by climate conditions of high temperature and moisture, lead to a more extensive break down of rocks. Therefore, regions with these climate conditions, such as tropical rainforests, experience the greatest degree of chemical weathering. This intensifies the development of soil characteristics and also promotes biological activity, a vital element of soil quality.

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Related Questions

Question Part Points Submissions Used If an object with mass m is dropped from rest, one model for its speed v after t seconds, taking air resistance into account, is v = mg c (1 − e−ct/m) where g is the acceleration due to gravity and c is a positive constant describing air resistance. (a) Calculate lim t→∞ v.

Answers

Answer:

[tex]\lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]

Explanation:

the velocity as a function of time is

[tex]v(t)=\frac{mg}{c}(1-e^{\frac{-ct}{m}})[/tex]

[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})[/tex]

[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-\frac{1}{\infty })\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-0)\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]

Final answer:

The limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance considered, is the terminal velocity mg/c. The exponential term in the equation approaches zero as time becomes very large, leading to the object reaching a constant terminal velocity.

Explanation:

To calculate the limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance taken into account, we use the provided equation v = mg/c (1 − e^{-ct/m}), where g is the acceleration due to gravity, which averages 9.80 m/s², and c is a positive constant representing air resistance. The limit represents the object's terminal velocity, which is the constant speed an object reaches when the force of gravity is balanced by the drag force of air resistance.

As t → ∞ (increases towards infinity), the exponential term e^{-ct/m} approaches zero. Hence, the limit of the speed v becomes:

∑ lim t→∞ v = lim t→∞ mg/c (1 − e^{-ct/m}) = mg/c.

The value mg/c is known as the terminal velocity, which is the maximum speed that the object will reach as it continues to fall.

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Julie and Eric row their boat (at a constant speed) 63 miles downstream for 7 hours helped by the current. Rowing at the same rate, the trip back against the current takes 9 hours. Find the rate of the boat in still water.

Answers

Answer:

Boat speed = 8 miles/hr

Explanation:

Let the speed of the boat be U

Let the speed of the current be V.

Therefore, the downstream speed is U+V

and the upstream speed is U-V

Now we know that Speed = distance / time

Therefore, downstream speed, U+V = 63 / 7

                                                    U+V = 9 miles/hr   ------(1)

                  Upstream speed, U-V = 63 / 9

                                                U-V = 7 miles/hr      --------(2)

Therefore subtracting (2) from (1), we get

( U+V) - ( U-V ) = 9-7

2V = 2

V = 1 miles/hr

Therefore the speed of the current is V = 1 mile/hr

Now from (1) we get

U+V = 9

U+1 = 9

U = 8

Therefore, the speed of the boat is U = 8 miles/hr

You have a pick-up truck that weighed 4,000 pounds when it was new. you are modifying it to increase its ground clearance. when you are finished

Answers

Based on the provided information, the correct statement is:

b. The bottom of your front bumper must not be more than 28 inches above the pavement.

The modification to increase ground clearance in the pick-up truck involves specific regulations for the front bumper height.

The statement "The bottom of your front bumper must not be more than 28 inches above the pavement" indicates a legal restriction to maintain a certain elevation for safety and compliance reasons.

This limitation aims to prevent potential hazards, such as underride collisions, and ensures that the modified truck adheres to regulatory standards. By specifying the maximum height of the front bumper, authorities aim to strike a balance between vehicle customization and road safety.

This regulation emphasizes the importance of maintaining a reasonable height for front bumpers to mitigate risks and uphold public safety standards, reflecting the broader considerations in vehicular modifications within the legal framework.

The probable question may be:

You have a pick-up truck that weighed 4,000 pounds when it was new. You are modifying it to increase its ground clearance. When you are finished

a. The bottom of your front bumper can be up to 32 inches above the pavement.

b. The bottom of your front bumper must not be more than 28 inches above the pavement.

c. You will no longer be required to have a rear bumper.

Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice was so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light (3.00×108 m/s). Give your answer in thousands of km.

Answers

Answer: 768000km

Explanation:

Velocity is given by the relation between the distance [tex]d[/tex] and the time it takes to travel that distance [tex]t[/tex]:

[tex]V=\frac{d}{t}[/tex]   (1)

In this problem we are told the time it takes for radio wave to travel from the Earth to the Moon and back is the "echo":

[tex]t=2.56s[/tex]  (2)

In addition, we know radio waves are electromagnetic waves (light), and its velocity is:

[tex]V=3(10)^{8}m/s[/tex]   (3)

Substituting (2) and (3) in (1):

[tex]3(10)^{8}m/s=\frac{d}{2.56s}[/tex]   (4)

And finding [tex]d[/tex]:

[tex]d=(3(10)^{8}m/s)(2.56s)[/tex]   (5)

Finally we can obtain the distance:

[tex]d=768000000m=768000km[/tex]  

Answer:

384,000 km

Explanation:

Given

Velocity of radio wave [tex]v = 3.00 \times 10^{8}m/s[/tex]

Duration of echo T = 2.56 s

Solution

Time taken for the radio wave to travel to moon and to travel back to earth as it was picked up by the astronaut's microphone is 2.56 s

Since any time delays in the electronic equipment can be ignored

time taken for the radio wave to reach moon

[tex]t = \frac{T}{2}\\\\t = \frac{2.56}{2} \\\\t = 1.28 s[/tex]

[tex]v = \frac{d}{t}\\\\d = vt\\\\d = 3 \times 10^8 \times 1.28\\\\d = 3.84 \times 10^8 m\\\\d = 384,000 km[/tex]

According to the ___________________ hypothesis, each emotion comes with its own specific profile of autonomic activity (heart rate, skin conductance, and finger temperature).

Answers

Answer:

autonomic specificity

Explanation:

The hypothesis which states that the different emotions in our body involve different and unique physiological profiles is called autonomic specificity hypothesis.

Viewed from this hypothesis perspective, emotions in the human body can be seen as time-tested solutions to timeless problems and challenges. With emotions, evolution in the human body has provided the human with  at least one generalized response to tackle these problems.

When the temperature goes up 3^\circ on the Cantor scale, it goes up 8^\circ on the Frobenius scale. On both scales, 18^\circ is the same temperature. How many Frobenius degrees are equal to 30^\circ Cantor?

Answers

Answer:

50°

Explanation:

It is given that the both scales are the same at 18˚

Now,

at 30˚ on the Cantor scale, the temperature is 30˚ − 18˚ = 12˚ above

the 18˚ mark.

also, it is given that with every 3˚ increase in the temperature on the Cantor scale is there is an 8˚ increase on the Frobenius scale,

mathematically, we can write it as (by unitary method)

3˚ increase in the temperature on the Cantor =  8˚ increase on the Frobenius scale

or

1˚ increase in the temperature on the Cantor =  (8/3)˚ increase on the Frobenius scale

thus, for x˚ increase in the temperature on the Cantor =  ((8/3)˚ × x) increase on the Frobenius scale

hence, for 12° increase we have

12˚ increase in the temperature on the Cantor =  ((8/3)˚ × 12) increase on the Frobenius scale

or

12˚ increase in the temperature on the Cantor =  32° increase on the Frobenius scale

hence, the final reading on the Frobenius scale will be, 18˚ + 32˚ = 50˚.

Which type of energy is thermal energy a form of?

Answers

Answer:

Kinetic Energy

Explanation:

Heat energy is another name for thermal energy. Kinetic energy is the energy of a moving object. As thermal energy comes from moving particles, it is a form of kinetic energy.


kinetic or thermal but can i get brainliest please

A fireman clings to a vertical ladder and directs the nozzle of a hose horizontally toward a burning building. The rate of water flow is 6.31kg/s, and the nozzle speed is 12.5 m/s. The hose passes vertically between the fireman’s feet, which are 1.30 m below the nozzle. Choose the origin to be inside the hose between the fireman’s feet. What torque must the fireman exert on the hose? (This could also be stated, what is the rate of change of the angular momentum of the water?)

Answers

Answer:

A torque of 102.5375 Nm must be exerted by the fireman

Explanation:

Given:

The rate of water flow = 6.31 kg/s

The speed of nozzle  = 12.5 m/s

Now, from the Newton's second law we have  

The reaction force to water being redirected horizontally (F) = rate of change of water's momentum in the horizontal direction

thus we have,

F = 6.31 kg/s x 12.5m/s

or

F = 78.875 N  

Now,

The torque (T) exerted by water force about the fireman's will be

T = (F x d)

or

T = 78.875 N x 1.30 m

T = 102.5375 Nm

hence,

A torque of 102.5375 Nm must be exerted by the fireman

The torque must the fireman exert on the hose toward a burning building is 102.5375 Nm.

What is Newtons second law of motion?

Newtons second law of motion shows the relation between the force mass and acceleration of a body. It says, that the force applied on the body is equal to the product of mass of the body and the acceleration of it.

It can be given as,

[tex]F=ma[/tex]

Here, (m) is the mass of the body and (a) is the acceleration.

For the flow of water, the second law of motion can be given as,

[tex]F=Q\times v[/tex]

As he rate of water flow is 6.31 kg/s, and the nozzle speed is 12.5 m/s. Thus the force of this can be given as,

[tex]F=6.31\times12.5\\F=78.875\rm N[/tex]

The torque for the fireman exert on the hose is equal to the product of force applied and the distance traveled. Therefore the value of torque is,

[tex]\tau=78.875\times1.30\\\tau=102.5375\rm Nm[/tex]

Thus, the torque must the fireman exert on the hose toward a burning building is 102.5375 Nm.

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Displacement is the slope of a velocity v. time graph.

A. True
B. False

Answers

Answer:

B. False

Explanation:

Acceleration is the slope of a velocity vs. time graph.

Displacement is the area under a velocity vs. time graph.

The force of gravity on an object varies directly with its mass. The constant of variation due to gravity is 32.2 feet per second squared. Which equation represents F, the force on an object due to gravity according to m, the object’s mass?F = 16.1mF = F = 32.2mF =

Answers

Answer:

F = 32.2m

Explanation:

The force of gravity on an object is given by:

[tex]F=mg[/tex]

where

m is the mass of the object

g is the acceleration due to gravity

Here we have:

- An object of mass m

- The acceleration of gravity is expressed as [tex]g=32.2 ft/s^2[/tex]

Therefore, substituting into the formula above, we find that the force of gravity on the object is

[tex]F=m\cdot 32.2 = 32.2m[/tex]

Answer:

F = 32.2m

Explanation:

Which of the following best explains why snow predictions by meteorologists are sometimes incorrect?
A. Weather data are misinterpreted.

B. Weather instruments are extremely precise.

C. Interference from the sun causes data to be collected inaccurately.

D. Local variations in weather are too small for weather instruments to replicate.

Answers

Answer:

C. Interference from the sun causes data to be collected inaccurately.

Explanation:

Snow predictions by meteorologists are sometimes incorrect because from the sun causes data to be collected inaccurately.

A box of negligible mass rests at the left end of a 2.00-m, 25.0-kg plank (Fig. P11.43). The width of the box is 75.0 cm, and sand is to be distributed uniformly throughout it. The center of gravity of the nonuniform plank is 50.0 cm from the right end. What mass of sand should be put into the box so that the plank balances horizontally on a fulcrum placed just below its midpoint?

Answers

Answer:

Required mass of sand is 20 kg

Explanation:

Given:

Mass of the plank = 25 kg

Distance of the Center of gravity of the Plank from the fulcrum = [tex]\frac{2}{2}-0.50 = 0.5m[/tex]

Distance of the Center of gravity of the sand box from the fulcrum = [tex]\frac{2}{2}-\frac{0.75}{2}= 0.625m[/tex]

Balancing the torque due to the plank and the sand box with respect to the fulcrum

Torque = Force × perpendicular distance

thus, we get

(25 × g) × 0.5 = weight of sand × 0.625

where, g is the acceleration due to gravity

or

(25 × g) × 0.5 = (mass of sand × g) × 0.625

or

mass of sand = 20 kg

Hence, the required mass of the sand is 20 kg

Answer:

20 Kg mass of sand should be put into the box so that the plank balances horizontally on a fulcrum placed horizontally on a fulcrum placed just below its midpoint.

Explanation:

Use the second condition of equilibrium.

[tex]$\sum} \tau=0$[/tex]

[tex]MgL-$M g x_{c m}=0$[/tex]

[tex]$M=\frac{m x_{c m}}{L}[/tex]

[tex]=\frac{25(0.50)}{0.625}[/tex]

[tex]=20 \mathrm{~kg}$[/tex]

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A small charged ball lies within the hollow of a metallic spherical shell of radius R. For three situations, the net charges on the ball and shell, respectively, are (1) +4q, 0; (2) –6q, +10q; (3) +16q, –12q. Rank the situations according to their charge on (a) the inner surface of the shell and (b) the outer surface, most positive first

Answers

Answer:

Part a)

2)  >  1)  >  3)

Part b)

1)  =  2)  =  3)

Explanation:

Due to charge induction the magnitude of charge on the inner surface of the outer shell is having same charge as that of the small sphere inside but the sign of charge must be opposite.

So here we can say

1)+ 4q, 0

so inner surface has charge - 4q and outer surface charge is +4q

2) -6q , +10q

so inner surface charge is +6q, outer surface charge is +4q

3) +16q , -12q

so inner surface charge is -16q, outer surface charge is +4q

Part a)

situations in which inner surface charge is arranged in decreasing order is given as

2)  >  1)  >  3)

Part b)

Situations in which outer surface charge is arranged in decreasing order is given as

1)  =  2)  =  3)

Final answer:

Situation (1) has the most positive charge on the inner surface of the shell, while situation (2) has the most positive charge on the outer surface.

Explanation:

To rank the situations according to their charge on the inner and outer surfaces of the metallic spherical shell, we need to consider the net charges on the small charged ball and the shell. Let's analyze each situation:

For situation (1) with a net charge of +4q on the ball and 0 charge on the shell: (a) The inner surface of the shell has a charge of +4q, and (b) the outer surface has a charge of 0.For situation (2) with a net charge of -6q on the ball and +10q on the shell: (a) The inner surface of the shell has a charge of -6q, and (b) the outer surface has a charge of +10q.For situation (3) with a net charge of +16q on the ball and -12q on the shell: (a) The inner surface of the shell has a charge of +16q, and (b) the outer surface has a charge of -12q.

Therefore, ranking the situations according to the charge on the inner surface would be: (1), (3), (2) - from most positive to least positive. For the outer surface, the ranking would be: (2), (1), (3) - from most positive to least positive.

Suppose that on a hot (33.0°C) and sticky (80% humidity) afternoon in the spring, a tornado passes over the high school. If the air pressure in the lab (volume of 180.0 m³) was 1.00 atm before the storm and 0.800 atm after the storm, to what volume would the laboratory try to expand in order to make up for the large pressure difference outside?
1. 211 m3
2. 134 m3
3. 1800 m3
4. 7,150 m3

Answers

Answer:

   V₂  =  225 m^{3}

so no exact option is match with the calculated answer.

Explanation:

According to Boyle's Law, we have

                                     P_1 V_1  =  P_2 V_2    ----------- (1)

Data Given;

                  P_1  =  1.0 atm

                  V_1  =  180 m³

                  P_2  =  0.80 atm

                  V_2  =  ?

Solving equation 1 for V₂,

                  [tex]V_2= \frac{P_1 V_1}{P_1}[/tex]

Putting values,

                   [tex]V_2 = \frac{1.0*180}{0.80}[/tex]

                  [tex]V_2 = 225 m^{3}[/tex]

so no exact option is match with the calculated answer.

                 

Answer:

E

Explanation:

A box at rest on a ramp is in equilibrium, as shown.

What is the force of static friction acting on the box? Round your answer to the nearest whole number.
_______N

What is the normal force acting on the box? Round your answer to the nearest whole number.
_______ N

Answers

Answer:

Ffs = 251 N

Fn = 691 N

Explanation:

Take the y direction to be normal to the ramp and the x direction to be parallel to the ramp.

The angle of the ramp is 20°, so the angle that the weight vector makes with the normal is also 20°.  Therefore:

Fgx = Fg sin 20°

Fgy = Fg cos 20°

Sum of the forces in the x direction (parallel to the ramp):

∑F = ma

Ffs − Fgx = 0

Ffs = Fgx

Ffs = Fg sin 20°

Ffs = 735 sin 20°

Ffs ≈ 251

Sum of the forces in the y direction (normal to the ramp):

∑F = ma

Fn − Fgy = 0

Fn = Fgy

Fn = Fg cos 20°

Fn = 735 cos 20°

Fn ≈ 691

To answer that question we need to apply equations of movement ( from Newton´s laws )

In equilibrium:

∑F  = 0         or    ∑Fx = 0     ;  ∑Fy = 0

Solution is:

a) F(sf) = 251 [N]

b) Fn = 691 [N]

From the attached drawings we can see:  ( Body free diagram)

∑ Fₓ  = F(sf)  - Pₓ  = 0           where P = m×g  = 735 [N] ( the weigth)

and Pₓ = P× cos20°

Then     F(sf)  = Pₓ × sin 20°

F(sf) = m×g×cos20°  =  735× 0.34202 [N]

F(sf) = 251.3847

F(sf) = 251 [N]

rounding to the nearest number

F(sf) = 691 [N]

∑ Fy = 0

∑ Fy = Fn - Py  = 0                     Py = P×cos20°      Py = m×g×cos20°

Py = 735×0.939693 [N]     Py =   [N]

Fn = Py = 690.674 [N]

rounding to the nearest whole number

Fn = 691 [N]

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A sensor on a traffic light is most likely to produce electromagnetic waves at which of these frequencies?

Answers

Answer:

Option-(D): 10¹¹ waves per second.

Explanation:

Electromagnetic waves:

The electromagnetic waves is such form of a energy transfer or wave propagation through any space with or without having any medium(particles).As, the medium or particles inside a space are able to transfer the amount of energy from the origin towards the receiver, which makes it very easy for the wave propagation through a medium.Now, the electromagnetic waves are generated from the sensor on a traffic light when the frequency,f level of the wave generation is about 10¹¹ Hertz(Hz).

Answer:

answer is A

Explanation:

Bob can row 14 mph in still water. The total time to travel downstream and return upstream to the starting point is 4 hours. If the total distance downstream and back is 42 miles, determine the speed of the river (current speed).

Answers

Answer:

7 mph

Explanation:

Let the speed of river is c mph.

The effective downstream speed = 14 + c

The effective upstream speed = 14 - c

Total time = 4 hrs

Total distance = 42 miles

Time for upstream + time for downstream = 4 hrs

21 / (14 + c) + 21 / (14 - c) = 4

21 (14 - c + 14 + c) = 4 (196 - c^2)

21 x 28 = 4 (196 - c^2)

c^2 = 196 -147 = 49

c = 7 mph

The speed of the river is 7 miles/h.

Speed of Bob in still water is 14 miles/h.

Speed of Bob downstream = (14 + x) miles/h

Speed of Bob upstream = (14 - x) miles/h

here x = speed of the river.

The total distance downstream and back is 42 miles. Hence,

Distance upstream = Distance downstream = 21 miles

Given that total time of journey upstream and downstream is 4 hours. Then, using the formula:

[tex]t = \frac{distance}{speed}[/tex], we get:

[tex]\frac{21 \hspace{0.5mm} miles} {(14 + x) \hspace{0.5mm} miles/h} + \frac{21 \hspace{0.5mm} miles}{(14 - x)\hspace{0.5mm}miles/h} =4 \hspace{0.5mm} hours[/tex]

or, [tex]\frac{2 \times 21 \times 14 \hspace{0.5 mm} h}{14^{2} -x^{2} } = 4 \hspace{0.5mm} hours[/tex]

or, 21 × 7 h = 14² - x²

or, x² = 49

or, x = 7 miles/h

easy bio A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 438 N and have a center of gravity that is 1.28 m above the floor. His upper legs weigh 144 N and have a center of gravity that is 0.760 m above the floor. Finally, his lower legs and feet together weigh 87 N and have a center of gravity that is 0.250 m above the floor. Relative to the floor, find the location of the center of gravity for his entire body.

Answers

Answer:

1.034 m above the floor

Explanation:

The location of center of body for a compound body, when the weights are given is calculated as:

[tex]\bar x = \frac{W_1x_1+W_2x_2+W_3x_3+W_4x_4+W_5x_5+.......+Wnx_n}{W_1+W_2W_3W_4W_5+.....+W_n}[/tex]

where,

[tex]\bar x[/tex] is the center of gravity of the entire body

W = weight of the individual body

x = center of gravity of the individual body

Thus on substituting the values we get,

[tex]\bar x = \frac{438\times 1.28+144\times 0.760+87\times 0.250}{438+144+87}[/tex]

or

[tex]\bar x = \frac{691.83}{669}[/tex]

or

[tex]\bar x =1.034m[/tex]

Hence, the center of gravity of the entire body lies 1.034 m above the floor

A rock is thrown into a still pond. The circular ripples move outward from the point of impact of the rock so that the radius of the circle formed by a ripple increases at the rate of 3 feet per minute. Find the rate at which the area is changing at the instant the radius is 4 feet. When the radius is 4 ​feet, the area is changing at approximately nothing square feet per minute. ​(Type an integer or a decimal. Round to the nearest thousandth as​ needed.)

Answers

Answer:

Rate of change of area is [tex]75.398ft^{2}/sec[/tex]

Explanation:

[tex]Area=\pi r^{2}\\\\\frac{d(Area)}{dt}=\frac{\pi r^{2}}{dt}=2\pi r\frac{dr}{dt}[/tex]

Applying values we get [tex]\frac{d(Area)}{dt}=2\pi 4\times 3=75.398ft^{2}/sec[/tex]

A 55.6-kg skateboarder starts out with a speed of 2.44 m/s. He does 80.4 J of work on himself by pushing with his feet against the ground. In addition, friction does -244 J of work on him. In both cases, the forces doing the work are non-conservative. The final speed of the skateboarder is 7.24 m/s. (a) Calculate the change (PEf - PE0) in the gravitational potential energy. (b) How much has the vertical height of the skater changed? Give the absolute value.

Answers

Final answer:

Applying the work-energy theorem, which states that the work done by nonconservative forces equates to the change in kinetic plus potential energy, the skateboarder's change in potential energy results in -18.4J, and the absolute change in height becomes 0.032 m.

Explanation:

The student is asking about the concept of conservation of energy in physics, particularly in the context of the work-energy theorem. This theorem states that work done by nonconservative forces (like friction) is equal to the change in the kinetic and potential energy of the system. According to the given problem, the initial kinetic energy of a 55.6-kg skateboarder is KE0 = 0.5 * m * v², and after performing 80.4J of work on himself, while friction does -244J of work on him, the final kinetic energy becomes KEf = m * v² / 2 where 'v' is the final speed of 7.24 m/s.

(a) Considering that there are no conservative forces, the work-energy theorem states that Wnc = ΔKE + ΔPE. From here, we can calculate that ΔPE or PEf - PE0 results in -18.4J.

(b)The change in vertical height can be obtained by Δh = ΔPE / (m*g), where 'g' is the gravitational constant. Thus, Δh =-18.4J divided by (55.6 * 9.8), which results in -0.032 m, but since we are asked for the absolute value, the height change is 0.032 m.

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Model rocket engines are rated by their thrust force and by the impulse they provide. You can use this information to determine the time interval at which the engines fire. Two rocket engines provide the same impulse. The first engine provides 6 N of thrust for 2 s; the second provides 4 N of thrust.

For how long does this second engine fire?

Answers

Answer: 3 seconds

Explanation:

Since they provide the same impulse,

Impulse= Ft

F1 = 6N

t1 = 2s

F2= 4N

t2= ?

F1= Force of first engine

t1= time elapsed by first engine

F2= force of second engine

t2= time elapsed by second engine

F1t1 = F2t2

6 × 2 = 4t2

t2= 12/4

t2 = 3 seconds

This second engine fires for 3 s

[tex]\texttt{ }[/tex]

Further explanation

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

thrust of first engine = F₁ = 6 N

elapsed time of first engine = t₁ = 2 s

thrust of second engine = F₂ = 4 N

Asked:

elapsed time of second engine = t₂ = 2 s

Solution:

We will use Newton's Law of Motion to solve this problem as follows:

[tex]\Sigma F = ma[/tex]

[tex]\Sigma F = m \Delta v \div t[/tex]

[tex]\Sigma F = I \div t[/tex]

[tex]\boxed {I = \Sigma F \times t}[/tex] → Impulse Formula

[tex]\texttt{ }[/tex]

Two rocket engines provide the same impulse :

[tex]I_1 = I_2[/tex]

[tex]\Sigma F_1 \times t_1 = \Sigma F_2 \times t_2[/tex]

[tex]6 \times 2 = 4 \times t_2[/tex]

[tex]12 = 4 \times t_2[/tex]

[tex]t_2 = 12 \div 4[/tex]

[tex]\boxed {t_2 = 3 \texttt{ s}}[/tex]

[tex]\texttt{ }[/tex]

Learn moreVelocity of Runner : https://brainly.com/question/3813437Kinetic Energy : https://brainly.com/question/692781Acceleration : https://brainly.com/question/2283922The Speed of Car : https://brainly.com/question/568302Average Speed of Plane : https://brainly.com/question/12826372Impulse : https://brainly.com/question/12855855Gravity : https://brainly.com/question/1724648

[tex]\texttt{ }[/tex]

Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

A lens collects light and focuses it into a small spot. This increases the ________ of the light wave.A lens collects light and focuses it into a small spot. This increases the ________ of the light wave.

Answers

Answer:

intensity.

Explanation:

when the light collected by the lens is focused into a small spot it tends to increase the intensity of the light.

as different path of light with different intensity combines from passing through the lens it tends to make the light path and intensity coherent and after being coherent there intensity increases.

Car A

Mass: 1,500 kg

Velocity: 10 m/s


Car B

Mass: 1,500 kg

Velocity: 25 m/s


Car C

Mass: 1,000 kg

Velocity: 10 m/s


which order shows decreasing momentum?

~A, B, C

~B, A, C

~C, B, A

Answers

Answer:

B,A,C

Explanation:

Hope this helps!!!

Answer:

B, A, C

Explanation:

Momentum:

[tex]P = m*v[/tex]

[tex]P = momentum[/tex]

[tex]m = mass[/tex]

[tex]v = velocity[/tex]

Momentum of car A

[tex]P_{A} = 1500kg*10m/s = 15,000kg*m/s[/tex]

Momentum of car B

[tex]P_{B} = 1500kg*25m/s = 37,500kg*m/s[/tex]

Momentum of car C

[tex]P_{C} = 1000kg*10m/s = 10,000kg*m/s[/tex]

In decreasing order:

B, A, C

A loop of radius r = 3.0 cm is placed parallel to the xy-plane in a uniform magnetic field = 0.75 T . The resistance of the loop is 18 Ω. Starting at t = 0, the magnitude of the field decreases uniformly to zero in 0.15 seconds. What is the magnitude of the electric current produced in the loop during that time?

Answers

Answer:

i = 7.777 × 10⁻⁴ A = 0.77 mA

Explanation:

Given:

loop radius, r = 3.0 cm = 0.03 m

Area, A = π x r² = π x 0.03² = 0.0028 m²

Magnetic Field, B = 0.75 T

Loop resistance, R = 18 Ω

time, t = 0.15 seconds

Now,

the induced emf is given as:

EMF = [tex]-\frac{BA}{t}[/tex]

also

EMF = i x R

Where, i is the current flowing

equating both the formulas for EMF, we get

[tex]{i}{R}=-\frac{BA}{t}[/tex]

or

[tex]{i}=-\frac{BA}{tR}[/tex]

substituting the values in the above equation we get

[tex]{i}=-\frac{0.75\times 0.0028}{0.15\times 18}[/tex]

or

the magnitude of the current, i = 7.777 × 10⁻⁴ A = 0.77 mA

Final answer:

The electric current produced in a loop due to a changing magnetic field can be calculated using Faraday's law of electromagnetic induction. Based on the given data, the magnitude of the induced current in the loop is approximately 1.18 mA.

Explanation:

The scenario described in the question depicts a situation where a magnetic field decreases uniformly in magnitude, thus inducing an electric current within a circular loop according to Faraday's law of electromagnetic induction.

The electric current produced in the loop as a result of the change in the magnetic field can be calculated using the expression, I = ΔΦ/ Δt * R, where ΔΦ is the change in magnetic flux, Δt is the time interval, and R is the resistance of the loop.

In this case, the initial magnetic flux, Φ1 = B1 * A, where B1 is the initial magnetic field and A is the area of the loop. Since the magnetic field decreases to zero, the final magnetic flux, Φ2, is zero. So, ΔΦ = Φ2 - Φ1 = - B1 * A. Here, A = πr², where r is the radius of the loop.

So, I = - [ (B1* πr²) / ( Δt * R ) ]. Substituting the given values into the equation, we get I = (0.75 T * π * (0.03 m)²) / ( 0.15 s * 18 Ω ) = 0.00118 A or 1.18 mA.

So, the magnitude of the induced current is approximately 1.18 mA.

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Two airplanes leave an airport at the same time and travel in opposite directions. One plane travels 87 km/h faster than the other. If the two planes are 11,865 km apart after 7 hours, what is the rate of each plane?

Answers

Answer:

Speed of A = 891 km/h

Speed of B = 804 km/h

Explanation:

Let the speed of aeroplane B is v, the speed of aeroplane is A is 87 km/h faster than B.

So, the speed of aeroplane A is 87 + v.

Distance traveled by A after 7 hours, d1 = (87 + v) x 7

Distance traveled by B after 7 hours, d2 = v x 7

Total distance traveled = 11865 km

So, d1 + d2 = 11865

(87 + v) x 7 + v x 7 = 11865

609 + 14 v = 11865

14 v = 11256

v = 804 km/h

So, the speed of A = 87 + 804 = 891 km/h

Speed of B = 804 km/h

How to tell if something is an electrolyte

Answers

Answer: If it has ions, it is an electrolyte

Explanation:

Let's start by explaining that electrolytes are compounds that contain charged particles or ions, which can be cations (positive ions) or anions (negative ions).

So, it is this composition that makes an electrolytic material conduct electricity.

In this sense, the way to identify if a material is an electrolyte or not, is knowing whether it is composed of ions or not.

A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object moving to the left at 1.0 m/s. What is the total kinetic energy after the collision?

Answers

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

[tex]E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2[/tex]

[tex]E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2[/tex]

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

A gas is compressed by an adiabatic process that decreases its volume by a factor of 2.In this process, the pressurea) increases by a factor of more than 2.b) increases by a factor of 2.c) does not change.d) increases by a factor of less than 2.

Answers

Answer:

a) Increase by a factor of more than 2

Explanation:

The compression process is an adiabatic one. The opposite of this kind of process is the isothermal process, where the heat flow makes the temperature to be constant.

Let's consider this kind of process by means of an equation of state; for example, the ideal gas law:

[tex]P=\frac{RT}{v}[/tex]

Be [tex]v_{0}[/tex] the initial volume. And [tex]v_{f}=2v_{0}[/tex] the final volume.

If we consider the ideal gas law, it is evident that if the temperature remains constant (isothermal process), the pressure increases by a factor of 2; but in an adiabatic process the temperature of a gas tends to increase its temperature, so the pressure will be a higher than the resultant for the isothermal process.

A small frictionless cart is attached to a wall by a spring. It is pulled 14 cm from its rest​ position, released at time tequals​0, and allowed to roll back and forth for 5 seconds. Its position at time t is s equals 14 cosine left parenthesis pi t right parenthesis. a. What is the​ cart's maximum​ speed? When is the cart moving that​ fast? Where is it​ then? What is the magnitude of the acceleration​ then? b. Where is the cart when the magnitude of the acceleration is​ greatest? What is the​ cart's speed​ then?

Answers

Answer:given below

Explanation:

Cart is pulled 14 cm from mean position

and its position is given by

[tex]x=14cos\left ( \pi t\right )[/tex]

therefore its velocity is acceleration is given by

[tex]v=-14\pi sin\left ( \pi t\right )[/tex]

[tex]a=-14\pi ^2cos\left ( \pi t\right )[/tex]

[tex]\left ( a\right ) cart\ max.\ speed\ is[/tex]

[tex]v_{max}=14\pi at\ t=0.5sec[/tex]

and its position is x=0

acceleration at t=0.5sec

a=0

[tex]\left ( b\right )[/tex]

[tex]a_{max}=14\pi ^2 cm/s^2[/tex]

at t=0,1,2 sec

at t=0

x=14 cm

v at t=0

v=0 cm/s

for t=1 sec

x=-14 cm i.e. 14 cm behind mean position

v=0 m/s

Final answer:

The cart's maximum speed is 14π m/s and occurs at the equilibrium point in its oscillation at integer multiples of the half period. At these moments, the magnitude of the acceleration is 14π² m/s². The magnitude of the acceleration is greatest when the cart is at the extreme points of its oscillation, where its speed is zero.

Explanation:

The cart, undergoing simple harmonic motion, has a maximum speed when it passes through equilibrium - the midpoint of its oscillating path. From the equation for the straightforward harmonic motion, we know that the speed v of the cart is given by the derivative of s(t) = 14cos(πt). The derivative of this function is v(t) = -14πsin(πt), and the maximum speed occurs when sin(πt) = ±1, which gives a maximum speed of ±14π m/s.

The maximum speed is attained at integer multiples of the half period (1/2, 3/2, 5/2 seconds, and so on). At this moment, the cart is at the equilibrium position, s=0. Acceleration a(t) is given by the second derivative of the position function s(t), which is a(t) = -14π²cos(πt). The magnitude of the acceleration at the time of maximum speed is 14π² m/s².

In the case of the maximum magnitude acceleration, this is attained at the turning points of the oscillation when cos(πt) = ±1. At these moments, the speed of the cart is zero.

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Which ion channels mediate the falling phase of an action potontial?

Answers

Answer:

Voltage-gated K+ channels

The correct answer is that potassium ion channels mediate the falling phase of an action potential.

During the falling phase of an action potential, the membrane potential of the neuron must return to its resting state after being depolarized. This is primarily achieved through the efflux of potassium ions (K^+) out of the cell. The opening of voltage-gated potassium channels allows for this efflux, which repolarizes the membrane potential back towards the resting membrane potential.

 Here is the sequence of events during an action potential:

 1. Resting potential: The neuron is at its resting membrane potential ,typically around -70 mV, due to the concentration gradients of ions across the membrane and the selective permeability of the membrane to potassium ions via leak channels.

 2. Rising phase (Depolarization): When a stimulus reaches the threshold level, voltage-gated sodium channels open, allowing sodium ions (Na^+) to rush into the cell. This influx of positive charges depolarizes the membrane potential towards +30 mV.

3. Peak of the action potential: The membrane potential reaches its peak when the sodium channels become inactivated, stopping the influx of sodium ions.

4. Falling phase (Repolarization): Voltage-gated potassium channels open, and potassium ions move out of the cell, restoring the membrane potential towards the resting potential. The movement of potassium ions out of the cell is slower than the initial sodium influx, which is why the falling phase is slower than the rising phase.

5. Overshoot: Sometimes, the membrane potential temporarily overshoots the resting potential, becoming more negative than at rest, due to the continued efflux of potassium ion.

6. Return to resting potential: The potassium channels close, and the sodium-potassium pump restores the original ion distribution across the membrane, bringing the membrane potential back to the resting potential.

In summary, the falling phase of an action potential is mediated by the opening of voltage-gated potassium channels, which allows for the efflux of potassium ions, thereby repolarizing the neuron's membrane potential."

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