What is the reactive force when a fish swims through water

Answers

Answer 1

The reactive force when a fish swims through water is the force of the water on the fish. This is an example of Newton’s third law of motion which states that for every action, there is an equal and opposite reaction. The active force is the fish pushing against the water, so the reactive force would be the reverse, the equal force of the water pushing back on the fish.


Related Questions

A boy slides a book across the floor, using a force of 5 N over a distance of 2
m. What is the kinetic energy of the book after he slides it? Assume there is
no friction.
A. 5 J
B. 10 J
C. 20 J
D. 2.5 J
SUBMIT

Answers

The kinetic energy of the book after it is slids a distance of 2 meters will be 10 Joules.

How to determine the kinetic energy of an object?

The work-energy theorem states that "the work done on an object is the change in its kinetic energy".

Hence;

Kinetic energy = work done

Note that: work-done is expressed as:

Work done = f × d

Where f is force applied and d is distance traveled.

Given that:

Force applied f = 5 newton

Distance d = 2 meters

Work done = ?

Plug these values into the above formula and solve for the workdone.

Work done = f × d

Work done = 5N × 2m

Work done = 10Nm

Work done = 10 Joules

Therefore, the kinetic energy is 10 Joules.

Option B) 10 J is the correct answer.

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Part C
Increase the value of the Applied Force to 150 N. Pause the simulation and observe the magnitudes and directions of the applied force, the friction
force, and the sum of forces. Why do you think the crate moves this time?

Answers

The crate moves this time because the applied force of 150 N is greater than the maximum static friction force that opposes the motion of the crate. Static friction is a force that opposes the relative motion between two objects in contact. The maximum static friction force is determined by the normal force and the coefficient of static friction between the two surfaces in contact. The harder the surfaces are pushed together, the more force is needed to move them. When the applied force exceeds the maximum static friction force, the crate will start to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force.

How can graphs help demonstrate the qualitative relationship that may exist in a set of data to readers?

Answers

Graphs can help to demonstrate the qualitative relationship between a set of data by identifying patterns among these data.

What other information can the data show?Comparison between groups.Proportional relationships.Variation and dispersion.

Graphs can allow readers to identify a lot of information among a set of data. The most common information to be evaluated through the graphs is the existence of patterns between the data.

For example, a line graph can show the change in a variable over time, allowing readers to see if there is a consistent increase or decrease.

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A ball is allowed to fall freely from certain height it covers a distance of 1st sec equal to?

Answers

The distance covered by a ball in the first second of free fall is approximately 4.9 meters.

When an object falls freely under the influence of gravity, it experiences constant acceleration. In the case of Earth's gravity, the acceleration due to gravity is approximately 9.8 m/s². This means that the velocity of the falling object increases by 9.8 meters per second every second.

To determine the distance covered by the ball in the first second, we can use the equations of motion for uniformly accelerated motion.

The equation that relates distance (d), initial velocity (u), acceleration (a), and time (t) is:

d = ut + (1/2)at²

In this case, the initial velocity is zero (as the ball starts from rest), the acceleration is 9.8 m/s², and we want to find the distance covered in the first second (t = 1 second).

Plugging in the values:

d = 0 * 1 + (1/2) * 9.8 * (1)^2

d = 0 + (1/2) * 9.8

d = 0 + 4.9

d = 4.9 meters

Therefore, the ball covers a distance of approximately 4.9 meters in the first second of free fall.

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PLEASE HELP! Thank you!
Chloe and Sarah are driving bumper cars. Chloe, who is traveling west at 3.9 m/s, is behind Sarah, who is traveling west at 1.6 m/s. The total mass of Chloe’s car is 163 kg, and total mass of Sarah’s car is 179 kg. Immediately after Chloe collides with Sarah, Chloe’s velocity reduces to 0.95 m/s west. What is Sarah’s velocity immediately after the collision?

A. 5.2 m/s
B. 4.0 m/s
C. 4.3 m/s
D. 4.6 m/s

Answers

Sarah’s velocity immediately after the collision is 4.3 m/s west.

option C is the correct answer.

What is Sarah’s velocity immediately after the collision?

Sarah’s velocity immediately after the collision is calculated by applying the principle of conservation of linear momentum as follows;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

m₁ is the mass of Sarah's carm₂ is the mass of Chloe's caru₁ is the initial velocity of Sarahu₂ is the initial velocity of Chloev₁ is the final velocity of Sarahv₂ is the final velocity of Chloe

Sarah’s velocity immediately after the collision is calculated as;

179 (1.6) + 163(3.9) = 179v₁ + 163(0.9)

922.1 = 179v₁ + 146.7

179v₁ = 775.4

v₁ = 4.3 m/s

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A wave has an amplitude of 0.0800 m
and is moving 7.33 m/s. One oscillator
in the wave takes 0.230 s to go from
one crest to the next crest. Find the
wavelength of the wave.
(Unit = m)

Answers

If a wave has an amplitude of 0.0800 m and is moving 7.33 m/s. The

wavelength of the wave is 1.69m.

What is the wavelength?

The wavelength of a wave can be determined using the equation:

Wavelength = velocity / frequency

To determine the frequency we need to calculate the reciprocal of the time it takes for one complete oscillation.

frequency = 1 / time

frequency = 1 / 0.230

frequency ≈ 4.35 Hz

Substitute the values into the wavelength equation:

wavelength = velocity / frequency

wavelength = 7.33 / 4.35

wavelength ≈ 1.69m

Therefore the wavelength of the wave is approximately 1.69 meters.

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Trace the decay of U-238 to Ra-226 as shown in Figure 39.15 in the textFigure out what particles must be emitted in each step, and write the reaction for that step in terms of symbols

Answers

The reaction equations for the steps involved in the decay of U-238 to Ra-226 are;

[tex]^{238}_{92}U\ \rightarrow \ ^{234}_{90}Th \ + \ ^{4}_{2}He[/tex]

[tex]^{234}_{90}Th \ \rightarrow \ ^{230}_{88}Ra \ + \ ^{4}_{2}He[/tex]

[tex]^{230}_{88}Ra \ \rightarrow \ ^{226}_{86}Ra \ + \ ^{4}_{2}He[/tex]

What is the radioactive equation for the decay of U-238?

The radioactive equation for the decay of U-238 to Ra-226 is calculated as follows;

First the uranium atom (U-238) will decay thorium by emitting alpha particle as shown in the equation below;

[tex]^{238}_{92}U\ \rightarrow \ ^{234}_{90}Th \ + \ ^{4}_{2}He[/tex]

The second stage is, the thorium will decay to radium by emitting alpha particles again as shown below;

[tex]^{234}_{90}Th \ \rightarrow \ ^{230}_{88}Ra \ + \ ^{4}_{2}He[/tex]

The third, and final stage, the radium will decay to an isotope of radium again, by emitting alpha particle as shown below;

[tex]^{230}_{88}Ra \ \rightarrow \ ^{226}_{86}Ra \ + \ ^{4}_{2}He[/tex]

Thus, the reaction equations for the steps involved in the decay of U-238 to Ra-226 are;

[tex]^{238}_{92}U\ \rightarrow \ ^{234}_{90}Th \ + \ ^{4}_{2}He[/tex]

[tex]^{234}_{90}Th \ \rightarrow \ ^{230}_{88}Ra \ + \ ^{4}_{2}He[/tex]

[tex]^{230}_{88}Ra \ \rightarrow \ ^{226}_{86}Ra \ + \ ^{4}_{2}He[/tex]

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Match the elements with the number of neutrons. (4 points)
1.
88 38 Sr
2.
39 19 Ka
3.
14 6 Carbon
4.
234 90 Thorium plus 4 2 Helium
a.
20
b.
50
c.
8
d.
74

Answers

The following elements are matched with their number of neutrons

1. 88 38 Sr - b. 50

2. 39 19 Ka - a. 20

3. 14 6 Carbon - c. 8

4. 234 90 Thorium plus 4 2 Helium - d. 74

To determine the number of neutrons for each element, we need to subtract the atomic number (which represents the number of protons) from the mass number (which represents the total number of protons and neutrons).

1. 88 38 Sr:

The atomic number of strontium (Sr) is 38, which means it has 38 protons. The mass number is 88. Subtracting the atomic number from the mass number: 88 - 38 = 50. Therefore, the number of neutrons in strontium is 50.

2. 39 19 Ka:

The atomic number of potassium (K) is 19, indicating 19 protons. The mass number is 39. Subtracting the atomic number from the mass number: 39 - 19 = 20. Hence, the number of neutrons in potassium is 20.

3. 14 6 Carbon:

The atomic number of carbon (C) is 6, representing 6 protons. The mass number is 14. Subtracting the atomic number from the mass number: 14 - 6 = 8. Thus, the number of neutrons in carbon is 8.

4. 234 90 Thorium plus 4 2 Helium:

The atomic number of thorium (Th) is 90, which means it has 90 protons. The mass number is 234. Subtracting the atomic number from the mass number: 234 - 90 = 144. However, there is also an additional helium (He) isotope mentioned with a mass number of 4 and atomic number of 2. Adding the number of neutrons from thorium and helium: 144 + 2 = 146. Hence, the number of neutrons in thorium plus helium is 146.

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how many seconds makes 20 years (show all workings)​

Answers

There are 630,720,000 seconds in 20 years.

To calculate the number of seconds in 20 years, we need to consider the number of seconds in a minute, minutes in an hour, hours in a day, and days in a year.

1 minute consists of 60 seconds.

1 hour contains 60 minutes (60 minutes × 60 seconds = 3600 seconds).

1 day consists of 24 hours (24 hours × 3600 seconds = 86,400 seconds).

1 year typically consists of 365 days (365 days × 86,400 seconds = 31,536,000 seconds).

To find the number of seconds in 20 years, we multiply the number of seconds in one year by 20:

20 years × 31,536,000 seconds = 630,720,000 seconds.

Therefore, there are 630,720,000 seconds in 20 years.

This calculation assumes a non-leap year with 365 days.

If the 20 years span a leap year, the total number of seconds would be slightly higher, accounting for the extra day in the leap year.

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Select the correct answer.
A boat moves 60 kilometers east from point A to point B. There, it reverses direction and travels another 45 kilometers toward point A. What are the total
distance and total displacement of the boat?
O A.
OB.
O C.
O D.
The total distance is 105 kilometers and the total displacement is 45 kilometers east.
The total distance is 60 kilometers and the total displacement is 60 kilometers east.
The total distance is 105 kilometers and the total displacement is 15 kilometers east.
The total distance is 60 kilometers and the total displacement is 45 kilometers east.

Answers

The total distance is 105 kilometers and the total displacement is 15 kilometers east. Option C

How to solve for the  total distance

To calculate the total distance, we add the distances traveled in each leg of the journey: 60 kilometers (from A to B) + 45 kilometers (from B back to A) = 105 kilometers.

However, displacement refers to the change in position of an object in a straight line from its starting point to its ending point. In this case, since the boat starts and ends at the same point (A), the total displacement is zero.

Hence The total distance is 105 kilometers and the total displacement is 15 kilometers east.

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Refer to the picture!

Answers

(a) The work done by the donkey on the cart is 59,721.9 J.

(b) The work done by the force of gravity on the cart is -48,434.87 J.

(c) The work done on the cart by friction during this time is 11,315.12 J.

What is the work done by the donkey on the cart?

(a) The work done by the donkey on the cart is calculated as follows;

Wd = Fd cosθ

where;

F is the applied force by the donkeyd is the displacementθ is the angle of inclination

Wd = 375 N x 163 m x cos(12.3)

Wd = 59,721.9 J

(b) The work done by the force of gravity on the cart is calculated as;

Wg =  Fg x d x cosθ

Where;

Fg is the force of gravityd is the displacementθ is the angle between the force of gravity and displacement

θ = 90⁰ + 4.03⁰ = 94.03⁰

Wg = (431 kg x 9.81 m/s²) x 163 m x cos (94.03)

Wg = -48,434.87 J

(c) The work done on the cart by friction during this time is calculated as;

Wf = Ff x d x cosθ

where;

Ff is the force of friction;

Ff = μmg cosθ

Ff = 0.0165 x 431 kg x 9.81 x cos (4.03)

Ff = 69.59 N

Wf = 69.59 x 163 x cos (4.03)

Wf = 11,315.12 J

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An apartment has the dimensions 18 m by 9 m by 6 m. The temperature is 25°C, and the relative humidity is 59 percent. What is the total mass (in kg) of water vapor in the air in the apartment?

Answers

The total mass of water vapor in the air in the apartment can be calculated using the formula for absolute humidity. The total mass is approximately 5.04 kg.

To calculate the total mass of water vapor in the air, we need to determine the volume of the apartment and then use the formula for absolute humidity.

Calculate the volume of the apartment: The volume of the apartment can be calculated by multiplying its dimensions: Volume = length x width x height = 18 m x 9 m x 6 m = 972 cubic meters.

Determine the vapor pressure: The relative humidity of 59 percent indicates that the air is holding 59 percent of the maximum amount of water vapor it can hold at the given temperature.

Find the saturation vapor pressure: The saturation vapor pressure is the maximum pressure of water vapor that air can hold at a given temperature. At 25°C, the saturation vapor pressure is approximately 3.17 kPa.

Calculate the actual vapor pressure: The actual vapor pressure can be calculated by multiplying the saturation vapor pressure by the relative humidity: Actual vapor pressure = Relative humidity x Saturation vapor pressure = 0.59 x 3.17 kPa = 1.867 kPa.

Calculate the mass of water vapour: The mass of water vapor can be determined using the formula for absolute humidity: Mass = Absolute humidity x Volume = (Actual vapor pressure / (0.287 kJ/(kg·K) x Temperature)) x Volume. In this case, the temperature is 25°C, which is 298 K.

Mass = (1.867 kPa / (0.287 kJ/(kg·K) x 298 K)) x 972 m^3 = 5.04 kg.

Therefore, the total mass of water vapor in the air in the apartment is approximately 5.04 kg.

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