A meteorologist who has been analyzing weather data from the last fifty years found evidence of climate change. An ecologist has been studying deposits of pollen from the last fifty years to track the types of plants that grew in certain regions. She has also found evidence of climate change. What does this imply about science?
A. Meteorology and ecology are the same type of science.
B. When scientific disciplines overlap, researchers should choose new topics.
C. Different scientific disciplines can be used to investigate similar questions.
D. There should be fewer sciences, as they seem to answer the same questions.
Explain

Answer:

    d₃ = 37,729 km,     θ=  5.1º North of West

Explanation:

This is a velocity addition problem, the easiest way to solve it is to decompose the velocities in a Cartesian system, the x-axis coincides with the West-East direction and the y-axis with the South-North direction

* first displacement is

           d₁ₓ = 11 km

* second offset is

          cos 6 = d₂ₓ / d₂

          sin 6 = d_{2y} / d₂

          d₂ₓ = d₂ cos 6

          d_{2y} = d₂ sin 6

          d₂ₓ = 6 cos 6 = 5.967 km

          d_{2y} = 6 sin 6 = 0.6272 km

* third displacement is unknown

* fourth and last displacement

          cos (-11) = d₄ₓ / d₄

          sin (-11) = d_{4y} / d₄

          d₄ₓ = d₄ cos (-11)

          d_{4y} = d₄ sin (-11)

          d₄ₓ = 21 cos (-11) = 20.61 km

          d_{4y} = 21 sin (-11) = -4.007 km

They tell us that at the end of the tour you are back on the island, so the displacement must be zero

X axis

           x = d₁ₓ + d₂ₓ + d₃ₓ + d₄ₓ

           0 = 11 +5.967 + d₃ₓ + 20.61

           d₃ₓ = -11 - 5.967 - 20.61

           d₃ₓ = -37.577 km

Y axis  

          y = d_{1y} + d_{2y} + d_{3y} + d_{4y}

          0 = 0 + 0.6272 + d_{3y} -4.007

          d_{3y} = 4.007 - 0.6272

          d_{3y} = 3.3798 km

This distance can be given in the form of module and angle

Let's use the Pythagorean theorem for the module

           d₃ = [tex]\sqrt{d_{3x}^2 + d_{3y}^2}[/tex]

           d₃ = [tex]\sqrt{37.577^2 + 3.3798^2}[/tex]

           d₃ = 37,729 km

Let's use trigonometry for the angle

            tan θ = d_{3y} / d₃ₓ

            θ = tan⁻¹ [tex]\frac{d_{3y}}{d_{3x}}[/tex]

            θ = tan-1 (-3.3798 / 37.577)

            θ = 5.1º

Since the y coordinate is positive and the x coordinate is negative, this angle is in the second quadrant, so the direction given in the form of cardinal coordinates is

            θ=  5.1º North of West

Answer:

Lol

Explanation:

Answer:

The force must be applied during 8 seconds to reach trhe same final speed.

Explanation:

By Impulse Theorem, a change in the magnitude of linear momentum of a system with constant mass can be done by applying a force during a given time. That is:

[tex]m\cdot (v_{f}-v_{o}) = F \cdot \Delta t[/tex]

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speed, measured in meters per second.

[tex]F[/tex] - Net external foce, measured in newtons.

[tex]\Delta t[/tex] - Time, measured in seconds.

We can eliminate mass and speeds by constructing the following relationship:

[tex]F_{1}\cdot \Delta t_{1} = F_{2}\cdot \Delta t_{2}[/tex] (2)

If we know that [tex]F_{1} = F[/tex], [tex]\Delta t_{1} = 4\,s[/tex] and [tex]F_{2} = \frac{F}{2}[/tex], then the time is:

[tex]4\cdot F = \frac{F\cdot \Delta t_{2}}{2}[/tex]

[tex]\Delta t_{2} = 8\,s[/tex]

The force must be applied during 8 seconds to reach trhe same final speed.

Answer:

I think it’s c

Explanation:

Answer:

pretty sure A is correct, as they are pushing in opposite directions, hence canceling each other

Explanation:

Answer:

a. spring constant = 125 N/m

b. This spring is less stiff than the one in Sample Problem A.

Explanation:

P.S - The Sample Problem A is as follows :

Given - Sample Problem A - A load of 45 N attached to a spring that is

            hanging vertically stretches the spring 0.14 m. What is the spring

             constant?

            Suppose the spring in Sample Problem A is replaced with a spring  

            that stretches 36 cm from its equilibrium position.

To find - a. What is the spring constant in this case?

            b. Is this spring stiffer or less stiff than the one in Sample Problem A.

Proof -

As given,

Load = 45 N

Amplitude = 0.14 m

Let the spring constant = k

As we know that,

Load = k (Amplitude)

⇒45 = k(0.14)

⇒k = [tex]\frac{45}{0.14}[/tex] = 321.43

∴ we get

Spring constant in Sample problem A = 321.43

Now,

a.)

Given, Amplitude = 36 cm = 0.36 m

Let the spring constant = k₁

⇒45 = k₁ (0.36)

⇒k₁ = [tex]\frac{45}{0.36}[/tex] = 125 N/m

b.)

AS we can see that k₁  < k

⇒ This spring is less stiff than the one in Sample Problem A.

a. spring constant = 125 N/m

b. This spring is less stiff than the one in Sample Problem A.

The spring constant generally shows the stiffness of the spring and is the ratio of the force applied to the deflection of the spring.

It is given in the question that:

Sample Problem A - A load of 45 N attached to a spring that is

hanging vertically stretches the spring 0.14 m.  

Suppose the spring in Sample Problem A is replaced with a spring  

that stretches 36 cm from its equilibrium position.

a. What is the spring constant in this case?

As given,

Load F = 45 N

Amplitude  x= 0.14 m

Let the spring constant = k

As we know that spring force will be

[tex]F=k\times x[/tex]

[tex]45=k\times 0.14[/tex]

⇒k = 321.43N/m

∴ we get

Spring constant in Sample problem A is   [tex]k=321.43\ \frac{N}{m}[/tex]

Now,

Given, Amplitude = 36 cm = 0.36 m

Let the spring constant = k₁

⇒45 = k₁ (0.36)

⇒k₁= 125 N/m

b.) Is this spring stiffer or less stiff than the one in Sample Problem A.

AS we can see that k₁  < k  new spring is less stiff than the one in Sample Problem A.

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Answer:

Are you sure it was soccer ball? Or meine hearts

Explanation:

Answer:

yes. that is how a baby is conceived.

Answer:

See explanation

Explanation:

Since the original count rate is 600 Bq,

i) after 1 half life, the count rate decreases to 1/2 of 600 Bq = 300 Bq

ii) after 2 half lives, the count rate decreases to 1/4 of 600 = 150 Bq

iii) after 3 half lives, the count rate decreases to 1/6 of 600 = 100 Bq

iv) after 4 half lives, the count rate decreases to 1/8 of 600 = 75 Bq

Answer:

The car skids in a distance of 61.275 meters.

Explanation:

Since the only force exerted on the car is the kinetic friction between the car and the horizontal road, deceleration of the vehicle ([tex]a[/tex]), measured in meters per square second, is determined by the following expression:

[tex]a = \mu_{k}\cdot g[/tex] (1)

Where:

[tex]\mu_{k}[/tex] - Coefficient of kinetic friction, no unit.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

If we know that [tex]\mu_{k} = 0.52[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], then the net deceleration of the vehicle is:

[tex]a = 0.52\cdot \left(-9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]a = -5.1\,\frac{m}{s^{2}}[/tex]

The distance covered by the car is finally calculated by this kinematic expression:

[tex]\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex] (2)

Where:

[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speed, measured in meters per second.

[tex]a[/tex] - Net deceleration, measured in meters per square second.

If we know that [tex]v_{o} = 25\,\frac{m}{s}[/tex], [tex]v = 0\,\frac{m}{s}[/tex] and [tex]a = -5.1\,\frac{m}{s^{2}}[/tex], then the distance covered by the car is:

[tex]\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(25\,\frac{m}{s} \right)^{2}}{2\cdot \left(-5.1\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\Delta s = 61.275\,m[/tex]

The car skids in a distance of 61.275 meters.

Answer:

See explanation

Explanation:

From the formula;

0.693/t1/2 = 2.303/t log (Ao/A)

t1/2 = half life of Sodium-24

Ao = initial activity of Sodium-24

A= activity of Sodium-24 at time = t

So,

0.693/15 = 2.303/15 log (800/A)

0.0462 = 0.1535  log (800/A)

0.0462/0.1535 =  log (800/A)

0.3 = log (800/A)

Antilog(0.3) =  (800/A)

1.995 =  (800/A)

A = 800/1.995

A = 401 Bq

ii) 0.693/15 = 2.303/30 log (800/A)

0.0462 = 0.0768 log (800/A)

0.0462/0.0768 =  log (800/A)

0.6 =  log (800/A)

Antilog (0.6) =  (800/A)

3.98 =  (800/A)

A = 800/3.98

A = 201 Bq

iii)

0.693/15 = 2.303/45 log (800/A)

0.0462 = 0.0512  log (800/A)

0.0462/0.0512  =  log (800/A)

0.9 = log (800/A)

Antilog (0.9) =  (800/A)

7.94 = (800/A)

A = 800/7.94

A= 100.8 Bq

iv)

0.693/15 = 2.303/60 log (800/A)

0.0462 = 0.038 log (800/A)

0.0462/0.038 = log (800/A)

1.216 = log (800/A)

Antilog(1.216) = (800/A)

16.44 = (800/A)

A = 800/16.44

A = 48.66 Bq

Answer:

1. D

Climate is generally defined as the weather condition that prevails in a particular region over a long period of time. Climate is usually measured by examining the pattern of variation in several climatic factors such as rainfall, temperature, relative humidity, wind, pressure, etc. While the weather of a place can change within a space of few hours, it takes years for a change in climatic condition to occur.  

2. d

3. c

4.a.

5. c

6. a.

7. c

Explanation:

The correct answers are (1) d. two of the above (average air temperature and humidity), (2)c. the relative humidity is 100 percent, (3)d. two of the above (water vapor condenses out of the air and air temperature decreases), (4)a. cirrocumulus, (5)c. cumulus, (6)c. steam fog, and (7)c. sleet.

Temperature is a physical quantity that measures the average kinetic energy of the particles in a substance or system. It is a measure of how hot or cold something is, and is typically measured in units such as degrees Celsius or Fahrenheit. Temperature can also be thought of as a measure of the direction in which heat energy flows, with heat energy naturally flowing from areas of higher temperature to areas of lower temperature.

Here in the Question,

1. Weather factors include d. two of the above (average air temperature and annual precipitation are two factors that affect weather, but humidity is also an important factor that can influence the feel of the air).

2. The dew point is the temperature at which b. water vapor starts to condense. When air cools, it can reach a point where it is unable to hold all of its moisture in the form of water vapor. At this point, the water vapor starts to condense into visible droplets, such as dew, and the temperature at which this happens is called the dew point. When the dew point is reached, the relative humidity is at 100 percent.

3. Relative humidity may decrease if d. two of the above (water vapor condenses out of the air and air temperature decreases) occur. If the air cools and reaches the dew point, water vapor will start to condense into droplets, which can reduce the amount of water vapor in the air and lower the relative humidity. Similarly, if the temperature drops without any change in water vapor content, the relative humidity will decrease because colder air can hold less moisture than warmer air.

4. The type of cloud that forms at high altitudes is a. cirrocumulus. These clouds are typically found at altitudes above 18,000 feet and are characterized by small, white, puffy clouds arranged in rows or ripples. They are often a sign of fair weather but can also indicate an approaching storm.

5. The type of cloud that forms when strong air currents carry warm air upward is c. cumulus. Cumulus clouds are large, puffy clouds that can develop vertically, forming a towering cloud with a flat top. They are often associated with thunderstorms and can produce heavy rain, hail, and lightning.

6. The type of fog that forms when cool air moves over a warm lake is c. steam fog. Steam fog, also called evaporation fog or sea smoke, occurs when cold, dry air moves over a warm, moist surface and causes water vapor to rise and condense into fog. This type of fog is often seen over bodies of water during the fall and winter.

7. Rain that passes through a layer of freezing air near the ground becomes c. sleet. Sleet is formed when raindrops fall through a layer of freezing air near the ground and freeze into small ice pellets before hitting the surface. It is different from hail, which forms in strong thunderstorms when updrafts carry raindrops upward into colder air where they freeze and then fall back to the ground, and snow, which forms in clouds when water vapor freezes directly into ice crystals. Glaze is a type of ice that forms when rain falls onto a surface that is below freezing, forming a layer of ice on top of the surface.

Therefore, The correct answers are:1. Weather factors include average air temperature, annual precipitation, and humidity. 2. The dew point is the temperature at which water vapor starts to condense. 3. Relative humidity may decrease if water vapor condenses out of the air or if the air temperature decreases. 4. The type of cloud that forms at high altitudes is cirrocumulus. 5. The type of cloud that forms when strong air currents carry warm air upward is cumulus. 6. The type of fog that forms when cool air moves over a warm lake is steam fog. 7. Rain that passes through a layer of freezing air near the ground becomes sleet, which is different from hail and snow.

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Answer:

The electric potential is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field with negligible acceleration of the test charge to avoid producing kinetic energy or radiation by test charge.

SI unit: volt

Other units: statvolt

In SI base units: V = kg⋅m^2⋅A^−1⋅s^−3

Dimension: M L^2 T^−3 I^−1

Explanation:

Hope it is helpful....

Explanation:

common symbol - V

SI unit - volt

other unit - statvolt

Answer:

Mass = 1133.33 kg (Approx.)

Explanation:

Given:

Momentum = 2.04 x 10⁴ kg[m/s]

Velocity = 18 m/s

Find:

Mass

Computation:

Mass = Momentum / Velocity

Mass = [2.04 x 10⁴] / 18

Mass = 1133.33 kg (Approx.)

Answer:

T = 0.375 s, f = 2.66 Hz and ω = 16.71 rad/s

Explanation:

Given that,

The mass of a harmonic oscillator, m = 0.5 kg

The force constant of the spring, k = 140 N/m

The frequency of a harmonic oscillator is given by :

[tex]f=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}[/tex]

Substitute all the values,

[tex]f=\dfrac{1}{2\pi }\sqrt{\dfrac{140}{0.5}} \\\\f=2.66\ Hz[/tex]

Time period is given by :

[tex]T=\dfrac{1}{f}\\\\T=\dfrac{1}{2.66}\\\\T=0.375\ s[/tex]

The angular frequency is given by :

[tex]\omega=2\pi f\\\\\omega=2\pi \times 2.66\\\\\omega=16.71\ rad/s[/tex]

Hence, this is the required solution.

Answer:

Uhh 2 one

Explanation

I'm assuming the question is supposed to be like this:

How  does the Orbital velocity of a satellite depends on the mass of the satellite?

Answer

It is independent of mass of satellite.

Answer:

Kinetic energy increases and potential energy decrease when velocity of an object increase.

Answer:

-the ratio of the speed of light

in air to the speed of light in the substance.

-speed of light in air 300,000 km/sec, which decreases when it passes through a transparent substance.

-e.g.. speed of light in substance = 200,000 km/sec, R.I. = 300,000/200,000 = 1.5

Explanation:

Is earth’s freshwater

Only about 3 percent of Earth’s water is fresh water.

Answer:

2.66×10⁻⁹ N.

Explanation:

From the question,

Applying newton's law of universal gravitation,

Fg = GMm/r²............................... Equation 1

Where Fg = gravitational force, G = universal constant, M = mass of the mercury, m = mass of the human, r = radius of Mercury

Given: M = 3.31023 kg, M = 70 kg, r = 2.4106

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute these values into equation 1

Fg = 6.67×10⁻¹¹(70×3.31023)/(2.4106²)

Fg = 2.66×10⁻⁹ N.

Answer:

(a) The volume of the liquid helium at 25 K is 5.13 L

(b)  The volume of the liquid helium at 293 K is 60.14 L.

Explanation:

Given;

mass of the liquid helium, m = 10 g

initial temperature of the liquid helium, T₁ = 4.2 K

pressure of the liquid helium, P = 1.00 atm

Atomic mass of Helium, = 4 g

number of moles of Helium, n  = 10 / 4 = 2.5 moles

The initial volume of the liquid helium is calculated as;

[tex]PV_1 = nRT_1\\\\V_1 = \frac{nRT_1}{P} \\\\[/tex]

where;

R is ideal gas constant,  = 0.08205 L.atm./mol.K

[tex]V_1 = \frac{2.5 \times 0.08205 \times 4.2}{1 } \\\\V_1 = 0.862 \ L[/tex]

(a) The volume of the liquid helium at 25 K.

Apply Charles law;

[tex]\frac{V_1}{T_1} =\frac{V_2}{T_2} \\\\V_2 = \frac{V_1T_2}{T_1} \\\\V_2 = \frac{0.862 \times 25 }{4.2} \\\\V_2 = 5.13 \ L[/tex]

(b)  The volume of the liquid helium at 293 K.

[tex]\frac{V_1}{T_1} =\frac{V_2}{T_2} \\\\V_2 = \frac{V_1T_2}{T_1} \\\\V_2 = \frac{0.862 \times 293 }{4.2} \\\\V_2 = 60.14 \ L[/tex]

Answer:

the altitude of these satellites above the surface of the earth;

35790 km ( SI units )

22243.63 miles ( U.S. customary unit )

Explanation:

Given the data in the question;

Time taken by the satellite to complete on revolution is 23.934 hours

= 23.934 × 60 × 60 = 86162.4 seconds

now, let h represent altitude, r represent Orbit radius, v represent Orbit speed.

we know that

v² = GM/r

= gR²/r

= (9.81m/s² × ( 6.37 × 10⁶ m)²) / r

v = 19.95 × 10⁶ / √r   ---------- let this be equation

also;

time t = 2πr/v

86162.4 s = 2πr/v -------- let this be equation 2

a) Determine the altitude of these satellites above the surface of the earth in both SI and U.S. customary units

we substitute v in equation into equation 2

so

86162.4 s = 2πr / (19.95 × 10⁶ / √r)

r = 42.16 × 10⁶ m

so, altitude h = r - R

h = 42.16 × 10⁶ m - 6.37 × 10⁶ m

h = 35790000 m

convert to kilometer

h = 35790000 / 1000

h = 35790 km

Convert to miles

h = 35790 / 1.609

h = 22243.63 miles

Therefore, the altitude of these satellites above the surface of the earth;

35790 km ( SI units )

22243.63 miles ( U.S. customary unit )

Answer:

the kinetic energy gained by the toy is 6J.

Explanation:

Given;

net applied to the toy by dog, F = 2 N

distance moved by the toy, d = 3 m

Apply the principle of work-energy theorem to determine the kinetic energy gained by the toy.

ΔK.E = W

         = F x d

         = 2 x 3

         = 6 J

Therefore, the kinetic energy gained by the toy is 6J.

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J