How do animals affect the amount of carbon earth’s atmosphere

A controlled burn removes dead vegetation that might otherwise help a wildfire start and spread.

The term controlled burn refers to setting up an area in which the fire is controlled in order to avoid wild fires. These are deliberately set up in order to avoid the bush from burning down.

Let us recall that a wild fire is able to blaze across a large causing damage to a buildings as well as life and other properties in the way of the fire and could cause huge looses including loss of habitat.

Thus, the National Park Service sometimes creates controlled burns to mitigate wildfires because  a controlled burn removes dead vegetation that might otherwise help a wildfire start and spread.

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The calculated charge density on each plate is 8*10⁻⁸ C/m²

The charge density is a measurement of the amount of electric charge present per unit of surface area, body volume, or field. How much charge is held in a specific field is determined by the charge density. It is possible to calculate charge density in terms of length, area, or volume.

L = 50cm = 0.5m for side length. Q = 4*10⁻⁸C, charge on the plate

= Q/A for surface charge density

Each component's surface charge density is then equal to half of the plate's overall charge density. Thus,

σ(face) = 1/2σ

(face = Q/2A)

(face = Q/2L2) Now that we have Q and L, we can plug in.

σ(face) = 4*10⁻⁸/ 2*0.5²

σ(face) = 4*10⁻⁸ / 0.5

(face) = 8 * 10 ⁻⁸ C/m²

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The value of electric flux for mentioned plane will be 657N⋅m²/C at 40° with the x axis.

For a uniform electric field passing through a plane surface,

electric flux(ϕ)=EAcosθ

where,

θ is the angle between the electric field and the normal to the surface,

E is the electric field

A is surface area

(a) The electric field is perpendicular to the surface so θ=0

​

=(3,50×10³ N/C)[(0.350m)(0.700m)]cos0

=858N⋅m²/C

(b)

The electric field is parallel to the surface θ=90

, so cosθ=0

therefore,

the flux is zero.

(c) For the mentioned plane,

electric flux(ϕ)=(3.50×10³N/C)[(0.350m)(0.700m)]cos40.0

=657N⋅m²/C

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The gravitational field strength is 51.6925 N/kg.

We need to know about the gravitational field to solve this problem. The gravitational field is the area where affected by gravitational force. The magnitude of the gravitational field can be calculated by this equation

g = G . m / r²

where g is the gravitational field, G is gravitational constant (6.67 x 10¯¹¹ Nm²/kg²), m is the mass of the planet and r is the radius of the planet

From the question above, we know that

m = 3.10 x 10²⁴ kg

r = 2.00 x 10⁶ m

By substituting the given parameter, we get

g = G . m / r²

g = 6.67 x 10¯¹¹ . 3.10 x 10²⁴/ (2.00 x 10⁶)²

g = 51.6925 N/kg

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Spectroscopy is the study of the absorption and emission of light and other radiation by matter.

Spectroscopy is the interaction of lights and atoms and the study of the emission and absorption of light.

Spectroscopy can be defined as the study of light absorption as well as its emission. It is also uses to study of structure of atoms.

It offers techniques that has to do with how a sample responds to the radiation of light.

Its applications are:

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The interference pattern remains the same if the rate of electrons going through the slits is decreased to one electron per hour during a diffraction process.

This is referred to as a condition which occurs when two waves meet and combine while traveling along the same medium such as air, water etc.

The interference pattern is the same if the the rate of electrons going through the slits is decreased to one electron per hour during a diffraction process. The decrease in the rate of electrons doesn't affect the pattern of the waves being a constructive or destructive one.

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The value of electric flux will be ​zero.

For a uniform electric field passing through a plane surface,

​electric flux(ϕ) = EAcosθ

where,

θ is the angle between the electric field and the normal to the surface

E is the electric field

A is surface area

(a)

The electric field is perpendicular to the surface,

so,

θ=0

​electric flux(ϕ) = EAcosθ

​electric flux(ϕ)   =(6.20×10⁵N/C)(3.20m²)cos0

​electric flux(ϕ) =1.98×10⁶Vm

(b)

The electric field is parallel to the surface

θ=90°

so,

cosθ=0 and the flux is zero For a uniform electric field passing through a plane surface,

therefore, ​electric flux(ϕ) =0

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The instantaneous velocity of the ball is 18ft/s.

The pace at which an object's position changes as perceived from a particular point of view and as measured by a particular unit of time (for example, 60 km/h northbound) is defined as its velocity. Velocity is a fundamental concept in kinematics, the branch of classical mechanics that deals with the motion of bodies.

In order to be defined, the physical vector quantity known as velocity needs to have both a magnitude and a direction. Speed is a coherent derived unit whose quantity is measured in metres per second (m/s or m/s1) in the SI. Speed is the scalar absolute value (magnitude) of velocity (metric system).

Calculations:

at t=1s, the height is y=50(1)-16(12)=34ft.

at t=1.1s, the height is y=50(1.1)-16(1.12)=35.64ft

Between these two heights, the change in time is 0.1s. The change in height is 35.64ft-34ft=1.64ft.

Therefore [tex]V_{avg}=\frac{del y}{del t} = \frac{1.64ft}{0.1s} =16.4ft/s[/tex]

at t=1.01s, the height is y=50(1.01)-16(1.012)=34.1784ft

between t=1.01s and t=1s the change in time is 0.01s and the change in height is 0.1784ft.[tex]V_{avg}=\frac{0.1784ft}{0.01s} =17.84ft/s[/tex]

at t=1.001, the height is y=50(1.001)-16(1.0012)=34.017984ft

between t=1s and t=1.001s, the change in time is 0.001s and the change in height is 0.017984ft

[tex]V_{avg} =\frac{0.017984ft}{0.001s}=17.984ft/s[/tex]

It appears that as the time interval approaches 0, the average velocity approaches 18ft/s. Therefore it is safe to assume the instantaneous velocity at t=1s is 18ft/s.

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565.2 cm trace of water left on the floor

Given that,

A wet bicycle tire leaves a trace of water on the floor

The tire has a radius of 30 cm. and the bicycle wheel makes 3 full rotations before stopping

The trace of water left on the floor is the circumference of cycle

The circumference of circle is:

r = 30 cm

Therefore,

Given that,

bicycle wheel makes 3 full rotations before stopping

trace of water = 3 x 188.4 = 565.2

Thus, 565.2 cm trace of water left on the floor.

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The force the earth exerts on a person that is 1 kg is 9.8 N.

The force the person exert on earth is 9.8 N.

If our planet were twice as far from the Sun as it is now, the Sun’s force on the Earth will be quarter the initial value.

The force the earth exerts on you is calculated as follows;

F = mg

where;

F = 1 kg x 9.8 m/s²

F = 9.8 N

Thus, the force the earth exerts on an object that is 1 kg is 9.8 N.

Also, based on Newton's third law of motion, the force the exerts on you is equal to the force you exert on Earth.

F = GmM/R²

where;

F = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ x 1.98 x 10³⁰) / (150.23 x 10⁹)²

F = 3.5 x 10²² N

F = GmM/(2R)²

F = GmM/(4R²)

F = ¹/₄GmM/R²

Thus, if our planet were twice as far from the Sun as it is now, the Sun’s force on the Earth will be quarter the initial value.

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Gauss law states that the electric flux is defined as the electric field multiplied by the area of the surface in a plane perpendicular to the field.

Mathematically,

Φ=Q ϵo

Where;

Q is enclosed charge

ϵo is the permittivity of the free space

If the charge is moved to another location inside the same cube, the flux in the electric field remains the same. This is because as long as the charge remains within the plane of the electric field, the flux is calculated in all directions and thus the flux remains unchanged. It only changes when the charge is moved outside the cube.

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Answer:

In the given graph,

Velocity is constant between 4 to 8 which is 2m/s

So, at t=5s

Instantaneous velocity=2m/s

Explanation:

hope this help u thank u

The sound power incident on the eardrum at the threshold of pain is,

[tex]5 \times 10^{-5} W[/tex]

In physics, the power transferred per unit area is known as the intensity or flux of radiant energy, where the area is measured on a plane perpendicular to the direction of the energy's propagation. Watts per square meter (W/m2) and kilograms per square meter (kg/s3) are the units used in the SI system. With waves like acoustic waves (sound) or electromagnetic waves like light or radio waves, intensity is most usually employed to describe the average power transfer across one period of the wave. Other situations where energy is exchanged can also be described in terms of intensity. One could, for instance, figure out how much kinetic energy each drop of water from a sprinkler is carrying.

Power =  intensity x area

Given:

Area, A = [tex]5 \times 10^{-5} \;m^{2}[/tex]

Intensity, I = [tex]1\;W/m^{2}[/tex]

We know that,

Sound Power, P = [tex]I \times A[/tex]

P = [tex]1 \times 5 \times 10^{-5}[/tex]

P = [tex]5 \times 10^{-5}\;W[/tex]

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At 13 seconds, the meteor's velocity is - 562.4 m / s, if it enters the earth's atmosphere at - 435 m / s

We know that,

v = u + at

where,

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

Given that,

u = - 435 m / s

a = - 9.8 m / s²

t = 13 s

v = - 435 - ( 9.8 * 13 )

v = - 562.4 m / s

Velocity is the rate of change of distance with respect to time. It is denoted by V.

Therefore, at 13 seconds, the meteor's velocity is - 562.4 m / s, if it enters the earth's atmosphere at - 435 m / s

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The average speed of the plane is 3,967.7km/h or 1,102.15m/s.

Speed is a scalar number in kinematics that quantifies how quickly an item is moving. Since the plane's speed might fluctuate, we refer to the plane's average speed when describing its rate of motion. The whole distance travelled is divided by the total amount of time required to go that distance to arrive at this pace.

[tex]speed = \frac{distance}{time}[/tex]

For, [tex]time (t_{1} ) = \frac{615}{918}[/tex]

as 1m/s = 3.6km/h , therefore,

[tex]time (t_{1} ) = \frac{615}{918*3.6}[/tex]

[tex]time (t_{1} ) = 0.186 hours[/tex]

For, [tex]time(t_{2} ) = \frac{615}{1377}[/tex]

as 1m/s = 3.6km/h , therefore,

[tex]time (t_{2} ) = \frac{615}{1377*3.6}[/tex]

[tex]time (t_{2} ) = 0.124 hours[/tex]

Thus, the average speed of plane is,

[tex]s = \frac{615+615}{0.124+0.186}[/tex]

s = 3967.7km/h

or, [tex]\frac{3967.7}{3.6}[/tex] m/s = 1,102.15m/s

Therefore, the average speed of the plane is 3,967.7km/h or 1,102.15m/s.

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The centripetal force exerted on the earth by the sun is 3.56775 × 10²² N.

Mass of the Earth, m = 6 × 10²⁴ kg

The average distance between the Sun and the Earth is, r = 1.5 × 10⁸ km                                                                                                           .                                                                                                                   r = 1.5 × 10¹¹ m

The path of the Earth around the Sun is circular, with a circumference:

c = 2πr

 c = 2 × 3.1416 ×  1.5 × 10¹¹ m

 c = 9.42478 × 10¹¹ m

The Earth takes time of T = 365.25 days to complete one revolution around the Sun, that is to travel the distance of c.

T = (365.25 days) × (24hr/1day) × (60min/1hr) × (60s/1min)

T = 3.15576 × 10 ⁷s.

Thus the velocity of the Earth is,

v = c/T

v = (9.42478 × 10¹¹ m) / (3.15576 × 10 ⁷s)

v = 2.98653 × 10⁴ m/s

The centripetal force exerted on the Earth by the Sun is given by,

F = 2mv/r

F = 2× (6 × 10²⁴ kg) × (2.98653 × 10⁴ m/s ) /(1.5 × 10¹¹ m)

F = 3.56775 × 10²² N

Therefore, the centripetal force exerted on the earth by the sun is 3.56775 × 10²² N

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The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

                  [tex]V_x'=0.8c\\V=0.6c\\m=4*10^5kg[/tex]

                  [tex]KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}[/tex]

                      [tex]V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V[/tex]

          [tex]V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c[/tex]

              [tex]KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J[/tex]

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

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To launch a 100 kg human so that he leaves a cannon moving at a speed of 4 m/s, you need a spring with an appropriate spring constant. This spring will be compressed 2. 0 m from its natural length to launch the person.400N/m spring constant is needed.

The spring constant, k, is a proportional constant. It gauges how firm the spring is. When a spring is compressed or extended to a length that differs by an amount x from its equilibrium length, it produces a force F = -kx that pushes it back towards its equilibrium position.

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The fundamental frequency will observe  the change by the multiple of [tex]\sqrt{2}[/tex]

The fundamental frequency of a standing wave which has

string of length L

mass per unit meu

has the tension T

is given by f = 1/2L[tex]\sqrt{T/meu}[/tex]

Now we are required to find if the tension is doubled, with all other factors held constant, what will be the effect on the fundamental frequency

According to the formula,

the fundamental frequency f is directly proportional to the factor of square root

Hence, when the tension of the string is doubled while other factors of the system remain constant ,the formula will become

f= 1/2L[tex]\sqrt{2T/meu}[/tex]

Hence the change in the fundamental frequency will be of the factor [tex]\sqrt{2}[/tex]

Therefore, the change will be of [tex]\sqrt{2}[/tex]

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We describe the motion in the graph as

Motion is the change in position of an object with time.

Since on the graph

From the graph, in the first 2 seconds, the motion of the object is described as the object is at rest.

From the time , t = 2 seconds to t = 4 seconds, we see that the graph has a positive slope. During this time interval, from t = 2 seconds to t = 4 seconds, the object is accelerating.

So, the motion of the object is described as the object is accelerating

For the time interval t = 4 seconds to t = 6 seconds, we see that the velocity is a straight line parallel to the x(time) axis. This implies that the velocity of the object is constant for t = 4 seconds to t = 6 seconds. So, for t = 4 seconds to t = 6 seconds, the motion of the object is described as object is moving with constant velocity

So, we describe the motion in the graph as

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Three-dimensional shape the presence of suggests that an object has weight and gravity, while the absence of it suggests lightness or airiness.

A solid figure, object, or shape with three dimensions—length, breadth, and height—is known as a three-dimensional shape in geometry. Three-dimensional shapes have height, which is equivalent to thickness or depth, unlike two-dimensional shapes. These objects are sometimes referred to as 3-D forms because three dimensions can also be written as 3-D. Every 3-D shape takes up space, which is quantified by volume. For instance, The fundamental 3-dimensional shapes we encounter every day include a cube, rectangular prism, sphere, cone, and cylinder. We can see 3-D shapes all around us. A square prism can be seen in a book and a box, a cube in a Rubik's Cube and a die, etc.

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Both the law and its application make advantage of closed loops and symmetries.

Comparing Ampere's law to Biot-Savart law, the former is simpler to apply. In more complicated situations when symmetry cannot be applied, the Biot-Savart law is used.

Ampere's law is more helpful and simpler to apply if there is symmetry.

Gauss's law is used to compute the electric field, while Ampere's law is used to calculate the magnetic field.

The significance of the Biot-Savart law is as follows:

Both Ampere's law and the Biot-Savart law are applied to determine the magnetic field generated by a conductor that is carrying current. When there is symmetry present, we often employ Ampere's law.  

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Answer:تاتا البال،ًـلاللبببلا

لولاليثقغاايصسضشطذدزلب

Explanation:ابرلرتهىغرلب

 The electric field at the center of uniformly charge ring is zero if the charge were placed at a point on the axis of the ring other than the center.

if the charge were placed at a point on the axis of the ring other than the center the electric field then there is no motion.

electric field - It is a force produced by any charge near its surrounding.

If the charge were placed at a point on the axis of the ring other than the center then there is no motion because at the center of the ring E.F ( Electric field is zero.

electric filter the point on the axis of uniformly charged ring depend on the distance of the point from the center of the ring.

.

E.F =  0 because each half cancels each other . the negative charge will be in equilibrium as every part of the ring is uniformly attracted by it.

The electric field at the center of uniformly charge ring is zero if the charge were placed at a point on the axis of the ring other than the center.

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The force required to pull the block will be 13.89 kN.

Let μ be the viscosity of oil, y be thickness, v be the velocity of the block, and A be the base area of the block. Then the force is given as,

F = μAv / y

An inclined surface with a coating of sae 10w oil at 100°C is dragged up by a block that weighs 10 lbf and measures 10 in on each edge. if the oil coating is 0.001 in. thick and the block is moving at a speed of 2 ft/s. Assume that the oil film's velocity distribution is linear. A 25° angle from the horizontal separates the surface from the horizontal.

The diagram is given below. Then the force is given as,

F - 10 sin 25° = 10 × (10 / 12)² × 2 / 0.001

F = 4.226 + 13888.89

F = 13893.1 N

F = 13.89 kN

13.89 kN of force will be required for pulling the block.

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1. The horizontal component of the velocity is 1.73×10³ m/s

2. The vertical component of the velocity is 1×10³ m/s

3. The range of the shell is 3.53×10⁵ m

4. The maximum height reached is 1.53×10⁵ m

1. How to determine the horizontal component of the velocity

Horizontal velocity = u × Cosθ

Horizontal velocity = 2×10³ × Cos30

Horizontal velocity = 1.73×10³ m/s

2. How to determine the vertical component of the velocity

Vertical velocity = u × Cosθ

Vertical velocity = 2×10³ × Sin30

Vertical velocity = 1×10³ m/s

3. How to determin the range

R = u²Sine(2θ) / g

R = (2×10³)² × Sine (2×30) / 9.8

R = 3.53×10⁵ m

4. How to determine the maximum height

H = u²Sine²θ / 2g

H = [(2×10³)² × (Sine 30)²] / (2 × 9.8)

H = 1.53×10⁵ m

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|E| = 16.67 N/C and the direction of this Electric field is in the downward direction.

We have a Charge particle present inside a Electric field.

We have to determine the magnitude (in n/c) and direction of an electric field that exerts a 2 x [tex]10^{-5}[/tex] Newton.

The electric field intensity at a point R meters away from a charge Q is the amount of force experienced by the test charge ([tex]$q_{o}[/tex]) of unit magnitude at a distance R meters away the Charge Q. Mathematically -

E = [tex]$\frac{F}{q}[/tex]

According to the question, we have -

F = 2 x [tex]10^{-5}[/tex] Newton

|Q| = - 1.2 [tex]\mu[/tex]C = - 1.2 x [tex]10^{-6}[/tex]  [tex]\mu[/tex]C = 1.2 x [tex]10^{-6}[/tex] [tex]\mu[/tex]C

Therefore -

E = [tex]$\frac{2\times 10^{-5} }{1.2\times 10^{-6} }= \frac{100}{6} = 16.67\;N/C[/tex]

Therefore -

|E| = 16.67 N/C and the direction of this Electric field is in the downward direction.

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The correct answe is South Africa - Economic Growth and Development:

After the formal end of the previous apartheid system two decades ago, South Africa can now boast having one of the richest economies in Africa and a well-functioning democracy. It is the largest economy in Africa, but it also has deeply ingrained structural issues that limit its ability to expand and flourish.

One of the largest economies on the African continent is that of South Africa. However, despite a period of rapid development from 2003 to 2007, its real GDP average yearly growth rate between 2001 and 2010 has been very weak and unquestionably significantly below the African average. A number of African nations have had substantially faster growth rates, which has improved a number of development-related indices.

By the standards of the recent growth records of several Euro Zone nations, South Africa's development is hardly sluggish! One crucial aspect of the economy is that South Africa has achieved relatively modest progress in meeting a number of important development targets and some of the Millennium Development Goals, but her economy does not appear to have achieved the "take-off" required to kick start significant development progress, especially against the backdrop of her deep social problems.

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