Which of the following is NOT an important phenomenon that commonly erodes and weathers exposed rock outcrops to form sedimentary material?
a) Lightning
b) Wind
c) Rain
d) Freezing/Thawing

Answer:

Anyway, space exploration absolutely does give us a direct benefit. When space technology has advanced far enough, we will be able to leave this planet in large numbers and live among the stars. This will solve our population/environment/resource/energy problems for a long, long time. Even a fraction of the money spent annually on space exploration could save millions of people in poverty-stricken countries, and improve living conditions for future generations. The foundations of the world we live in are largely based on science and it is indeed vital to extend our knowledge of the universe.

                                            Thank You

                                        Please mark me brainliest

Let v be the object's volume. The object displaces 0.4v cm³ of water, which, at a density of about 0.997 g/cm³, has a weight of

b = (0.000997 kg/cm³) (0.4 v cm³) g ≈ 0.000391v N

(and b is also the magnitude of the buoyant force). Then the net force on the object while it's floating in water is

∑ F = b - mg = 0

so that b = mg, where mg is the object's weight. This weight never changes, so the object feels the same buoyant force in each liquid.

(a) In methanol, we have

b = 0.000391v N = (0.00079 kg/cm³) (pv cm³) g

where p is the fraction of the object's volume that is submerged. Solving for p gives

p = (0.000391 N) / ((0.00079 kg/cm³) g) ≈ 0.0505 ≈ 5.05%

(b) In carbon tetrachloride, we have

b = 0.000391v N = (0.00158 kg/cm³) (pv cm³) g

==>   p ≈ 0.0253 ≈ 2.53%

Answer:

N = 167 Newtons

R = 727 Newtons

Explanation:

i) For static equilibrium, moments about any convenient point must sum to zero.

A moment is the product of a force and a moment arm length. Only the force acting perpendicular to a moment arm passing through the pivot point makes a moment.

ii) I will ASSUME the two moment arms are 0.05m and 0.15 m

CCW moments about the fulcrum are

190 N(0.2 m) + 280 N(0.05 m) = 52 N•m

CW moments are (N)N(0.15 m + 90 N(0.3 m) = 27 + 0.15N N•m

For static equilibrium, these must be equal

27 + 0.15N = 52

       0.15N = 25

              N = 166.6666666...

Sum moments about N to zero

(Same as saying CW and CCW moments must balance)

190(0.2 + 0.15) + 280(0.05 + 0.15) - R(0.15) - 90(0.3 - 0.15) = 0

R = 726.6666666...

We could verify this by summing vertical forces to zero.

R - 190 - 280 - 166.666666 - 90 = 0

R = 726.6666666...

Answer:

don't know what class are you which subject is this

The Moment of Inertia of the Disc is represented by [tex]I = \frac{15}{32}\cdot M\cdot R^{2}[/tex]. (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

[tex]I = I_{D} - I_{H}[/tex] (1)

Where:

Then, this formula is expanded as follows:

[tex]I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)[/tex] (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ([tex]m[/tex]):

[tex]\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}[/tex]

[tex]m = \frac{1}{2}\cdot M[/tex]

And the resulting equation is:

[tex]I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)[/tex]

[tex]I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}[/tex]

[tex]I = \frac{15}{32}\cdot M\cdot R^{2}[/tex]

The moment of inertia of the Disc is represented by [tex]I = \frac{15}{32}\cdot M\cdot R^{2}[/tex]. (Correct answer: A)

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Answer:

Help meh in this question

Explanation:

Air pressure or atmospheric pressure is the pressure as the force exerted by the collisions of particles in the air.

The key difference between air pressure and liquid pressure is that air pressure allows the gaseous state of matter to be compressible, whereas liquid pressure makes a liquid incompressible

Answer:

Atmospheric pressure, also known as barometric pressure, is the pressure within the atmosphere of Earth. The standard atmosphere is a unit of pressure defined as 101,325 Pa, which is equivalent to 760 mm Hg, 29.9212 inches Hg, or 14.696 psi.

Liquid pressure is the increase in pressure at increasing depths in a liquid. This pressure increases because the liquid at lower depths has to support all of the water above it. We calculate liquid pressure using the equation liquid pressure = mass x acceleration due to g density x depth in fluid.

Answer:

HOPE THIS ANSWER WILL HELP YOU

Answer:

An object in a motion when it is continuously changing its position based on a reference point and observed by a person or a device.

Answer:

Force is 2 × 10^-4 N

Explanation:

[tex]{ \bf{F = \frac{ \gamma_{o} I _{1} I _{2} l}{2\pi r} }}[/tex]

F is force

I is current

r is separation distance

gamma is a constant

l is wire length

[tex]F = \frac{(4\pi \times {10}^{ - 7}) \times (20) \times (10) \times 1 }{2\pi \times 0.2} \\ \\ F = 2 \times {10}^{ - 4} \: \: newtons[/tex]

The magnitude of the force on 1m length of the 20A wire is 2×10⁻⁴N.

To determine the answer, we need to know about magnetic field due to a current carrying wire and magnetic force.

The magnetic field due to a straight wire having current I₁ at a perpendicular distance (say) 'd'  is μ₀I₁ / 2[tex]\pi[/tex]d.

Magnetic force on a current carrying wire of length 'L' and having current I₂ is I₂(L×B). Where B is the magnetic field at that wire.

  ( Here L and B are perpenicular to each other so L×B= L·B)    

                              = 20×1×10⁻⁵

                              =2×10⁻⁴N

Thus, we can conclude that the magnitude of the force on 1m length of the 20A wire is 2×10⁻⁴N.

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Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

(c) The average force on the right sled is 922.625 N

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴  22.7 × v₁'  = -3.63 × 3.05

v₁' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law,  m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴  22.7 × v₂'  = -3.63 × 3.05

v₂' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = [tex]F_{average}[/tex] × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

[tex]F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}[/tex]

∴ [tex]F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s} = 922.625 \ kg\cdot m/s^2 = 922.625 \ N[/tex]  

The average force on the right sled applied by the cat while landing, [tex]\mathbf{F_{average}}[/tex] = 922.625 N

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(A) The total distance covered by the plane is 500,000 m

(B) The plane's total displacement is 300,000 m

(C) The average velocity of the plane is 100 m/s

The given parameters:

initial displacement of the plane = 400 km Dallas

final displacement of the train = 100 km Huntsville

the time taken for the initial displacement = 120 min

final time for the second trip = 70 min

A sketch of the plane's trip is as follows:

                                         Dallas

                              400 km ↑ ↓

                                            ↑  ↓ Huntsville  100 km

                                             ↑

                                             ↑

                                        Houston

(A) The total distance covered by the plane:

The total distance is the total path travelled by the plane.

Total distance = 400 km + 100 km = 500 km, = 500,000 m

(B) The plane's total displacement:

The total displacement is the change in the position of the plane.

The displacement = 400 km - 100 km = 300 km = 300,000 m

(C) The average velocity of the plane

[tex]Average \ velocity = \frac{change \ in \ displacement }{change \ in \ time} \\\\Average \ velocity = \frac{400 \ km - 100 \ km}{120 \ \min - \ 70 \ \min} \\\\Average \ velocity =\frac{300 \ km}{2\ hr - 1.167 \ hr} \\\\Average \ velocity = \frac{300 \ km}{0.833 \ hr} \\\\Average \ velocity = 360.14 \ km/hr\\\\Average \ velocity = \frac{360.14 \ km/h}{3.6 \times \frac{km/hr}{m/s} } = 100 \ m/s[/tex]

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Answer:

Below

Explanation:

To find the speed of an object you can use this formula

     speed = displacement / elapsed time

For the first one

24 m / 0.6 seconds = 40 m/s

For the second one

0.3 m / 0.01 seconds = 30 m/s

Hope that helped

Explanation:

Her cells will not work well when they have low levels of oxygen.

Answer:

Please mark brainliest

Answer:

[tex]\huge\boxed{\sf F = 147 N}[/tex]

Explanation:

Given:

Mass = m = 12 g = 0.012 kg

Final Velocity = Vf = 35 m/s

Initial Velocity = Vi = 0 m/s

Distance = S = 5 cm = 0.05 m

Required:

Force = F = ?

Formula:

3rd equation of motion

2aS = Vf²-Vi²

Solution:

2aS = Vf²-Vi²

a = (35)²- (0)² / 2S

a = 1225 / 2(0.05)

a = 1225 / 0.1

a = 12250 m/s²

Force = Mass x Acceleration

F = 0.012 x 12250

F = 147 N

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

Given

[tex]\vec F = (4xz+y^2)\,\vec\imath + 2xy\,\vec\jmath + 2x^2\,\vec k[/tex]

its curl would be

[tex]\nabla\times\vec F = \left(\dfrac{\partial(2xy)}{\partial z} - \dfrac{\partial(2x^2)}{\partial y}\right)\,\vec\imath - \left(\dfrac{\partial(4xz+y^2)}{\partial z} - \dfrac{\partial(2x^2)}{\partial x}\right)\,\vec\jmath + \left(\dfrac{\partial(4xz+y^2)}{\partial y} - \dfrac{\partial(2xy)}{\partial x}\right)\,\vec k[/tex]

which reduces to the zero vector. Since the curl is zero, and [tex]\vec F[/tex] doesn't have any singularities, [tex]\vec F[/tex] is indeed a gradient field.

To determine what it is a gradient of, we look for a scalar function f(x, y, z) such that [tex]\nabla f = \vec F[/tex]. This entails solving for f such that

[tex]\dfrac{\partial f}{\partial x} = 4xz+y^2 \\\\ \dfrac{\partial f}{\partial y} = 2xy \\\\ \dfrac{\partial f}{\partial z} = 2x^2[/tex]

Integrate both sides of the first equation with respect to x, which gives

[tex]f(x,y,z) = 2x^2z+xy^2 + g(y,z)[/tex]

Differentiate both sides of this with respect to y :

[tex]\dfrac{\partial f}{\partial y} = 2xy = 2xy+\dfrac{\partial g}{\partial y} \\\\ \implies \dfrac{\partial g}{\partial y} = 0 \implies g(y,z) = h(z)[/tex]

Differentiate f with respect to z :

[tex]\dfrac{\partial f}{\partial z} = 2x^2 = 2x^2 + \dfrac{\mathrm dh}{\mathrm dz} \\\\ \implies \dfrac{\mathrm dh}{\mathrm dz} = 0\implies h(z) = C[/tex]

So it turns out that [tex]\vec F[/tex] is the gradient of

[tex]f(x,y,z) = 2x^2z+xy^2+C[/tex]

Answer:

Submultiples of kilogram are decagram ,centigram etc. And multiples of kilogram is megagram(1 tonne) , gigagram etc.

hope this helps you:)

2: For a sample of data [tex]x_1,x_2,\ldots,x_N[/tex], the mean of this sample denoted by [tex]\overline x[/tex] is the sum of the data divided by the number of data points,

[tex]\overline x = \dfrac{x_1+x_2+\cdots+x_N}N = \displaystyle\frac1N\sum_{i=1}^N x_i[/tex]

As an example, consider [tex]x_1=-1[/tex], [tex]x_2=1[/tex], and [tex]x_3=3[/tex]. Then

[tex]\overline x = \dfrac{-1+1+3}3 = 1[/tex]

3: Standard deviation is a measure of how dispersed a given data sample is relative to the mean. Consult the plot: for a normal distribution, approximately 68% of it lies within 1 standard deviation of the mean, approx. 95% within 2 standard deviations, and approx. 99.7% within 3 standard deviations.

For instance, if the data is pulled from a normally distributed population with mean 0 and standard deviation 1, if you were to randomly select any data from the population, then 68% of the time it will fall in the range (-1, 1); 95% of the time it will fall within (-2, 2); 99.7% of the time it fall within (-3, 3).

To compute the standard deviation for a sample, for each [tex]x_i[/tex] in [tex]x_1,x_2,\ldots,x_N[/tex], you

• take the difference between [tex]x_i[/tex] and the mean [tex]\overline x[/tex]

• square this difference

• sum all the squared differences

• divide the sum by N - 1 (for a sample) or N (for a population)

• take the square root

Here the standard deviation is denoted [tex]\sigma^x_{N-1}[/tex], which I would read as "the sample standard deviation of the data x" - sample because of the N - 1 subscript.

Continuing with the previous example, we'd have

[tex]\sigma^x_{N-1} = \displaystyle \sqrt{\frac{\left(-1-1\right)^2+\left(1-1\right)^2+\left(3-1\right)^2}{3-1}} = \sqrt4 = 2[/tex]

4: Not much more to say here, the standard error is basically a measure of how accurate a given estimate is about the population based on the sample data. It's analogous to uncertainty in measuring length with a ruler, for instance.

In our example,

[tex]\alpha^x = \dfrac2{\sqrt3}[/tex]

5: If x, y, and z are random variables, then I suppose ρ is meant to denote a function of these random variables (so that ρ itself is just another random variable). For instance, you could have ρ = x + 3y - 2z. Then [tex]\overline\rho[/tex] is the sample mean of ρ.

I'm not entirely sure about the notation [tex]x(\overline x,\sigma^x_{N-1},\alpha^x)[/tex], but I suspect it's just referring to sample x with mean [tex]\overline x[/tex] and standard deviation [tex]\sigma^x_{N-1}[/tex] with standard error [tex]\alpha^x[/tex].

∆ρ is just the differential of ρ, essentially capturing how ρ changes with respect to small changes in x, y, and z. The expression you see here follows from the chain rule for differentiation.

The formula you see for [tex]\sigma^\rho_{N-1}[/tex] is the sample standard deviation of ρ. Think of ∆ρ as a vector with 3 components. Then [tex]\sigma^\rho_{N-1}[/tex] is the magnitude of this vector.

Similarly, [tex]\alpha^\rho[/tex] is the standard error for ρ, and corresponds to the magnitude of the vector whose components are the standard errors of x, y, and z.

In order for these statistics to make sense, each of x, y, and z must be samples of the same number of data. Say we take x as before [tex](x_1=-1,x_2=1,x_3=3)[/tex], along with [tex]y_1=0,y_2=4,y_3=-2[/tex] and [tex]z_1=-3,z_2=\frac12,z_3=10[/tex]. Suppose ρ = x + 3y - 2z. Then

• the sample means of y and z :

[tex]\overline y = \dfrac{0+4-2}3 = \dfrac23 \\\\ \overline z = \dfrac{-3+\frac12+10}3 = \dfrac52[/tex]

• the standard deviations of y and z :

[tex]\sigma^y_{N-1} = \sqrt{\dfrac{\left(0-\frac23\right)^2+\left(4-\frac23\right)^2+\left(2-\frac23\right)^2}{3-1}} = 2\sqrt{\dfrac73} \approx 3.06\\\\ \sigma^z_{N-1} = \sqrt{\dfrac{\left(-3-\frac52\right)^2+\left(\frac12-\frac52\right)^2+\left(10-\frac52\right)^2}{3-1}} = \dfrac{\sqrt{181}}2 \approx 6.73[/tex]

• the values of ρ :

[tex]\rho_1 = x_1+3y_1-2z_1 = -1+2\times0-2\times(-3) = 5 \\\\ \rho_2 = x_2+3y_2-2z_2 = 1+3\times4-2\times\dfrac12=12 \\\\ \rho_3 = x_3+3y_3-2z_3 = 3+3\times(-2)-2\times10 = -23[/tex]

• the sample mean of ρ :

[tex]\overline\rho = \dfrac{5+12-23}3 = -2[/tex]

• by the chain rule,

[tex]\Delta\rho = \Delta x+3\Delta y-2\Delta z[/tex]

so the standard deviation of ρ :

[tex]\sigma^\rho_{N-1} = \sqrt{\left(\sigma^x_{N-1}\right)^2 + \left(3\sigma^y_{N-1}\right)^2 + \left(-2\sigma^z_{N-1}\right)^2} \\\\\sigma^\rho_{N-1}= \sqrt{2^2 + 9\left(2\sqrt{\dfrac73}\right)^2 + 4\left(\dfrac{\sqrt{181}}2\right)^2} = \dfrac12\sqrt{\dfrac{703}3} \approx 7.65[/tex]

• the standard errors of y and z :

[tex]\alpha^y = \dfrac{2\sqrt{\frac73}}{\sqrt3} = \dfrac23\sqrt7 \approx 1.76 \\\\ \alpha^z = \dfrac{\frac{\sqrt{181}}2}{\sqrt3} = \dfrac12\sqrt{\dfrac{181}3} \approx 3.88[/tex]

• the standard error of ρ :

[tex]\alpha^\rho=\sqrt{\left(\alpha^x\right)^2+\left(3\alpha^y\right)^2+\left(-2\alpha^z\right)^2}\\\\\alpha^\rho=\sqrt{\left(\dfrac2{\sqrt3}\right)^2+9\left(\dfrac23\sqrt7\right)^2+4\left(\dfrac12\sqrt{\dfrac{703}3}\right)^2}=\sqrt{269}\approx16.40[/tex]

Answer:

I hope it's helpful for you ☺️

Answer:

Nó

Answer:

molecule, a group of two or more atoms that form the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of that substance

Explanation:

Here are examples of common molecules:

H2O (water)

N2 (nitrogen)

O3 (ozone)

CaO (calcium oxide)

C6H12O6 (glucose, a type of sugar)

NaCl (table salt)

Explanation:

molecule, a group of two or more atoms that form the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of that substance

Successive resonances/harmonics are observed to occur at 129, 215 and 301 Hz (Note: 129 Hz is not necessarily the fundamental frequency). the speed of sound is 76 added to 129 is 215

The pitch of the sound is a major property where the frequency of sound occur by human ear within the range of human hearing.

It is higher than the frequency of the sound where the higher is its pitch and a lower frequency means a lower pitch.

The loudness of sound is another property which can be determined by amplitude of the sound is a measure of the magnitude of the maximum disturbance of sound.

The speed is another property can be detrained as  the sound waves which can travel through the medium is called the speed of sound and it is different for different mediums, travels fastest in solids.

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Answer:

Explanation:

ai. com=(67*0+179*1.25)/(179+67)=0.91m

ii. com=(67*2.5+179*1.25)/(179+67)=1.59m

b. 0.91=(67(2.5-d)+179(1.25-d))/(179+67)

   d=0.68m

c. 1.5=(67*-3+67*-4+179*v)/(67+67+179)

    v=5.24m/s

The answers to your question are ;

A) Location of the center of the mass of system when origin of coordinate system is

i) On the original location of man = 0.91 m

ii) On back end of boat = 1.59 m

B) The boat is displaced by = 0.68 m

C) speed of boat after their jump ( V ) = 5.24 m/s

Given data :

mass of man = 67 kg

mass of boat = 179 kg

length of boat = 2.5 m

assumptions : No friction/drag force

Solutions

A) determine location of center of mass

i) On the man's original location = 0

= ( mass of man * 0 + mass * 1.25 ) / ( mass of boat + mass of man)

=  ( 0 + 179 * 1.25 ) /  ( 179 + 67 )

= ( 0 + 223.75 ) / ( 246 ) = 0.9096  ≈ 0.91 m

ii) On the back end of the boat

= ( mass of man * length of boat + mass of boat * 1.25 ) / ( mass of boat + mass of man )

= (  67 * 2.5 + 179 * 1.25 ) / ( 179 + 67 )

= ( 167.5 + 223.75 ) / ( 246 ) = 1.590 m

B) By How much is the boat displaced ( d )

0.91 ( location of center of mass from man's original location )

0.91  = (  67 *  (2.5 - d) +  179 (  1.25 - d ) ) / ( 179 + 67 )

        = ( 67 * ( 2.5 - d ) + 223.75 - 179d )) / ( 246 )

∴ d ≈ 0.68 m

C) Determine speed of boat after the jump

Initial speed of boat = 1.5 m/s

hence speed after the jump ( v )

speed of first man = 3 m/s

speed of second man = 4 m/s

1.5 m/s = ( 67 * -3 + 67 * -4 + 179 * v ) / ( 67 + 67 + 179 )

           = ( - 201 + -268 + 179v ) / ( 313 )

∴ v ≈ 5.24 m/s

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Answer:

I believe it is True

Explanation:

Un proyectil se define como cualquier objeto que traza una trayectoria parabólica. Los parámetros importantes en el movimiento de proyectiles son; Tiempo de vuelo, rango y altura máxima.

a) 1 segundo

b) 1,25 m

c) 10 m

Deje que el tiempo de vuelo sea T

T = 2usinθ / g

u = velocidad inicial

g = aceleración debida a la gravedad

θ = ángulo

T = 2 * 10 m / s * sin (30) / 10

T = 1 segundo

Sea la altura máxima H

H = u ^ 2 sin ^ 2 θ / 2g

H = (10) ^ 2 (sin30) ^ 2/2 * 10

Alto = 1,25 m

Sea el rango R

R = u ^ 2sin 2θ / g

R = (10) ^ 2 sin 2 (30) / 10

R = 8.66 m

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a) El tiempo que tarda el proyectil en llegar al piso es:

[tex]t_{vuelo}=1\: s[/tex]  

b) La máxima altura alcanzada por el proyectil es:

[tex]y_{max}=1.25\: m[/tex]  

c) La máxima distancia alcanzada por el proyectil es:

[tex]x_{max}=8.66\: m[/tex]

a)

Para calcular el tiempo que le toma al proyectil llegar al piso, se puede usar la siguiente ecuación de tiro parabólico.

[tex]y=y_{i}+v_{iy}t-0.5gt^{2}[/tex] (1)

Donde:

Recordemos que la componente de la velocidad incial (v(i)=10 m/s) en el eje y esta dada por:

[tex]v_{iy}=v_{i}sin(30)[/tex]

[tex]v_{iy}=10sin(30)[/tex]

Sabemos que y(i) = 0 y ademas haciendo que y = 0, sabremos el tiempo total de vuelo. De la ecuación (1):

[tex]0=0+10sin(30)t-0.5(10)t^{2}[/tex]

Despejando t.

[tex]10sin(30)=0.5(10)t[/tex]

[tex]t=\frac{10sin(30)}{5}[/tex]  

El tiempo de vuelo sera:

[tex]t_{vuelo}=1\: s[/tex]  

b)

Para calcular la máxima altura, usamos la siguiente ecuación.

[tex]v_{fy}^{2}=v_{iy}^{2}-2gy[/tex]

Si hacemos que veocidad final en y v(fy) sea 0, encontraremos la máxima altura.

[tex]0=v_{iy}^{2}-2gy_{max}[/tex]

[tex]0=(v_{i}sin(30))^{2}-2gy_{max}[/tex]

Despejamos y(max):

[tex]y_{max}=\frac{(v_{i}sin(30))^{2}}{2g}[/tex]

[tex]y_{max}=\frac{(10sin(30))^{2}}{2(10)}[/tex]  

La máxima altura alcanzada por el proyectil será:

[tex]y_{max}=1.25\: m[/tex]  

c)

Sabemo que la velocidad del proyectil en la dirección x es constante, por lo tanto, la ecuacion cinemática será:

[tex]x=vt[/tex]

La máxima distancia se determina con el tiempo de vuelo:

[tex]x_{max}=v_{ix}t_{vuelo}[/tex]

La componente de la velocidad en la direccion x es:

[tex]v_{ix}=v_{i}cos(30)[/tex]

Por lo tanto, el máximo desplazamiento será:

[tex]x_{max}=10cos(30)*1[/tex]

[tex]x_{max}=8.66\: m[/tex]

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Population composition is the description of a,population according to characteristics such as age and sex . These data are often compared over time using population pyramids.