why do you like the full moon ?

Answer:

   P = 10135.6 Pa

Explanation:

For this exercise we use that the pressure varies with the height

           P = P₀ + ρ g h

where h is the height from the head to the heart, which is approximately

h = 40 cm = 0.40m  and P₀ is the head pressure P₀ = 6000 Pa

          P = 6000 + 1055 9.8 0.40

          P = 6000 + 4135.6

          P = 10135.6 Pa

Answer:

The arrow will bury itself farther by 3S₁

Explanation:

lets assume; the Arrow shot by me has a speed twice the speed of the arrow fired by the younger shooter

Given that ; acceleration is constant , Frictional force is constant

                    A₂ =   A₁

Vf²₂ - Vi²₂ / 2s₂  = Vf₁² - Vi₁² / 2s₁ ---- ( 1 )

final velocities = 0

Initial velocities : Vi₂ = 2(Vi₁ )

Back to equation 1

0 - (2Vi₁ )² / 2s₂ =  0 - Vi₁² / 2s₁

hence :

s₂ = 4s₁

hence the Arrow shot by me will burry itself farther by :

s₂ - s₁ = 3s₁

Note :  S1 = distance travelled by the arrow shot by the younger shooter

Answer:

Average Velocity = 3.63 m/s

Explanation:

First, we will calculate the total displacement of the quarterback, taking forward direction as positive:

Total Displacement = 15 m - 3 m + 24 m = 36 m

Now, we will calculate the total time taken for this displacement:

Total Time = 3 s + 1.71 s + 5.2 s = 9.91 s

Therefore, the average velocity will be:

[tex]Average\ Velocity = \frac{Total\ Displacement}{Total\ Time}\\\\Average\ Velocity = \frac{36\ m}{9.91\ s}[/tex]

Average Velocity = 3.63 m/s

Answer:

The maximum speed of Bolt for the 100 m race is 14.66 m/s

Explanation:

Given;

initial distance covered by Bolt, d = 200 m

time of this motion, t = 19.3 s

The second distance covered by Bolt, = 100 m

Assuming Bolt maintained the same acceleration for both races.

His acceleration can be determined from the 200 m race.

d = ut + ¹/₂at²

where;

u is his initial velocity = 0

d =  ¹/₂at²

[tex]at^2 = 2d\\\\a = \frac{2d}{t^2} \\\\a = \frac{2\times 200}{19.3^2} \\\\a = 1.074 \ m/s^2[/tex]

Let the final or maximum velocity for the 100 m race = v

v² = u² + 2ad₂

v² =  2 x 1.074 x 100

v² = 214.8

v = √214.8

v = 14.66 m/s

The maximum speed of Bolt for the 100 m race is 14.66 m/s

Answer:

fair palybtgshsisuehdh

Answer:

201.5537 mph

Explanation:

Given the following data;

Speed = 90.1 m/s

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the formula;

Speed = distance/time

To convert this value into miles per hour;

Conversion;

1 meter = 0.000621 mile

90.1 meters = 90.1 * 0.000621 = 0.05595 miles

1 metre per second = 2.237 miles per hour

90.1 meters per seconds = 90.1 * 2.237 = 201.5537 miles per hour

90.1 m/s = 201.5537 mph

Answer:

0

Explanation:

a = dv/dt

if v is constant than the slope of the v graph will be 0, so dv/dt is 0

a= 0

Answer:

a

Explanation:

you take 32,000kg ÷2.0m

Answer:

a n c

Explanation:

Answer:

The frequency of the second wave is half of the frequency of first one.

Explanation:

The wavelength of the second wave is double is the first wave.

As we know that the frequency is inversely proportional to the wavelength of the velocity is same.

velocity = frequency x wavelength

So, the ratio of frequency of second wave to the first wave is

[tex]\frac{f_2}{f_1} =\frac{\lambda _1}{\lambda _2}\\\\\frac{f_2}{f_1} =\frac{\lambda _1}{2\lambda _1}\\\\\frac{f_2}{f_1} =\frac{1}{2}\\\\[/tex]

The frequency of the second wave is half of the frequency of first one.

Solution :

It is given that a car ran off from a cliff and it crashes on canyon floor. Now the system of a car as well as the earth together have a  [tex]\text{ net momentum of zero}[/tex] when the car crashes on the canyon floor, thus reducing the momentum of the car to zero. The earth also stops its upward motion and it also reduces the momentum to zero.

Answer:

Static friction depends on the mass of the object.

Explanation:

Friction is the force between two surfaces in contact. The force of friction between two surfaces in contact depends on;

1) nature of the object and the surface(how rough or smooth the surfaces are)

2)surface area of the object and the surface

3) mass of the object

Since;

F=μmg

Where;

μ= coefficient of static friction

m= mass of the object

g= acceleration due to gravity

Hence, as the mass of the object increases, the magnitude of static friction force between an object and a surface increases and vice versa.

Answer:

a)  T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] ,  b) T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c)  x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex],  d)  m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

      ∑ τ = 0

      T_A d - W₂ x -W₁ L/2 = 0

      T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]

b) the tension in string B

we write the expression of the translational equilibrium

       ∑ F = 0

       T_A - W₂ - W₁ - T_B = 0

       T_B = T_A -W₂ - W₁

       T_ B =   [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]  - g m₂ - g m₁

       T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

         T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

          g m₂ (d -x) -  g m₁  (0.5 L -d) + 0 = 0

          m₂ (d-x) = m₁ (0.5 L- d)

          m₂ x = m₂ d - m₁ (0.5 L- d)

          x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex]

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

           0 = d - \frac{m_1}{m_2} \  \frac{L}{2d}

         [tex]\frac{m_1}{m_2} \ (0.5L -d) = d[/tex]

          [tex]\frac{m_1}{m_2} = \frac{ d}{0.5L-d}[/tex]

          m₂ = m₁  [tex]\frac{0.5 L -d}{d}[/tex]

          m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

Answer:

The breakdown of air occurs at a maximum voltage of 3kV/mm.

Explanation:

The breakdown of air occurs at a maximum voltage of 3kV/mm.

At this level of voltage the air between the plates become highly ionised and breakdown occurs. Since, the distance held between the plates is 1mm , it can withstand a maximum voltage of 3 kV.

After this voltage the air will become conductive in nature and will form ions in the air between the plates and ultimately breakdown will take place with a flash.

Answer:

Scattering is an interaction that can happen when a given particle or wave, like an electron, impacts a target or material. Then the electron changes it's original path and leaves some energy in the process. (This is a really simplified explanation of scattering, this is a really complex phenomenon, but let's not dive into that path)

Particularly, Davisson and Germer used a beam of electrons against a target of nickel, and these scattered electrons were detected by a detector. All of that in a vacuum chamber.

Then the correct answer is a nickel target.

"After de Broglie proposed the wave nature of matter, Davisson and Germer demonstrated the wavelike behavior of electrons by observing an interference pattern from electrons scattering off a nickel target"

N = 52.0 N

Explanation:

Given: [tex]F_a= 20.0\:\text{N}=\:\text{applied\:force}[/tex]

[tex]m=6.00\:\text{kg}[/tex]

[tex]N = \text{normal force}[/tex]

The net force [tex]F_{net}[/tex] is given by

[tex]F_{net} = N + F_a\sin 20 - mg=0[/tex]

Solving for N, we get

[tex]N = mg - F_a\sin 20[/tex]

[tex]\:\:\:\:\:\:= (6.00\:\text{kg})(9.8\:\text{m/s}^2) - (20.0\:\text{N}\sin 20)[/tex]

[tex]\:\:\:\:\:\:= 52.0\:\text{N}[/tex]

Answer:

Look at explanation

Explanation:

I only know 1 way, there is another way you can rephrase this using derivatives but that's pretty much the same thing.

The slope is calculated by Δy/Δx so the slope of distance vs time graph is Δd/Δt which is the velocity

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]

Where:

[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s              

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Where:

[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s

[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]

Hence, the number of revolutions is:

[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]

Therefore, the number of revolutions is 44.6.

I hope it helps you!

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

[tex]W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m[/tex]

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

[tex]W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J[/tex]

So, the new work is more than 130 J.

Answer:

0.5

Explanation:

because the block is attached to the pulley of the string

Answer:

0.113 J

Explanation:

Applying,

w = ke²/2................. Equation 1

Where w = workdone in stretching the spring, k = spring constant, e = extension

make k the subject of the equation

k = 2w/e²................ Equation 2

From the question,

Given: w = 3 J, e = 49-32 = 17 cm = 0.17 m

Substitute these values into equation 2

k = (2×3)/0.17²

k = 6/0.17

k = 35.29 N/m

(a) if the spring from 37 cm to 45 cm,

Then,

w = ke²/2

Given: e = 45-37 = 8 cm = 0.08

w = 35.29(0.08²)/2

w = 0.113 J

Answer:

665.25 Pa

Explanation:

Given data :

Weight of the disk, w = 45 N

Diameter, d = 30 cm

                    = 0.30 m

Therefore, radius of the disk,

[tex]$r=\frac{d}{2}$[/tex]

[tex]$r=\frac{0.30}{2}$[/tex]

   = 0.15 m

Now, area of the cylindrical disk,

[tex]$A=\pi r^2$[/tex]

[tex]$A=3.14 \times (0.15)^2$[/tex]

   [tex]$=0.07065 \ m^2$[/tex]

∴ The gauge pressure at the top of the oil column is :

   [tex]$p=\frac{w}{A}$[/tex]

   [tex]$p=\frac{47}{0.07065}$[/tex]

      = 665.25 Pa

Therefore, the gauge pressure is 665.25 Pa.

The definition of pressure allows to find the result for the pressure at the top of the oil cylinder is:

The pressure is defined by the relationship between perpendicular force and area.

          [tex]P = \frac{F}{A}[/tex]

where P is pressure, F is force, and A is area.

They indicate that the wooden cylinder weighs W = 45.0 N and has a diameter of d = 30 cm = 0.30 m.

The area is:

        A = π r² = [tex]\pi \frac{d^2}{4}[/tex]  

In the attachment we see a diagram of the forces, where the weight of the cylinder and the thrust are equal.

         B-W = 0

          B = W

The force applied to the liquid is the weights of the cylinder. Let's replace.

          [tex]P= \frac{W}{A} \\P = W \frac{4}{\pi d^2 }[/tex]  

Let's calculate.

          [tex]P = \frac{45 \ 4 }{\pi \ 0.30^2 }[/tex] P = 45 4 / pi 0.30²

          P = 636.6 Pa

In conclusion using the definition of pressure we can find the result for the pressure at the top of the oil cylinder is:

Learn more about pressure here: brainly.com/question/17467912

Answer:

The force on the second piston is 5246.5 N .

Explanation:

Area of first piston, a = 3.002 cm^2

Area of second piston, A = 315 cm^2

Force on first piston, f = 50 N

let the force of the second piston is F.

According to the Pascal's law

[tex]\frac{f}{a} = \frac{F}{A}\\\\\frac{50}{3.002}=\frac{F}{315}\\\\F = 5246.5 N[/tex]

Answer:

periodic

Explanation:

Answer:

The height is the same

Explanation:

Because they were at the same height but they fell at different velocities

Answer:

Reverse Osmosis

Explanation:

Reverse osmosis is a type of filtration that involves passing a solvent through a semipermeable membrane in the opposite direction that natural osmosis does. Separation is always enforced through the use of pressure in this process. Ions, fine dust particles, molecules, and larger particles are typically removed from solvents using this method. The technique is particularly popular in the treatment and purification of water.

Answer:

filtration is used to separate

Answer:

  K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

        Em₀ = K = ½ mv²

final point. When it is at tea = 50º

        Em_f = K + U

        Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

        h = L - L cos 50

        Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

        Em₀ = Em_f

         ½ mv² = ½ m v_b² + mg L (1 - cos50)

         K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

          K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

          K_b = 36 +42.0

          K_b = 78 J

Answer:

18

Explanation:

I'm pretty sure I got it right