An organ pipe is 1.6 m long and open at both ends. What are the first three harmonic frequencies of this pipe?

Answer:

1920Joules

Explanation:

The formula for calculating the kinetic energy of a body is expressed as;

KE = 1/2 mv²

m isthe mass

V is the speed

For the two masses, the combined KE is expressed as;

KE  = 1/2(m1+m2)v²

KE = 1/2(45+15)(8)²

KE  = 1/2 * 60 * 64

KE  = 30 * 64

KE  = 1920J

Hence the combined kinetic energy of the boy and the bicycle is 1920Joules

The combined kinetic energy of the boy and the bicycle is of 1920 J.

Given data:

The mass of boy is, m = 45.0 kg.

The mass of bicycle is, M = 15.0 kg.

The speed of bicycle is, v = 8.00 m/s.

The kinetic energy of an object is defined as the energy possessed by an object by virtue of motion of object. The combined kinetic energy of the boy-bicycle system is given as,

[tex]KE = \dfrac{1}{2}(m+M)v^{2}[/tex]

Solve by substituting the values as,

[tex]KE = \dfrac{1}{2}(45+15) \times 8^{2}\\\\KE = 1920 \;\rm J[/tex]

Thus, we can conclude that the combined kinetic energy of the boy and the bicycle is of 1920 J.

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Answer:

Explanation:

because it is godhood

Answer:

Following are the responses to this question:

Explanation:

The small current passes thru the capacitor of the strain gauge and the current is generated throughout the resistor. For the very first time,  in contrast to what we calculate, its resistance of the multimeter is quite high and therefore the small stream flowing through the bulb would have very little impact on the measure. Thus, as the current flows through the flashbulb, this same calculation is of excellent price, its material is heated and resistance varies with increase. Therefore, when the bulb will be on, sensitivity is greater.

Answer:

They will fall at the same time.  This is because gravity accelerates all objects at the same speed, Earth's gravity being approximately 9.8m/s²

They'll both fall at the same time. This is because gravity accelerates everything at the same rate, with Earth's gravity being approximately 9.8 m/s2.

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Answer:

Absorbed sunlight is balanced by heat radiated from Earth's surface and atmosphere. ... The atmosphere radiates heat equivalent to 59 percent of incoming sunlight; the surface radiates only 12 percent. In other words, most solar heating happens at the surface, while most radiative cooling happens in the atmosphere

Answer:

-75 cm

Explanation:

At l ; F = 350 Hz

At l + 15 cm ; F = 280 Hz

I = 350

I + 15 = 280

280I = 350(I + 15)

280I = 350I + 5250

280I - 350I = 5250

-70I = 5250

I = - 75cm

The length is - 75 cm

Answer: Closer to the Roche limit, the body is deformed by tidal forces. Within the Roche limit, the mass' own gravity can no longer withstand the tidal forces, and the body disintegrates. hope this helps can u give me brainliest

Explanation:

Answer:

he times are getting closer as we use each filter, in the expression n would be the number of filters  

t = [tex]\sqrt{ \frac{2(y_o - (n-1) h)}{g} }[/tex]

Explanation:

For this exercise let's use the kinematics relations

          y = y₀ + v₀ t - ½ g t²

When the first filter reaches the ground, its height is y = 0, as they release its initial velocity is zero

for the 1st filter

          0 = y₀ - ½ g t²

          t² = 2y₀ / g

          t = [tex]\sqrt{ \frac{2y_o}{g} }[/tex]

when we release the second filter upon arrival it has a height y = h where h is the height of each filter

         h = y₀ - ½ g t²

         t = [tex]\sqrt{ \frac{2(y_o- h)}{g} }[/tex]

when we release the third filter it reaches y = 2h

         2h = y₀ - ½ g t²

          t = [tex]\sqrt{ \frac{2(y_o -2h)}{g} }[/tex]

we can write the terms of this succession

          (n-1) = y₀ - [tex]\frac{1}{2}[/tex] g t²

           t = [tex]\sqrt{ \frac{2(y_o - (n-1) h)}{g} }[/tex]

therefore we see that the times are getting closer as we use each filter, in the expression n would be the number of filters

Answer:

water vapor

Explanation:

did assignment on edge

Answer:

Ocean-floor sediments can also be used to determine past climate. ... Ice cores contain more 16O than ocean water, so ice cores have a lower 18O/ 16O ratio than ocean water or ocean-floor sediments. Water containing the lighter isotope 16O evaporates more readily than 18O in the warmer subtropical regions

Explanation:

Solution :

The volume of the rigid cylinder = [tex]$100 \ cm^3 = 100 \times 10^{-6} \ m^3$[/tex]

Initial pressure inside the cylinder, [tex]$P_i = 800 \ kPa$[/tex]

Initial temperature inside the cylinder, [tex]$T_i = 35^\circ C= 308 \ K$[/tex]

Final temperature inside the cylinder, [tex]$T_f = 5^\circ C= 278 \ K$[/tex]

Final pressure inside the cylinder, [tex]$P_f = 100 \ kPa$[/tex]

Area of the hole, A = [tex]$0.3 \ mm^2 = 3 \times 10^{-7} \ m^2$[/tex]

Velocity of the air through the hole, V = 100 m/s

The final pressure and the temperature inside the cylinder will be the condition same as the ambient conditions.

At initial state, from the equation of state,

PV = mRT,    where R = 287 J/kg-K for air

[tex]$800 \times 10^3 \times100 \times10^{-6} = m_1 \times 287 \times 308$[/tex]

∴  [tex]$m_1=9.1 \times 10^{-4} \ kg$[/tex]

Since the exit condition does not change with time, we have ,

At ambient condition, [tex]$P_f = 100 \kPa$[/tex] and [tex]$T_f= 278 \ K$[/tex].

Therefore, we can find the density of the air

[tex]$P=\rho R T$[/tex]

[tex]$\rho = \frac{P}{RT}$[/tex]

  [tex]$=\frac{100 \times 10^3}{287 \times 278}$[/tex]

 [tex]$= 1.25 \ kg/m^3$[/tex]

Mass flow rate of air from the cylinder = [tex]$\dot m$[/tex]

[tex]$\dot m$[/tex] can be written as [tex]$\dot m$[/tex] [tex]$=\rho_{f} \times A \times v$[/tex]

[tex]$\dot m$[/tex] = [tex]$1.25 \times 3 \times 10^{-7} \times 100$[/tex]

[tex]$\dot m$[/tex] = [tex]$3.75 \times 10^{-5}$[/tex] kg/s

Mass escaped from the cylinder in 5 seconds

[tex]$m=3.75 \times 10^{-5} \times 5$[/tex]

   [tex]$= 1.875 \times 10^{-4} \ kg$[/tex]

Mass of air remaining in the cylinder after 5 seconds :

[tex]$m_2 = m_1 - m$[/tex]

[tex]$m_2 = 9.1 \times 10^{-4} - 1.875 \times 10^{-4}$[/tex]

[tex]$m_2 = 7.225 \times 10^{-4} \ kg$[/tex]

    = 0.7225 grams

Answer:

1.B, 2.A

Explanation:

Answer:

Break down small pebbles and sediments, like sand

Break down large rocks like mountains

Explanation:

Answer:

Mass = 18 kg

Explanation:

Formula for force in centripetal motion is;

F = mv²/r

We have;

Mass; m.

Speed; v = 3 m/s

radius; r = 1.8 m

Force; F = 90 N

Thus;

Making m the subject;

m = Fr/v²

m = 90 × 1.8/3²

m = 18 kg

The force applied on the block of  mass 4 kg is 17 N and a frictional force of 5 N is opposing this force. Hence, the net force is 8N and the acceleration is 2 m/s².

Force is a vector quantity thus, characterized by a magnitude and direction.   If two same or different forces are acting on a body from the same direction, the net force will be the sum of the magnitude of the two forces.

If two forces are acting from opposite directions, they will cancel each other and the net force is the substracted value of their magnitudes.

Given that, mass of the block = 4 kg

force applied on the block = 17 N

frictional force  = 5 N

Net force =  17 - 5 = 8 N.

Force = mass × acceleration. (by Newton's second law of motion)

acceleration = force/ mass

                     = 8 N / 4 Kg = 2 m/s²

Therefore, the acceleration of the block is 2 m/s².

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Answer:

Vf = 69.61 m/s

Explanation:

We will use the third equation of motion to solve this problem:

[tex]2gh = V_{f}^2 - V_{i}^2\\[/tex]

where,

g = acceleration due to gravity = 9.81 m/s²

h = height of cliff = 247 m

Vf = final velocity = ?

Vi = initial velocity = 0 m/s (boulder breaks loose from rest)

Therefore,

[tex](2)(9.81\ m/s^2)(247\ m) = V_{f}^2 - (0\ m/s)^2\\V_{f} = \sqrt{4846.14\ m^2/s^2}\\[/tex]

Vf = 69.61 m/s

Answer: 1.135 L/s; 1.35 kg/s, 22.57 m/s

Explanation:

Given

Volume of bucket [tex]V=15\ gal\approx 56.78\ L[/tex]

time to fill it [tex]t=50\ s[/tex]

Volume flow rate

[tex]\dot{V}=\dfrac{56.78}{50}=1.135\ L/s\approx 1.135\times 10^{-3}\ m^3/s[/tex]

The inner diameter of the hose [tex]D=1.5\ cm[/tex]

diameter of the nozzle exit [tex]d=0.8\ cm[/tex]

we can volume flow rate as

[tex]\Rightarrow \dot{V}=Av\quad \quad \text{v=average velocity through nozzle exit}\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}d^2\times v\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}(0.8\times 10^{-2})^2\times v\\\\\Rightarrow v=\dfrac{4\times 1.135\times 10^{-3}}{\pi \times 64\times 10^{-6}}=22.57\ m/s[/tex]

Mass flow rate

[tex]\Rightarrow \dot{m}=\rho \times \dot{V}\\\Rightarrow \dot{m}=1\ kg/L\times 1.135\ L/s=1.35\ kg/s[/tex]

Answer:

C. Nucleic Acids.

Explanation:

Nucleic acids are thus uniquely capable of directing their own self-replication, allowing them to function as the fundamental informational molecules of the cell. The information carried by DNA and RNA directs the synthesis of specific proteins, which control most cellular activities.

Did some simple research, if that's fine and I hope this is correct as it says on the sites. Have a good one :)

Answer:

[tex]\boxed{\text{\sf \Large 2.0 m/s^2 $}}[/tex]

Explanation:

Use acceleration formula

[tex]\displaystyle \text{$ \sf acceleration=\frac{change \ in\ velocity}{change \ in \ time} $}[/tex]

[tex]\displaystyle a=\frac{65-47}{12.0-3.0} =2.0[/tex]

Answer:

A) Because light is due to the electromagnetic fields (fields moving at the speed of light)

Note: Electrical energy is also used to describe the energy usage

in RLC circuits

I think the answer would be A.

After all, it is 0 which is technically a dead center number meaning that the net should be balanced and still.

Hope this helps and have a nice day.

-R3TR0 Z3R0

Answer:2m/s^2

Explanation:

a=f/m

Answer:

The correct answer is:

(a) 0

(b) 0.5 m/s

(c) 7740 N

(d) 0

Explanation:

The given values are:

mass,

m = 3000 kg

Tension,

T = 7,200 N

Angle,

= 30°

(a)

Even as the block speed becomes unchanged, the kinetic energy (KE) will adjust as well:

⇒ [tex]\Delta K =0[/tex]

By using the theorem of energy, the net work done will be:

⇒  [tex]\Delta K =0[/tex]

(b)

According to the question, After 0.25 m the block is moving with the constant speed

= 0.5 m/s.

(c)

The given kinetic friction coefficient is:

u = 0.3

The friction force will be:

= [tex]u(mg-Tsin30^{\circ})[/tex]  

On substituting the values, we get,

= [tex]0.3[(3000\times 9.8)-(7200\times 0.5)][/tex]

= [tex]0.3[29400-3600][/tex]

= [tex]0.3\times 25800[/tex]

= [tex]7,740 \ N[/tex]

(d)

On including the friction,

The net work will be:

⇒ [tex]\Delta K=0[/tex]

Answer:

M = 0.31 kg

Explanation:

This exercise must be done in parts, let's start by finding the speed of the set arrow plus apple, for this we define a system formed by the arrow and the apple, therefore the forces during the collision are internal and the moment is conserved

let's use m for the mass of the arrow with velocity v₁ = 20.4 m / s and M for the mass of the apple

initial instant. Just before the crash

          p₀ = m v₁ + M 0

instant fianl. Right after the crash

          p_f = (m + M) v

          p₀ = p_f

          m v₁ = (m + M) v

          v =[tex]\frac{m}{m+M} \ v_1[/tex]                          (1)

now we can work the arrow plus apple set when it leaves the child's head with horizontal speed and reaches the floor at x = 8 m. We can use kinematics to find the velocity of the set

          x = v t

          y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

when it reaches the ground, its height is y = 0 and as it comes out horizontally, [tex]v_{oy} = 0[/tex]

          0 = h - ½ g t²

          t² = 2h / g

For the solution of the exercise, the height of the child must be known, suppose that h = 1 m

            t = [tex]\sqrt{ \frac{ 2 \ 1}{9.8} }[/tex]

            t = 0.452 s

let's find the initial velocity

             v = v / t

             v = 8 / 0.452

             v = 17.7 m / s

From equation 1

              v = m / (m + M) v₁

              m + M = [tex]m \ \frac{v_1}{v}[/tex]

              M = m + m \  \frac{v_1}{v}

we calculate

              M = 0.144 + 0.144  [tex]\frac{20.4}{17.7}[/tex]

              M = 0.31 kg

Answer:

Focal length(f)= -15 cm, magnification= -3 (image is real). So, -v/u=-3 ; or v=3u {v=image distance and u=object distance}. Using mirror formula,

1/f= 1/v + 1/u

-1/15 = 1/3u + 1/u

4/3u = -1/15

u= -20 cm

v =3u= 3×(-20)= -60 cm

So object is 20 cm and image is 60 cm in front of the mirror.

Explanation: