A mass undergoes SHM with amplitude of 4 cm. The energy is 8.0 J at this time. The mass is cut in half, and the system is again set in motion with amplitude 4.0 cm. What is the energy of the system now?

Hi there!

We can use the following equation to find the frequency of each harmonic:

[tex]f_n = \frac{n}{2L} \sqrt{\frac{T}{\lambda}}[/tex]

n = nth harmonic

L = length of string (m)

T = Tension of string (N)

λ = linear density (kg/m)

Begin by converting the linear mass density to kg:

2.00g /m · 1 kg / 1000g = 0.002 kg/m

Now, we can use the equation to find the first three harmonics.

First harmonic:

[tex]f_1 = \frac{1}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{131.76 Hz}[/tex]

Second harmonic:

[tex]f_2 = \frac{2}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{263.52Hz}[/tex]

Third harmonic:

[tex]f_3 = \frac{3}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{395.28Hz}[/tex]

Answer:

Approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex], assuming that no external force (e.g., gravitational pull) was acting on this rocket.

Explanation:

Assume that no external force is acting on this rocket. The system of the rocket and the fuel on the rocket would be isolated (an isolated system.) The momentum within this system would be conserved.

Let [tex]v_{0}\; \rm m\cdot s^{-1}[/tex] be the initial velocity of the rocket.

The velocity of the exhaust gas would be [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] since the gas is ejected away from the rocket.

Let [tex]\Delta v\; \rm m\cdot s^{-1}[/tex] denote the increase in the velocity of the rocket. The velocity of the rocket after ejecting the gas would be [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex].

The momentum [tex]p[/tex] of an object of velocity [tex]v[/tex] and mass [tex]m[/tex] is [tex]p = m \cdot v[/tex].

The combined mass of the rocket and the fuel was [tex]10000\; \rm kg[/tex]. The initial momentum of this rocket-fuel system would be:

[tex]\begin{aligned}p_{0} &= m \cdot v\\ &= 10000\; {\rm kg} \times v_{0}\; {\rm m \cdot s^{-1}} \\ &= (10000\; v_{0})\; \rm {kg \cdot m\cdot s^{-1}}\end{aligned}[/tex].

The momentum of the [tex]5.0\; \rm kg[/tex] of fuel ejected at [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] would be:

[tex]\begin{aligned} & 5.0 \; {\rm kg} \times (v_{0} - 5000)\; {\rm m\cdot s^{-1}}\\ =\; & (5.0\, v_{0} - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

After ejecting the [tex]5.0\; \rm kg[/tex] of the fuel, the mass of the rocket would be [tex]10000\; \rm kg - 5.0\; \rm kg = 9995\; \rm kg[/tex]. At a velocity of [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex], the momentum of the rocket would be:

[tex]\begin{aligned} & 9995 \; {\rm kg} \times (v_{0} + \Delta v)\; {\rm m\cdot s^{-1}}\\ =\; & (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

Take the sum of these two quantities to find the momentum of the rocket-fuel system after the fuel was ejected:

[tex]\begin{aligned}p_{1} &= (5.0\, v_{0} - 25000)\; {\rm kg \cdot m\cdot s^{-1}\\ &\quad\quad + (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}} \\ &= (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

The momentum of the rocket-fuel system would be conserved. Thus [tex]p_{0} = p_{1}[/tex].

[tex](10000\, v_{0})\; {\rm kg \cdot m\cdot s^{-1}} = (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Solve this equation for [tex]\Delta v[/tex], the increase in the velocity of the rocket.

[tex]10000\, v_{0} = 10000\, v_{0} + 9995\, \Delta v - 25000[/tex].

[tex]9995\, \Delta v = 25000[/tex].

[tex]\begin{aligned}\Delta v &= \frac{25000}{9995} \approx 2.5\end{aligned}[/tex].

Thus, the velocity of the rocket would increase by approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex] after ejecting the [tex]5.0\; \rm kg[/tex] of fuel.

The mechanical advantage of ramp A is greater than others and it will require the least force to move the load to greater distance.

The mechanical advantage of the ramp is calculated as follows;

[tex]M.A = \frac{L}{h}[/tex]

The mechanical advantage of the ramp B, C, D and E is calculated as;

[tex]M.A = \frac{L}{h} \\\\[/tex]

The mechanical advantage of the ramp A is calculated as follows;

[tex]M.A = \frac{2L}{h} \\\\M.A = 2(\frac{L}{h} )[/tex]

Since the length of the ramp A is greater than other ramps, the mechanical advantage will be greater and it will require the least force to move the load to greater distance.

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20 seconds

Explanation:

Let [tex]x_a[/tex] be the distance traveled by the accelerating car and [tex]x_c[/tex] be the distance traveled by the car moving with a constant velocity. When they cover the same distance, we can write

[tex]x_a = x_c \Rightarrow v_{0a}t + \frac{1}{2}at^2 = v_ct[/tex]

where [tex]v_c[/tex] is the velocity of car moving at a constant rate and a is the acceleration of the accelerating car. Since the accelerating car started from rest, then [tex]v_{0a}[/tex] is zero so our equation above simplifies to

[tex]\frac{1}{2}at^2 = v_ct[/tex]

Note that the variable t cancels out so solving for t, we get

[tex]\frac{1}{2}at = v_c \Rightarrow t = \dfrac{2v_c}{a}[/tex]

Plugging in the given values,

[tex]t = \dfrac{2(20\:\text{m/s})}{2\:\text{m/s}^2} = 20\:\text{s}[/tex]

Answer:

Explanation:

Plane 2 is moving north at

8.5sin20 = 2.9 m/s

Plane 2 is moving west at

8.5cos20 = 8.0 m/s

Part A

v = √((13 - 2.9)² + 8.0²) = 12.876... 13 m/s

Part B

θ = arctan((13 - 2.9) / 8.0) = 51.617... 52° N of E

Part C

13 m/s  52° S of W

relative velocity magnitude is independent of reference frame

Answer:

Explanation:

The answer:

https://www.chegg.com/homework-help/questions-and-answers/race-car-starting-rest-travels-around-circular-turn-radius-247-m-certain-instant-car-still-q402991

Total acceleration of car is 148.31 m/s².

Centripetal acceleration is a characteristic of an object's motion along a circular path. Centripetal acceleration applies to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.

Given parameters:

Radius of the circular path: r = 22.5 m.

Angular speed: ω = 0.541 rad/s.

So, centripetal acceleration; α = ω²r = (0.541)² × 22.5 m/s² = 6.58 m/s².

Tangential acceleration: [tex]\alpha_t[/tex] = αr = 6.58 × 22.5 = 148.16 m/s².

Hence, total acceleration = √(α² + [tex]\alpha_t[/tex]²) = √(6.58² +148.16²) = 148.31 m/s².

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#SPJ2

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

[tex]P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}[/tex]

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi[tex](\frac{6894.76\ Pa}{1\ psi})[/tex] = 379212 Pa

[tex]\rho[/tex] = density of water = 1000 kg/m³

Therefore,

[tex]v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}[/tex]

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

[tex]\frac{V}{t} = Av\\\\t =\frac{V}{Av}[/tex]

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = [tex]\frac{\pi (0.015875\ m)^2}{4}[/tex] = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = [tex][\frac{\pi (9.144\ m)^2}{4}][1.524\ m][/tex] = 100.1 m³

Therefore,

[tex]t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\[/tex]

t = 18353.5 s = 305.9 min = 5.1 hours

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-cant travel through space since there's no molecules to travel through

-sound travels 4.3 times faster in water than air

-sounds are waves that pass through our ears via vibrations and travel by vibrations of molecules

-different types of sound like audible, inaudible, infrasonic, ultrasonic,

-sounds waves are either longitudinal, mechanical and pressure waves

-sound travels at 767 miles per hour

Answer:

Explanation:

Δx means a change in the magnitude of the x variable, often used in reference to a number line on the horizontal axis of a plot.

Answer:

TRUE

FALSE

TRUE

TRUE

TRUE

TRUE

Explanation:

1. Most EM waves has the same speed. True

2. EM waves travel at the speed of 4x108 m/s. False

3. Electromagnetic waves are transverse waves consisting of changing electric fields and changing magnetic fields. True

4. Electromagnetic waves transfer energy through a vacuum. True

5. A wave is a disturbance that transfers energy. True

Explanation:

Fgravity = G*(mass1*mass2)/D²

so, if you double one of the masses, what does that do to our equation ?

Fgravitynew = G*(2*mass1*mass2)/D²

due to the commutative property of multiplication

Fgravitynew = 2* G*(mass1*mass2)/D² = 2* Fgravity

so, the correct answer will be 2×45 = 90 units.

In 60 minutes or 3600 seconds, the tip of the minute hand traverses the circumference of a circle with radius 3.00 cm, so it moves with a tangential speed of

(3.00 cm)/(3600 s) ≈ 0.00083 cm/s = 8.3 μm/s

At the highest attitude, the  velocity of  the ball is 0 m/s, so the kinetic energy is 0 as well.

Hence the answer is potential energy because it doesn't depend on velocity .

Explanation:

These prefixes are very commonly used in naming chemical compounds.

Di- means two.

For example, carbon dioxide's formula is be written as [tex]\text{CO}_2,[/tex] and it has 2 oxygen atoms, hence "di-oxide."

Tetra - means four.

For example, carbon tetrachloride's chemical formula is written as [tex]\text{CCl}_4[/tex], and there are four chlorine atoms

Deca- means ten

For example, lanthanum decahydride's chemical formula is written as [tex]\text{LaH}_{10}.[/tex] In this case there are 10 hydrogen atoms for every lanthanum atom.

Hepta - means seven

For example, iodine heptafluoride is written as [tex]\text{IF}_7[/tex]. Note the seven fluorine atoms attached to the iodine atom, hence the name "hepta-fluoride."

Answer:

200 feet

Explanation:

Answer:

im here\

Explanation:

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

According to above question ~

Let's find the charge (q) by using formula ~

Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds

Answer:

C

Explanation:

Definition of drug: a medicine or other substance which has a physiological effect when ingested or otherwise introduced into the body

Answer:

The data suggest that the true mean HAZ depth is larger when the current setting is higher.

Answer:

Small particles of dust and pollution in the air can contribute to (and sometimes even enhance) these colors, but the primary cause of a blue sky and orange/red sunsets or sunrises is scattering by the gas molecules that make up our atmosphere. Large particles of pollution or dust scatter light in a way that changes much less for different colors.

Explanation:

a.

The components of the force are Fx = 2.2163 N, Fy = -0.5445 N and Fz = 0 N

The force on a current carrying conductor in a magnetic field is given by F = iL × B where i = current = 9.00 A, L = 25.0 cmk = 0.25 mk (since the conductor is along the z-direction). B = magnetic field. Since B has component Bx = -0.242T, By= -0.985, and Bz = -0.336, B = -0.242i + (-0.985j) + (-0.336)k = -0.242i - 0.985j - 0.336)k.

So, F = iL × B

F = 9.00 A{(0.25 m)k × [-0.242Ti + (-0.985Tj) + (-0.336T)k]T}

F = 9.00 A{(0.25 m)k × (-0.242T)i + (0.25 m)k × (-0.985Tj) + (0.25 m)k × (-0.336T)k]}

F = 9.00 A{-0.0605mT)k × i + (-0.24625 mT)k × j + (-0.084 m)k × k]}

F = 9.00 A{-0.0605mT)j + (-0.24625 mT) × -i + (-0.084 mT) × 0]}

F = 9.00 A{-0.0605mT)j + (0.24625 mT)i + 0 mT]}

F = -0.5445 AmT)j + (2.21625 AmT)i + 0 AmT]}

F = -0.5445j + 2.21625i + 0 k

F = (2.2163i - 0.5445j + 0 k) N

So, the components of the force are Fx = 2.2163 N, Fy = -0.5445 N and Fz = 0 N

b.

The magnitude of the net force on the wire is 2.282 N

The net force F = √(Fx² + Fy² + Fz²)

F = √[(2.2163 N)² + (-0.5445 N)² + (0 N)²)

F = √[(4.912 N)² + 0.2964 N)² + (0 N)²)

F = √[5.2084 N)²

F = 2.2822 N

F ≅ 2.282 N

So, the magnitude of the net force on the wire is 2.282 N

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Answer:

The 7 states of matter are solid, loquid, gas, fermionoc condensate, quark gluton plasm, bose einetein condensate amd ionised plasm but its usually only 3 they teach you

Answer:

7

Explanation:

solid, liquid,gas,fermionoc condensate,quark glutton plasm,bose einetein condensate amd ionised plasm.

Km values are standardized because half the Vmax (Vmax/2) is more informative than Vmax. This value (Km) can be used to calculate the affinity of the enzyme by a given substrate.

The Km (Michaelis constant) of the enzyme refers to the value in which the concentration of substrate is equal to half of its maximum velocity (Vmax/2).

This value (Km) is inversely proportional to the affinity of an enzyme by a given substrate.

An enzyme showing a high Km also exhibits a low affinity for its specific substrate, and thereby this enzyme requires a high concentration of the substrate to reach its maximum velocity (Vmax).

In consequence, the Km value is a more informative value than the maximum velocity (Vmax), which only indicates the concentration of an enzyme catalyzing a reaction under ideal conditions.

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Answer:

When you are wet or you are sweating current passes easily.

Explanation:

wetness increases conductivity. therefore 'when it is wet'.

Answer:

40 volts

Explanation:

Use the equation [tex]V=IR[/tex]

[tex]V=(4)(10)[/tex]

[tex]V=40[/tex]

Answer:

Hey mate.....

Explanation:

This is ur answer.....

Jelly is a colloid where colloidal particles are solids which are dispersed in the liquid. Jelly is a fluid which can not be considered as a particular one kind of state. It is in the category of 'gel' which is a colloidal form.

Hope it helps!

Brainliest pls!

Follow me! ;)

Explanation:

The orbiting period of a satellite at a height h from earth' surface is

T=2πr32gR2

where r=R+h.

Then, T=2π(R+h)R(R+hg)−−−−−−−−√

Here, R=6400km,h=1600km=R/4

T=2πR+R4−−−−−−√R(R+R4g)−−−−−−−−−⎷=2π(1.25)32Rg−−√

Putting the given values,

T=2×3.14×(6.4×106m9.8ms−2)−−−−−−−−−−−−√(1.25)32=7092s=1.97h

Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution around the earthh is equal to the period of revolution of the earth up around its own axis whichh is 24h. Let us find the height h of such a satellite above the earth's suface in terms of the earth,'s radius.

Let it be nR.Then

T=2π(R+nR)R(R+nRg)−−−−−−−−−−√

=2π(Rg)−−−−−√(1+n)32

=2×3.14(6.4×106m/s9.8m/s2)−−−−−−−−−−−−−−−⎷(1+n)32

(5075s)(1+n)32=(1.41h)(1+n)32

For T=24h, we have (24h)=(1.41h)(1+n)32

or (1+n)32=241.41=17

or 1+n(17)23=6.61

or n=5.61

The height of the geostationary satellite above the earth's surface is nR=5.61×6400km=6.59×104km.