A dart with mass md is launched toward a block of mass mb that is suspended from a string of length L. The dart is moving horizontally with speed v immediately before it strikes the block and remains embedded. The dart-block system then swings up to a point at which its center of mass reaches a maximum height H above its starting position, as shown at right above. The block’s mass mb is greater than the dart’s mass md.
(a) Indicate which object, the dart or the block, if either, experiences an impulse of larger magnitude during the collision. If the impulse is the same magnitude for both objects, state this explicitly. Briefly explain your reasoning.
(b) If the speed of the dart as it embeds itself into the block is greater than v, how would the maximum height reached by the center of mass of the dart-block system compare to H ? Explain your response without deriving or manipulating equations.

Answer:

the required distance is 89.125 m

Explanation:

Given the data in the question;

we know that, sound intensity B in decibels of sound is;

β(dB) = 10log₁₀( [tex]I[/tex] / [tex]I_0[/tex] )

where intensity [tex]I[/tex] = power / area carried by wave

[tex]I_0[/tex] = 10⁻¹² W/m² { minimum threshold intensity }

Now,

intensity [tex]I[/tex] = power / area carried by wave = P/A = P/4πr² { spherical  }

given that; β = 80.0 dB and P = 10 W

so

β(dB) = 10log₁₀( [tex]I[/tex] / [tex]I_0[/tex] )

we substitute

80 = 10log₁₀( P / 4πr²× [tex]I_0[/tex])

80 = 10log₁₀( 10 / 4πr²× 10⁻¹² )

8 = log₁₀(10) - log₁₀( 4πr²× 10⁻¹² )  

8 = 1 - log₁₀( 4πr²× 10⁻¹² )

8 - 1 = -log₁₀( 4πr²× 10⁻¹² )

7 = -log₁₀( 1.2566 × 10⁻¹¹ × r² )

7 = -[ log₁₀( 1.25 × 10⁻¹¹) + log₁₀( r² ) ]

7 = -[ -10.9 + log₁₀( r² ) ]

7 = 10.9 - log₁₀( r² )

-log₁₀( r² ) = 7 - 10.9

-log₁₀( r² )  = - 3.9

log₁₀( r² ) = 3.9

2log₁₀r = 3.9

log₁₀r  = 3.9 /2

log₁₀r = 1.95

r = 89.125 m

Therefore, the required distance is 89.125 m

Answer:

Option C: Third Class

Explanation:

This is third class because the effort or the input force is in the middle between the fulcrum and the load.

To solve this, we must be knowing each and every concept related to lever.  Therefore,  due to the effort or input force being in the center between the edge as well as the load, the supplied lever is of 3rd class. The correct option is option C.

A lever is a basic device that consists of a beam or stiff rod located at a set tight, or edge. It is used to apply force to a load and often offers a mechanical advantage.

A stiff object that can rotate about a spot on itself is called a lever. The three types of levers are categorized according to where the edge, weight, and effort are located. Due to the effort or input force being in the center between the edge as well as the load, the supplied lever is of  3rd class.

Therefore,  the correct option is option C.

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Answer:

In fission, energy is gained by splitting apart heavy atoms, for example uranium, into smaller atoms such as iodine, caesium, strontium, xenon and barium, to name just a few. However, fusion is combining light atoms, for example two hydrogen isotopes, deuterium and tritium, to form the heavier helium.

Explanation:

I hope this helped you

(Sorry If it didn't)

Answer: hi your question is incomplete attached below is the complete question

answer :

Explanation:

Using the information provided in the question attached I will match each term with its corresponding meaning

Answer: [tex]3\ m/s,\ 20,00\ J[/tex]

Explanation:

Given

Mass of the first car [tex]m_1=5000\ kg[/tex]

Mass of the second car [tex]m_2=5000\ kg[/tex]

The velocity of the first car is [tex]v_1=5\ m/s[/tex]

The velocity of the second car is [tex]v_2=1\ m/s[/tex]

Conserving momentum, take [tex]v_o[/tex] as the velocity after coupling

[tex]\Rightarrow m_1v_1+m_2v_2=\left( m_1+m_2\right)v_o\\\Rightarrow 5000\times 5+5000\times 1=\left( 10,000\right)v_o\\\\\Rightarrow v_o=\dfrac{25,000+5000}{10,000}\\\\\Rightarrow v_o=\dfrac{30,000}{10,000}\\\Rightarrow v_o=3\ m/s[/tex]

[tex]\text{Initial kinetic Energy }K_1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2\\\\\Rightarrow K_1=\frac{1}{2}\times 5000\left( 5^2+1^2\right)\\\\\Rightarrow K_1=65,000\ J\\\\\\\text{Final Kinetic Energy}\ K_2=\frac{1}{2}\left(m_1+m_2\right)v_o^2\\\\\Rightarrow K_2=\frac{1}{2}\times 10,000\times 3^2\\\\\Rightarrow K_2=45,000\ J\\\\\text{Kinetic energy lost is equivalent to change in Initial and final energy i.e.}\\\\\Rightarrow K_1-K_2=65,000-45,000\\\\\Rightarrow K_1-K_2=20,000\ J[/tex]

Answer:

Negative electric charges

a) The speed of the bead at B is 11.79m/s.

b) The change in mechanical energy due to friction as it  moves from B to C is 36.064 J.  

If we drop the object from the shelf or release the spring, that potential energy is converted back into kinetic energy. Kinetic energy can also be transferred from one body to another in a collision, which can be elastic or inelastic.

a)

Potential energy is converted  to kinetic energy

[tex]\frac{1}{2} mv^2=mgh\\\\v=\sqrt{2gh} \\\\v_B=\sqrt{2(9.8)(7.1-0)} \\\\v_B=11.79m/s[/tex]

The speed of the bead at B is 11.79m/s.

b)

[tex]W = mg(h_A) - mg(h_B) = 0.8(9.8)(7.1 - 2.5) = 36.064 J\\\\W=36.064J[/tex]

The change in mechanical energy due to friction as it  moves from B to C is 36.064 J.

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Answer:

D

Explanation:

D ok ok ok ok bye

The correct statement is that the alcohol has lower boiling point than water. The correct option is B.

At the boiling point, the pressure exerted by the surroundings on a liquid equals the pressure exerted by the liquid's vapor, under this condition, adding heat outcomes in the transformation of the liquid into its vapor without raising the temperature.

The boiling point of alcohol is low. Because there is less hydrogen bonding between ethanol molecules than between water molecules.

The ethanol requires less energy to vaporize than water, and water has a higher boiling point than ethanol.

The boiling point of a liquid can be known by utilizing the capillary method, that encompasses placing an inverted capillary in the liquid of interest as well as heating it.

Thus, the correct option is B.

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Answer:

    vₐ = v_c  [tex]( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )[/tex]

Explanation:

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

          Em₀ = K + U = ½ m v_c² - G Mm / R

final point. At a very distant point

         Em_f = U = - G Mm / R₂

energy is conserved

           Em₀ = Em_f

           ½ m v_c² - G Mm / R = - G Mm / R₂

           v_c² = 2 G M (1 /R -  1 /R₂)

if we consider the speed so that it reaches an infinite position R₂ = ∞

           v_c = [tex]\sqrt{\frac{2GM}{R} }[/tex]

now indicates that the mass and radius of the planet changes slightly

            M ’= M + ΔM = M ( [tex]1+ \frac{\Delta M}{M}[/tex] )

            R ’= R + ΔR = R ( [tex]1 + \frac{\Delta R}{R}[/tex] )

we substitute

           vₐ = [tex]\sqrt{\frac{2GM}{R} } \ \frac{\sqrt{1+ \frac{\Delta M}{M} } }{ \sqrt{1+ \frac{ \Delta R}{R} } }[/tex]

let's use a serial expansion

           √(1 ±x) = 1 ± ½ x +…

we substitute

         vₐ = v_ c ( [tex](1 + \frac{1}{2} \frac{\Delta M}{M} ) \ ( 1 - \frac{1}{2} \frac{\Delta R}{R} )[/tex])

we make the product and keep the terms linear

        vₐ = v_c  [tex]( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )[/tex]

The minimum escape speed, vc , for an object launched from the surface of the planet will be    [tex]v_a=v_c(1+\dfrac{1}{2}(\dfrac{\Delta M}{M}-\dfrac{\Delta R}{R})[/tex]

The escape velocity is defined as the velocity required to send the object out of the gravitational influence of the earth.

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

    [tex]E_{mo} = K + U = \dfrac{1}{2} m v_c^2 - \dfrac{G Mm} { R}[/tex]

final point. At a very distant point

      [tex]E_{mf} = U = \dfrac{- G Mm }{ R_2}[/tex]

energy is conserved

         [tex]E{mo} = E{mf}[/tex]

          [tex]\dfrac{1}{2}m v_c^2 - \dfrac{G Mm} {R} = \dfrac{- G Mm }{ R_2}[/tex]

[tex]v_c^2 = 2 G M (\dfrac{1} {R} - \dfrac{ 1 }{R_2})[/tex]

if we consider the speed so that it reaches an infinite position R₂ = ∞

[tex]v_c = \sqrt{\dfrac{2GM}{R}[/tex]

now indicates that the mass and radius of the planet changes slightly

[tex]M ’= M + \Delta M = M(1+\dfrac{\Delta M}{M})[/tex]  

[tex]R ’= R + \Delta R = R (1+\dfrac{\Delta R}{R} )[/tex]

we substitute

   [tex]vₐ = \sqrt{\dfrac{2GM}{R} }\dfrac{\sqrt{1+\dfrac{\Delta M}{M}}} {\sqrt{1+\dfrac{\Delta R}{R}}}[/tex]

let's use a serial expansion

√(1 ±x) = 1 ± ½ x +…

we substitute

      [tex]v_a=v_c(1+\dfrac{1}{2}\dfrac{\Delta M}{M})(1-\dfrac{1}{2}\dfrac\Delta R}{R})[/tex]

Hence the minimum escape speed, vc , for an object launched from the surface of the planet will be    [tex]v_a=v_c(1+\dfrac{1}{2}(\dfrac{\Delta M}{M}-\dfrac{\Delta R}{R})[/tex]

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Answer:

fffcdcffffdxxfxxtdyfycycycycy

Explanation:

It will be B.Its particles have no kinetic energy.

Answer:

O C. Its particles vibrate slightly.

Explanation:

At absolute zero, atoms would occupy the lowest energy state.They move much less than at higher temperatures, but they still have small vibrations at absolute zero.

Answer:

The answer is B its particles have no kinetic energy

Explanation:

trust the big brain

Answer:

mmmm im going to say B but i’m not sure

Explanation:

Electromagnetic waves can travel through air and unlike ocean waves or sound waves they do not need physical molecules for like a medium to travel

Answer: deceleration of [tex]1.904\ rad/s^2[/tex]

Explanation:

Given

Car is traveling at a speed of u=20 m/s

The diameter of the car is d=70 cm

It slows down to rest in 300 m

If the car rolls without slipping, then it must be experiencing pure rolling i.e. [tex]a=\alpha \cdot r[/tex]

Using the equation of motion

[tex]v^2-u^2=2as\\[/tex]

Insert [tex]v=0,u=20,s=300[/tex]

[tex]0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2[/tex]

Write acceleration as [tex]a=\alpha \cdot r[/tex]

[tex]-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2[/tex]

So, the car must be experiencing the deceleration of [tex]1.904\ rad/s^2[/tex].

The angular accelaration of a car traveling at 20.0 m/s on tires with a diameter of 70.0 cm, when car slows down to rest after traveling 300.0 m is 1.91 m/s

Angular acceleration: This can be defined as the rate of change of angular velocity. The s.i unit of angular acceleration is  rad/s²

In other to solve the problem above, we will use the formula for calculating angular acceleration

a = αr.............. Equation 1

Where a = acceleration of the car's tires, α = angular acceleration of the car's tires, r = radius of the car's tires

Therefore,

α = a/r................. Equation 2

But, we need to calculate a using the equations of motion

v² = u²+2as................ Equation 3

Where v = final velocity = 0 m/s (rest), u = initial velocity= 20m/s, s = distance = 300 m

Substitute these values into equation 3

0² = 20²+2(300²)a

-400 = 600a

a = -400/600

a = -0.67 m/s²

going back to equation 2,

Given: r = 70/2 = 35 cm = 0.35 m

Substitute this values into  equation 2,

α = -0.67/0.35

α = -1.91 rad/s²

Hence, the magnitude of the angular acceleration is 1.91 rad/s²

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Answer:

[tex]\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}[/tex]

Explanation:

We need to find the acceleration of the 2 kilogram object. Let's complete this in 2 steps.

First, we can find the force of the first, 8 kilogram object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

[tex]F=m \times a[/tex]

The mass of the object is 8 kilograms and the acceleration is 1.6 meters per square second.

Substitute these values into the formula.

[tex]F= 8 \ kg * 1.6 \ m/s^2[/tex]

Multiply.

[tex]F= 12.8 \ kg*m/s^2[/tex]

Now,  use the force we just calculated to complete the second part of the problem. We use the same formula:

[tex]F= m \times a[/tex]

This time, we know the force is 12.8 kilograms meters per square second and the mass is 2 kilograms.

Substitute the values into the formula.

[tex]12.8 \ kg*m/s^2= 2 \ kg *a[/tex]

Since we are solving for the acceleration, we must isolate the variable (a). It is being multiplied by 2 kg. The inverse of multiplication is division. Divide both sides of the equation by 2 kg.

[tex]\frac {12.8 \ kg*m/s^2}{2 \ kg}= \frac{2\ kg* a}{2 \ kg}[/tex]

[tex]\frac {12.8 \ kg*m/s^2}{2 \ kg}=a[/tex]

The units of kilograms cancel.

[tex]\frac {12.8}{2}\ m/s^2=a[/tex]

[tex]6.4 \ m/s^2=a[/tex]

The acceleration is 6.4 meters per square second.

Answer:

Wavelength?

Explanation:

Answer:

Decreases

Explanation:

I think this is right tell me if it's wrong.

Answer:

Increases.

Explanation:

The density of the second medium increases, the angle of refraction also increases because of higher density of the medium. In more denser medium, there is more resistance for the light ray so it bends more as compared to less denser medium in which less bending of light ray occur. The angle of refraction increases when the density of medium is more than the previous medium so we can say that the angle of refraction increases due to increase in density of another medium.

Answer:

34cm

Explanation:

Explanation:

Extrusive rocks may have a few grains that are large enough to see, but most of them will be too small to see individual minerals. ... The individual mineral grains are almost too small to see. Some extrusive rocks cool so quickly that they do not form any grains. Instead, they form a natural glass.

Answer:

A&B

Explanation:

Answer:

Chemical to mechanical(b) and Chemical to Thermal(a)

Answer:

It can travel in various wavelengths. brainliest?

Explanation:

The evidence that reveals about the light is that  It can travel in various wavelengths.

It is the wave property that should be blending of the wave related to the corners.

At the time when the wave is bending so it passed via the openings.

Also, the light diffracted via the small slits and it should be shown to arise in the water waves and the sound waves.

hence, the first option is correct.

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Answer: shorter

Explanation:

Answer:

c

Explanation:

the gas pedal gives the engine the fuel it needs to give the car the power to go

Answer:

pipes

Explanation:

Answer:

force on the electron is upwards

Explanation:

The magnetic force is given by the relation

        F = q v x B

the bold indicates vectors, we can write the module of this expression

        F = q v  B sin θ

the direction of the force is given by the right hand rule.

If we have a positive charge, the flea points in the direction of velocity, in this case towards the inside of the page.

fingers extended in the direction of the magnetite field, in our case to the right and

the palm gives the direction of the force for a positive charge, for a negative charge it is in the opposite direction. The palm points downwards, so the force on the electron is upwards

Answer:

The same current flows through each resistor in series. Individual resistors in series do not get the total source voltage, but divide it. The total resistance in a series circuit is equal to the sum of the individual resistances: RN(series)=R1+R2+R3+…

Answer:

one beta particle

Explanation:

Since the atomic number of 131_I_53 increases by one to 131_Xe_54, we have the release of a beta particle since only a beta particle can cause a change in atomic number without a change in mass number. Since a beta particle is one electron, ₋₁⁰e, our nuclear equation is

¹³¹I₅₃ → ¹³¹X₅₄ + ₋₁⁰e

The atomic and mass numbers on both sides are balanced

mass number 131 = 131 + 0 = 131

atomic number 53 = 54 - 1 = 53

Thus a beta particle is emitted