A tennis ball bouncing on a hard surface compresses and then rebounds. The details of the rebound are specified in tennis regulations. Tennis balls, to be acceptable for tournament play, must have a mass of 57.5 g. When dropped from a height of 2.5 m onto a concrete surface, a ball must rebound to a height of 1.4 m. During impact, the ball compresses by approximately 6 mm.
How fast is the ball moving when it hits the concrete surface? (Ignore air resistance.)

Answer:

121 Joules

6.16717 m

Explanation:

m = Mass of the rocket = 2 kg

k = Spring constant = 800 N/m

x = Compression of spring = 0.55 m

Here, the kinetic energy of the spring and rocket will balance each other

[tex]\frac{1}{2}mu^2=\frac{1}{2}kx^2\\\Rightarrow u=\sqrt{\frac{kx^2}{m}}\\\Rightarrow u=\sqrt{\frac{800\times 0.55^2}{2}}\\\Rightarrow u=11\ m/s[/tex]

The initial velocity of the rocket is 11 m/s = u.

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² = g

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11^2}{2\times -9.81}\\\Rightarrow s=6.16717\ m[/tex]

The maximum height of the rocket will be 6.16717 m

Potential energy is given by

[tex]P=mgh\\\Rightarrow P=2\times 9.81\times \frac{0^2-11^2}{2\times -9.81}\\\Rightarrow P=121\ J[/tex]

The potential energy of the rocket at the maximum height will be 121 Joules

Answer: 2/3 of the total volume

Explanation:

See attachment for details

Answer:

(B) be extremely unlikely and therefore violate the 2nd law of thermodynamics.

Explanation:

This process is highly unlikely and if happens would violet the 2nd law of thermodynamics.

The laws of motion do not rule out a move to one side of the room of all the air molecules. But it would be extremely unlikely to happen that. The second law of thermodynamics states that the universe responds to circumstances that become more and more likely, not improbable.

Answer:

The micturition reflex can be voluntarily controlled by the relaxation of the external urethral sphincter.

The micturition reflex, or the process of urination, can be voluntarily controlled by the external urethral sphincter and the sacral micturition center. The external urethral sphincter is a skeletal muscle that we can consciously control, while the sacral micturition center is a group of neurons that normally act reflexively unless allowed for voluntary control by the brain. As the bladder fills, signals are sent through the sacral pelvic nerves which activate parasympathetic neurons to proceed with urination.

The micturition reflex, which is another term for urination or voiding, can be voluntarily controlled by the external urethral sphincter and the sacral micturition center. The external urethral sphincter is a skeletal muscle that can be consciously controlled to maintain urinary continence, while the sacral micturition center, a group of neurons located in the sacral region of the spinal cord, acts reflexively unless its action is modified by higher brain centers for voluntary urination. In response to a filled bladder, these centers trigger the relaxation of both the internal and external urethral sphincters, the contraction of the detrusor muscle, and inhibit the somatic motor neurons for urination to occur.

Children learn to voluntarily control the urination process as they mature, thereby overriding the micturition reflex and delay voiding, a process known as potty training. Voluntary micturition requires (1) an intact spinal cord and (2) a functional pudendal nerve arising from the sacral micturition center. The sacral pelvic nerves play a crucial role in bladder control. Upon receiving signals of bladder stretch, they activate the parasympathetic neurons to release acetylcholine, which triggers detrusor muscle contraction and bladder emptying.

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The interference principle is responsible for alternating light and dark bands when light passes through two or more narrow slits.

The phenomenon in which two or more waves combine to generate a new wave with a larger, smaller, or equal amplitude. It depends on the alignment of the peaks and troughs of the overlapping waves.

When two or more waves collide, this is known as interference. They may add up, or they may partially or completely cancel each other,

When two waves with the same frequency. We will see varying intensities of light at different points on the screen owing to their superposition.

Hence interference principle is responsible for alternating light and dark bands when light passes through two or more narrow slits.

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Answer:

Magnetometer

Explanation:

Magnetometer technique is not using by scientists for studying the ocean floor.The scientists currently is using SONAR ( sound navigation and ragging) technique for studying the ocean floor.SONAR is used sound waves sound waves for studying the ocean floor or we can say that SONAR is based on sound propagation.

Therefore answer is Magnetometer

Answer:27.35 m/s

Explanation:

Given

Time of Flight of Projectile T=4.08 s

Range of Projectile =76.2 m

Time Of Flight of Projectile is given by

[tex]T=\frac{2u\sin \theta }{g}----------1[/tex]

where u=initial Velocity

[tex]\theta =[/tex]Launch angle

g=acceleration due to gravity

Range is given by [tex]R=\frac{u^2\sin 2\theta }{g}------2[/tex]

divide 1 and 2

[tex]\frac{R}{T}=\frac{u^2\sin 2\theta }{g}\times \frac{g}{2u\sin \theta }[/tex]

[tex]\frac{R}{T}=u\cos \theta ------3[/tex]

[tex]u\sin \theta =\frac{Tg}{2}------4[/tex]

squaring and adding 3 & 4 we get

[tex]u^2(\cos ^2\theta +\sin ^2\theta )=(\frac{R}{T})^2+(\frac{Tg}{2})^2[/tex]

[tex]u^2=(19.992)^2+(18.676)^2[/tex]

[tex]u=\sqrt{748.49}[/tex]

[tex]u=27.35 m/s[/tex]                              

The initial speed is the speed of the object at the beginning of the measurement or the starting speed.

The initial speed of the projectile is 27.35 m/s.

The initial speed is the speed of the object at the beginning of the measurement or the starting speed.

Given information-

The time taken by projectile to returns its original height is 4.08 s.

The distance traveled by it is 76.2 m.

Air resistance can be neglected.

The time of flight of a projectile motion can be given as,

[tex]T=\dfrac{2u\sin \theta}{g}[/tex]

Let the above equation is equation 1.

Here, [tex]u[/tex] is the initial velocity, and [tex]\theta[/tex] is the angle of launch.

Rewrite the above equation as,

[tex]u\sin \theta =\dfrac{Tg}{2}[/tex]  

Let the above equation is equation 2.

Now the range of the projectile motion can be given as,

[tex]R=\dfrac{u^2\sin (2\theta) }{g}\\R=\dfrac{u^22\cos \theta\sin \theta }{g}\\[/tex]

Divide this equation by the equation 1 as,

[tex]\dfrac{R}{T}=u\cos \theta[/tex]

Square and add the the above equation and equation 2 as,

[tex]u^2(\cos^2 \theta+\sin^2 \theta)=\dfrac{R}{T}+\dfrac{Tg}{2}\\u^2(1)=\dfrac{76.2}{4.08}+\dfrac{4.08\times9.81}{2}\\u=27.35\rm m/s[/tex]

Hence, the initial speed of the projectile is 27.35 m/s.

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Answer:

Vo = 4m/s

Explanation:

By conservation of energy:

[tex]1/2*m*Vf^2-1/2*m*Vo^2-m*g*h=0[/tex]

Solving for the initial speed:

[tex]Vo = \sqrt{2*(Vf^2/2-g*h)}[/tex]

[tex]Vo = \sqrt{2*(6^2/2-10*1)}[/tex]

Vo=4m/s

Answer:

The orbital velocity of the moon is, V = 1018 m/s

Explanation:

Given data,

The radius of the moon's orbit, R = 3.85 x 10⁸ m

The mass of the Earth, M = 5.98 x 10²⁴ kg

The formula for orbital velocity is,

                              V = √(GM/R²)

Substituting the values,

                               V = √(6.673 x 10⁻¹¹ x 5.98 x 10²⁴ / 3.85 x 10⁸ )

                                  = 1018 m/s

Hence, the orbital velocity of the moon is, V = 1018 m/s

The orbital speed of the Moon can be calculated using a specific formula that takes into account the mass of the Earth, the mass of the Moon, and the radius of the Moon's orbit.

The orbital speed of an object in orbit around another object can be calculated using the formula:

v = √[G * (M+E) / r]

Where v is the orbital speed, G is the gravitational constant (approximately 6.67 × 10^-11 N.m^2/kg^2), M is the mass of the larger object (in this case, the Earth), E is the mass of the orbiting object (in this case, the Moon), and r is the radius of the orbit.

Substituting the given values into the formula, we can calculate the orbital speed of the Moon.

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Answer:

15.65 °C

Explanation:

cold temperature (Tc) = -15.5 degree C = 273.15 - 15.5 = 257.65 kelvin

minimum coefficient of performance (η) = 8.25

find the maximum hot reservoir temperature of such a generator (Th)

η = \frac{Tc}{Th-Tc}

Th = Tc x (\frac{1}{η} + 1)

Th = 257.65 x (\frac{1}{8.25} + 1)

Th = 288.8 K

Th = 288.8 - 273.15 = 15.65 °C

Answer:

Ground state

Explanation:

At ground state all electrons are at the lowest energy level. At this level all the electrons, molecules or ions are said to be ground level. When electron get enough energy to jump they move to higher level. Any level higher than ground level is known as excited level. And energy of electron at excited state is higher than ground state. So the state at which all the electrons at their lowest possible energy level is the ground state.

Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

[tex]m_{1}[/tex] = mass of first body = 2.7 kg

[tex]m_{2}[/tex] = mass of second body = ?

[tex]v_{1i}[/tex] = initial velocity of the first body before collision = [tex]v[/tex]

[tex]v_{2i}[/tex] = initial velocity of the second body before collision = 0 m/s

[tex]v_{1f}[/tex] = final velocity of the first body after collision =

using conservation of momentum equation

[tex]m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}[/tex]

Using conservation of kinetic energy

[tex]m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35[/tex]

b)

[tex]m_{1}[/tex] = mass of first body = 2.7 kg

[tex]m_{2}[/tex] = mass of second body = 1.35 kg

[tex]v_{1i}[/tex] = initial velocity of the first body before collision = 4 ms⁻¹

[tex]v_{2i}[/tex] = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

[tex]v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67[/tex] ms⁻¹

Explanation:

Levitt and Dubner’s argument effectively uses logical, concrete evidence to arrive at conclusions about morality and cheating practices.the realities they present about sumo and the bagel business Their utilization of factual evidence in various illustrations shows morality and cheating practices are more common in high incentive situations where telling lies or an act of fooling people was rewarded awesomely. They use statistical evidence and different examples support the fact that Levitt and Dubner have arrived at a generalisation on moral grounds.

Answer:

Levitt and Dubner’s argument uses logic to present evidence to arrive at their final conclusions about morality and human cheating practices. The examples  they used were of Sumo wrestler's in Japan and the bagel business of a self-employed man who provided bagel's and cream cheese to office worker's on the "honor system". Levitt and Dubner used factual evidence as well as some illustrations to present their claims. They concluded that although cheating exists when there is high incentive, most people are  inherently honest when incentives are not a factor.

Explanation:

Answer:

The age of the universe would be 9.9 billion years

Explanation:

We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.

Thus, the time it has taken for the galaxies to reach their current separations is:

[tex]\displaystyle{t=D/v}[/tex]

and from Hubble's Law:

[tex]v =H_0D[/tex]

Therefore:

[tex]\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}[/tex]

With the given value for the Hubble's constant we have:

[tex]H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s[/tex]

and thus,

[tex]t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years[/tex]

The free-fall acceleration at the surface of Jupiter is approximately 24.79 m/s², which is more than two and two thirds times the gravitational pull experienced on Earth. A person weighing 150 pounds on Earth would weigh around 400 pounds on Jupiter.

The free-fall acceleration at the surface of Jupiter is approximately 24.79 m/s². This value is derived from using Newton's Law of Universal Gravitation and considering Jupiter's mass and radius. Jupiter has a mass about 300 times that of Earth and a radius approximately 11 times larger. The gravitational acceleration (g) at a planet's surface is given by the formula:

g = G x (mass of the planet) / (radius of the planet)²,

where G is the gravitational constant. Since the mass of Jupiter is much greater than that of Earth, and despite its larger radius, the acceleration due to gravity would be significantly higher on Jupiter. Therefore, an astronaut entering Jupiter's atmosphere would fall faster compared to falling through the Earth's atmosphere. If an astronaut who weighs 150 pounds on Earth were to stand on a scale on Jupiter, she would weigh approximately 400 pounds, which is more than two and two thirds times her weight on Earth. However, this value may vary slightly depending on whether she is near Jupiter's pole or equator, due to its oblateness.

Answer:

a) v=1.81 m/s; B) v=4.95 m/s

Explanation:

using momentum conservation

m1v1+m2v1=m1v2+m2v2

A)

5.2*672+700*0=5.2*428+700v2

The initial velocity of the block is 0, solving for v2

v2=1.81 m/s

B) Now both the bullet and the block travel together

m1v1+m2v1=(m1+m2)v2

5.2*672+700*0=(5.2+700)v

v=4.95 m/s

a. To find the resulting speed of the block after the bullet strikes it, we can use the law of conservation of momentum. b. The speed of the bullet-block center of mass can be found by calculating the weighted average of the speeds of the bullet and block.

a. To find the resulting speed of the block, we can use the law of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.

The initial momentum of the bullet is given by m1 * v1, and the final momentum is given by (m1 + m2) * vf, where m1 is the mass of the bullet, v1 is the initial velocity of the bullet, m2 is the mass of the block, and vf is the final velocity of the bullet and block combined.

Using these values, we can solve for vf and then find the resulting speed of the block.

b. The speed of the bullet-block center of mass can be found by calculating the weighted average of the speeds of the bullet and block. Since the mass of the bullet is much smaller than the mass of the block, the center of mass will be closer to the speed of the block.

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Answer:

T = 2.4 s

Explanation:

given,

angle at which ball is thrown = 29°

speed relative to ground = 24.3 m/s

ball is caught at a distance = 51.0463

acceleration due to gravity = 9.8 m/s²

time for which ball was in the air = ?

now,

velocity of ball in x-direction

V_x =  v cos θ

V_x =  24.3 x cos 29°

V_x = 21.25 m/s

velocity in y direction

V_y =  v sin θ

V_y =  24.3 x sin 29°

V_y = 11.78 m/s

distance on the ground when ball will reach maximum height

x = 51.0463/2 = 25.52 m

at top most point velocity is equal to zero

time for which the ball was in air

v = u + a t

0 = 11.78 - 9.8 t

[tex]t = \dfrac{11.78}{9.8}[/tex]

t = 1.20

this time is taken to travel half distance

total time = 2 x 1.20

T = 2.4 s

time for which ball was in air is T = 2.4 s

Because they perform specific tasks repeatedly throughout your program, as needed

Answer:

Because they perform specific tasks repeatedly throughout your program, as needed

Explanation:

Answer:

Lack of proper roof ventilation

Explanation:

The icicles hanging off the eaves of a roof are due to lack of proper roof ventilation. This means after snowfalls the attic of the roof receives warm air which melts the snow above. This is caused because the upper part of roof is above 32⁰C (which melts the snow) and lower part of your roof is at a lower temperature which refreezes the snow dripping down to form icicles.

Answer:

1964.8 J

[tex]5.12894\times 10^{-6}\ m^3[/tex]

1150 Joules

Explanation:

m = Mass of bullet = 0.5 kg

[tex]\Delta T[/tex] = Change in temperature = (327-20)

c = Specific heat of lead = 128 J/kg °C

[tex]\beta[/tex] = [tex]84\times 10^{-6}\ /^{\circ}C[/tex]

[tex]L_f[/tex] = Latent heat of fusion of lead = [tex]23000\ J/kg^{\circ}C[/tex]

(Values taken from properties of lead table)

Heat is given by

[tex]Q=mc\Delta T\\\Rightarrow Q=0.05\times 128\times (327-20)\\\Rightarrow Q=1964.8\ J[/tex]

The heat needed to raise the bullet to its final temperature is 1964.8 J

Change in volume is given by

[tex]\Delta V=V_0\beta \Delta T\\\Rightarrow \Delta V=5\times 10^{-6}\times 84\times 10^{-6}\times (327-20)\\\Rightarrow \Delta V=1.2894\times 10^{-7}\ m^3[/tex]

[tex]V=V_0+\Delta V\\\Rightarrow V=5\times 10^{-6}+1.2894\times 10^{-7}\\\Rightarrow V=5.12894\times 10^{-6}\ m^3[/tex]

The volume of the bullet when it comes to rest is [tex]5.12894\times 10^{-6}\ m^3[/tex]

Heat needed for melting

[tex]Q=mL_f\\\Rightarrow Q=0.05\times 23\times 10^3\\\Rightarrow Q=1150\ J[/tex]

The additional heat needed to melt the bullet is 1150 Joules

Final answer:

The calculation involves determining the heat required to increase the bullet's temperature, calculating the change in volume due to thermal expansion, and the additional heat required to melt the bullet, using the specific heat capacity, coefficient of thermal expansion, and latent heat of fusion for lead.

Explanation:

To solve this problem, we will use the specific heat capacity formula and the properties of lead to calculate the heat needed for temperature change and melting.

Heat required to raise the bullet's temperature: The heat (Q) needed can be calculated using the specific heat capacity equation Q = mcΔT, where 'm' is mass, 'c' is specific heat capacity, and ΔT is the change in temperature. For lead, the specific heat capacity (c) is approximately 128 J/(kg·°C).

Volume: To find the volume at the final temperature, we use the formula V = V_0(1+ αΔT), where V_0 is the initial volume, α is the coefficient of thermal expansion for lead, and ΔT is the change in temperature.

Additional heat to melt the bullet: To calculate the additional heat (Q_m) required to melt the bullet, we use the formula Q_m = mL_f, where 'm' is the mass of the bullet and L_f is the latent heat of fusion for lead.

To calculate these values, we also need the initial temperature (T_i), final temperature (T_f), and coefficient of thermal expansion for lead, which is 29.3 × 10⁻⁶ °C-1. For melting lead, the latent heat of fusion (L_f) is 24.7 kJ/kg.

Answer:

Opposite spin neutralizes the magnetic fields.

Explanation:

The reason for the significance of the unpaired electrons with respect to the magnetic properties is because electrons have opposite spin and when the electrons are paired then as a result their opposite spins neutralizes the effect of their magnetic field thus resulting in no field effect.

Thus single electrons which are unpaired contributes to the magnetic properties of the material as compared to the paired electrons.

Nuclear fusion would not occur in stars of any mass if hydrogen had the smallest mass per nuclear particle.

This is the process in which nuclear reactions between light elements form heavier elements.

Hydrogen won't be able to undergo nuclear fusion in stars of any mass if hydrogen had the smallest mass per nuclear particle due to the sub=atomic particles not being favorable in this condition.

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Answer:

R= - 3388.74

Explanation:

Given that

V₁= 76 k  ( in z-direction)

θ = 48°

V₂ = 60 cos48° i - 60 sin48°  k

The dot product of two vector given as

We know that dot product of two vector  is scalar and cross product of two vector is vector.

R= V₁ . V₂

We have to remember

i.i= j.j = k.k = 1

i.j = j.k = k.i = 0

Now

R= V₁ . V₂

R= (76 k ).(  60 cos48° i - 60 sin48°  k)

R= 0 - 60 x 76  sin48°

R= - 3388.74

The scalar product of the vectors is - 3388.74

Given information:

V₁= 76k  since it is in z-direction

Now vector V₂  makes an angle θ = 48° with x-axis so, it can be resolved as follows:

V₂ = 60 cos48°i - 60 sin48° k

The scalar product of vectors is the product of the projection of one vector with the other vector.

The scalar product or the dot product of two vectors is given as

V= V₁ . V₂

The dot product of the x,y,and z direction components follow the below mentioned rule:

i.i= j.j = k.k = 1

i.j = j.k = k.i = 0

So, the required scalar product

V = V₁ . V₂

V = (76k ).(60cos48° i - 60sin48°  k)

V = 0 -60 x 76sin48°

V = - 3388.74

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Answer:

(a) [tex]K = \frac{I}{2\pi a}[/tex]

(b) [tex]J = \frac{I}{2\pi as}[/tex]

Explanation:

(a) The surface current density of a conductor is the current flowing per unit length of the conductor.

                                   [tex]K = \frac{dI}{dL}[/tex]

Considering a wire, the current is uniformly distributed over the circumferenece of the wire.

                                   [tex]dL = 2\pi r[/tex]

The radius of the wire = a

                                    [tex]dL = 2\pi a[/tex]

The surface current density [tex]K = \frac{I}{2\pi a}[/tex]

(b) The current density is inversely proportional

                                     [tex]J \alpha  s^{-1}[/tex]    

                                     [tex]J = \frac{k}{s}[/tex]           ......(1)

k is the constant of proportionality

                                     [tex]I = \int\limits {J} \, dS[/tex]

                                     [tex]I = J \int\limits \, dS[/tex]     ........(2)

substituting (1) into (2)

                                     [tex]I = \frac{k}{s} \int\limits\, dS[/tex]

                                     [tex]I = k \int\limits^a_0 \frac{1}{s}  {s} \, dS[/tex]

                                     [tex]I = 2\pi k\int\limits\, dS[/tex]

                                     [tex]I = 2\pi ka[/tex]

                                     [tex]k = \frac{I}{2\pi a}[/tex]

substitute [tex]J = \frac{k}{s}[/tex]

                                     [tex]J = \frac{I}{2\pi as}[/tex]

In a wire with current I and radius a, the surface current density K with uniform distribution is I/(2πaL). If volume current density varies inversely with the radius, the current is more dense at the center and less dense towards the surface, described by I/(πa²L).

The subject here relates to the physical properties of the current flowing down a wire with a certain radius. (a) If the current I is uniformly distributed over the surface of the wire with a radius a, then the surface current density K can be given by the total current (I) divided by the surface area of the wire. In this context, the surface area of a cylindrical wire can be calculated by the formula 2π * a * L, where L is the length of the wire. Therefore, the surface current density K is I/(2πaL).

(b) If the current is distributed such that the volume current density J is inversely proportional to the radius, it implies that the current is more dense at the center of the wire and less dense as you approach the surface. The volume current density J can be described by the formula I/(πa²L). This complex distribution would likely require calculus to derive an exact relationship between current density and radius.

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Answer:

The magnitude of torque is τ = 24.522 N*m^2

Explanation:

To find the magnitude of the torque can use the equation of the force produce by the airplane so:

τ = F * d

τ = 0.804 N * 30.5 m

τ = 24.522 N*m

Check:

Find the acceleration

I = m*r^2=0.741kg*(30.5m)^2

I = 689.32 kg*m^2

τ = I*a_c

a_c = τ /I = 24.522 N*m^2 / 689.32 kg*m^2

a_c = 0.0355 m/s^2

τ = 0.0355 m/s^2 * 689.32 kg*m^2

τ = 24.522 N*m^2

Answer:

(a) 7 m

(b) 1  m

Explanation:

Given:

The magnitude of displacement  vector 'a' is 3 m

The magnitude of displacement vector 'b' is 4 m.

The vector 'c' is the vector sum of vectors 'a' and 'b'.

(a)

Now, when the angle between the vectors is 0°, it means that the vectors are in the same direction. When vectors are in the same direction, then their resultant magnitude is simply the sum of their magnitudes.

So, magnitude of 'c' when 'a' and 'b' are in same direction is given as:

[tex]|\overrightarrow c|=|\overrightarrow a|+|\overrightarrow b|\\\\|\overrightarrow c|=3 + 4 = 7\ m[/tex]

Therefore, the magnitude of vector 'c' is 7 m when angle between 'a' and 'b' is 0°.

(b)

When the angle between the vectors is 180°, it means that the vectors are exactly in the opposite direction. When the vectors are in opposite direction, then their resultant magnitude is the subtraction of their magnitudes.

So, magnitude of 'c' when 'a' and 'b' are in opposite direction is:

[tex]|\overrightarrow c|=||\overrightarrow a|-|\overrightarrow b||\\\\|\overrightarrow c|=|3 - 4| = 1\ m[/tex]

Therefore, the magnitude of vector 'c' is 1 m when angle between 'a' and 'b' is 180°.

Final answer:

The force you exert on the rock is equal in magnitude but opposite in direction to the force the rock exerts on you, in accordance with Newton's Third Law of Motion.

Explanation:

When you push a rock along level ground and make it speed up, the force you exert compares with the force the rock exerts on you in accordance with Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. Therefore, the force you exert on the rock is equal in magnitude but opposite in direction to the force the rock exerts on you.

The concept can be understood by considering various scenarios where different forces are applied to objects of differing masses, resulting in different levels of acceleration. For example, pushing a basketball produces a noticeable acceleration due to the basketball's low mass, whereas pushing a stalled SUV with the same force results in a much smaller acceleration due to the SUV's greater mass.

Friction can also affect how much force is needed to move an object. When pushing a heavy crate, you must overcome static friction, which matches the force you apply up to the point where the crate begins to move. Once in motion, dynamic friction takes over, which is generally lower than static friction, making it easier to keep the object moving.

Answer:

  λ₂ = 1,219 10⁻⁷ m , λ₃ = 1.028 10⁻⁷ m ,   λ₄ = 0.9741 10⁻⁷ m , λ₅ = 0.9510 10⁻⁷ m and  λ₆ = 0.9395 10⁻⁷ m

Explanation:

To calculate the lines of the hydrogen liman series, we can use the Bohr atom equation

           En = -13.606 / n²       [eV]

n       En

1       -13,606

2       -13.606 / 4 =    -3.4015

3       -13.606 / 9 =    -1.5118

4       -13.606 / 16 =  -0.8504

5       -13.606 / 25 = -0.5442

6       -13.606 / 36 = -0.3779

The lyma series are transitions where the state is fundamental (E1), let's calculate the first five transitions

State

initial final energy

6           1      -0.3779 - (- 13.606) =  13.23 eV

5           1      -0.5442 - (- 13.606) =  13.06 eV

4           1      -0.8504- (-13.606) =   12.76 eV

3            1      -1.5118 - (- 13.606) =   12.09 eV

2            1      -3.4015 - (- 13.606) = 10.20 eV

Let's use the relationship between the speed of light and the wavelength and the frequency

      c = λ  f

      f = c / λ  

Planck's relationship for energy

     E = h f

     E = h c / λ

    λ = hc / E

We calculate for each energy

E = 10.20 eV

      λ  = 6.63 10⁻³⁴ 3 10⁸ / (10.20 1.6 10⁻¹⁹)

      λ  = 12.43 10⁻⁷ / 10.20

      λ₂ = 1,219 10⁻⁷ m

E = 12.09 eV

     λ₃ = 12.43 10⁻⁷ / 12.09

     λ₃ = 1.028 10⁻⁷ m

E = 12.76 eV

      λ₄ = 12.43 10⁻⁷ /12.76

      λ₄ = 0.9741 10⁻⁷ m

E = 13.06 ev

      λ₅=  12.43 10⁻⁷ /13.06

       λ₅ = 0.9510 10⁻⁷ m

E = 13.23 eV

      λ₆ = 12.43 10⁻⁷ / 13.23

      λ₆ = 0.9395 10⁻⁷ m

Answer:

h=20.66m

Explanation:

First we need the speed when the cord starts stretching:

[tex]V_2^2=V_o^2-2*g*\Delta h[/tex]

[tex]V_2^2=-2*10*(-31)[/tex]

[tex]V_2=24.9m/s[/tex]   This will be our initial speed for a balance of energy.

By conservation of energy:

[tex]m*g*h+1/2*K*(h_o-l_o-h)^2-m*g*(h_o-l_o)-1/2*m*V_2^2=0[/tex]

Where

[tex]h[/tex] is your height at its maximum elongation

[tex]h_o[/tex] is the height of the bridge

[tex]l_o[/tex] is the length of the unstretched bungee cord

[tex]800h+21*(79-h)^2-63200-24800.4=0[/tex]

[tex]21h^2-2518h+43060.6=0[/tex] Solving for h:

[tex]h_1=20.66m[/tex]  and [tex]h_2=99.24m[/tex]  Since 99m is higher than the initial height of 79m, we discard that value.

So, the final height above water is 20.66m

Answer: using the conservation of potential energy stored in spring giving that at maximum amplitude velocity becomes zero.

Mgd= 1/2k(d-l)^2..... equation 1

M= 80kg=mass , g= 10m/s^2 =gravity, d=?=length of fully extended bungee rope, l=31m= length of bungee rope before extension, k=42N/m= spring constant

Simplifying equation above gives

2Mgh/k= d^2 - 2dl + l^2 ....eq 2

Substituting figures into the equ above gives

0 = d^2 - 100.1d +961 ...equ 3

Equ3 can be solved since it is a quadratic equation

d= (-b +or- square root (b^2 - 4ac))/2a ....equa4

Where a=1, b= -100.1, c= 961

Substituting figures into eequa4

d= 89.34m

So therefore the height above the river to me when bungee is fully extended is= 110 - 89.34

= 20.66

Explanation: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k.

Assuming that it performs simple harmonic motion.

Sound waves in air and water are longitudinal waves characterized by pressure differences. Sound in solids can have both longitudinal and transverse components.

Sound waves in air and water are longitudinal waves characterized by pressure differences. When sound waves propagate through a fluid like air or water, the disturbances are periodic variations in pressure, resulting in compressions (high-pressure regions) and rarefactions (low-pressure regions).

Fluids do not have appreciable shear strength, so the sound waves in them must be longitudinal or compressional. On the other hand, sound in solids can have both longitudinal and transverse components. For example, seismic waves generated by earthquakes have both longitudinal (compressional or P-waves) and transverse (shear or S-waves) components.