Standing 38.6 m away from a rock wall, you yell. How much time in seconds will it take you to hear your echo to two significant digits? Make sure to account for the travel from you to the wall and from the wall back to you.

The orbital speed of an ice cube in the rings of saturn is approximately 3.56 * 10^6 m/s

The law of gravitation states that the force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between the masses. Mathematically;

F = GMm/r²

where

M and m are the mass of ice cube and

Recall that;

s = Gm1/r^2

Also;

F = sm²

Substitute to have;

s = m²/F

For the centripetal acceleration

a = v²/r

Such that;

v²/r = Gm/r²

v² = Gm/r

v = √Gm/r

Substitute the given parameters into the formula to have:

V =  √6.67×10^-11 *  5.68 x 10^26 / 3.00 x 10^5

V = 355358.97m/s = 3.56 * 10^6 m/s

Therefore the orbital speed of an ice cube in the rings of saturn is approximately 3.56 * 10^6 m/s

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Answer: The answer is A.

Explanation:

Just because u have gay parents doesnt mean theyre bad lol XD

The voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V

The resistance of a wire is the opposition to the flow of current. It depends on the following;

Given that the resistivity of a 1.0 m long wire is 1.72 × 10-8 ωm and its cross sectional area is 2.0 × 10-6 m2. if the wire carries a current of 0.20A

The given parameters are;

The formula to use to get R will be

R = ρL / A

Substitute all the necessary parameters into the formula

R = 1.72 x [tex]10^{-8}[/tex] x 1 / 2 x [tex]10^{-6}[/tex]

R = 8.6 x [tex]10^{-3}[/tex]  Ω

From Ohm's law, V = IR

Substitute all the necessary parameters into the formula

V = 0.2 x 8.6 × [tex]10^{-3}[/tex]

V = 1.72 x [tex]10^{-3}[/tex] V

Therefore, the voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V

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The speed of transverse waves on the wire is 1.555.

Given: Length= 1.150 m

Mass= 4.80 g

Speed =mass/length

U=M/L formula

U=1.150 *10^(-3)/4.80

v=√t/u

v=√580*4.80/1.150*10^(-3)=1.555

1.555

The derivative of the displacement with respect to time gives the transverse velocity in the y direction, as velocity is the rate of change of position: Where k = 2/ is referred to as the wave number and = 2f as before, v(x,t) = y(x,t)/t = -A cos (k x - t + ).The linear density and tension v=FT can be used to determine the wave's speed. v = F T μ . According to the equation v=FT, v = F T, the tension would need to be increased by a factor of 20 if the linear density were to be doubled by almost as much.

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Faraday's law is applicable in the mechanism behind electric generators, credit cards, metal detectors, computer hard drives and electronic drum

Faraday's law states that a changing magnetic field through an area or equivalently, a changing area with constant field will cause a voltage.

So therefore, Faraday's law is applicable in the mechanism behind electric generators, credit cards, metal detectors, computer hard drives and electronic drum

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Kinetic energy of pieces A and B are 2724 Joule and 5176 Joule respectively.

Mass of piece B = Mb

Ma×Va = Mb×Vb

So, 1.9Mb × Va = Mb×Vb

=> 1.9Va = Vb

=>1/2 × Ma × Va² + 1/2 × Mb × Vb² = 7900

=> 1/2 × Ma × Va² + 1/2 × (Ma/1.9) × (1.9Va)² = 7900

=> 1/2 × Ma × Va² ×(1+1.9) = 7900 j

=> 1/2 × Ma × Va² = 7900/2.9 = 2724 Joule

Thus, we can conclude that the kinetic energy of piece A and B are 2724 Joule and 5176 Joule respectively.

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In hydrogen atom the electronic transition from n= ∞ to n=1 corresponds to largest energy emission.

Electronic transition is the jump of an electron from one energy level to another energy level .

E = hc/λ

h= plank constant =6.62*10^(-34)

c=speed of light

λ= wavelength of emited electron .

Thus , we can conclude that the electronic transition from infinity to n=1 (ground state) corresponds to largest energy emission .

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Angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.

The principle of conservation of angular momentum states that the total angular momentum acting on an object is constant, provided there is no  external torque acting on the object.

Angular momentum of a system is conserved as long as there is no net external torque acting on the system.

Thus, angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.

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Explanation:

India. Total wealth: $8.9 trillion | Wealth per capita: $6,440 | India, which is the fifth-largest economy in the world, is home to 3,57,000 HNWIs and 128 billionaires.

Answer:

50505050128128128128

Explanation:

Answer:

Kinetic energy of the projectile at the vertex of the trajectory: [tex]900\; {\rm J}[/tex].

Work done when firing this projectile: [tex]2500\; {\rm J}[/tex].

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity [tex]v_{x}[/tex] of this projectile would stay the same (at [tex]30\; {\rm m\cdot s^{-1}}[/tex]) throughout the flight.

The vertical velocity [tex]v_{y}[/tex] of this projectile would be [tex]0\; {\rm m\cdot s^{-1}}[/tex] at the vertex (highest point) of its trajectory. (Otherwise, if [tex]v_{y} > 0[/tex], this projectile would continue moving up and reach an even higher point. If [tex]v_{y} < 0[/tex], the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be [tex]v = 30\; {\rm m\cdot s^{-1}}\![/tex] when it is at the top of the trajectory. The kinetic energy [tex]\text{KE}[/tex] of this projectile (mass [tex]m = 2.0\; {\rm kg}[/tex]) at the vertex of its trajectory would be:

[tex]\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}[/tex].

Apply the Pythagorean Theorem to find the initial speed of this projectile:

[tex]\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Hence, the initial kinetic energy [tex]\text{KE}[/tex] of this projectile would be:

[tex]\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}[/tex].

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be [tex]2500\; {\rm J}[/tex].

(a) The angular acceleration will be 26.035 rev/[tex]min^{2}[/tex].

(b) The final angular velocity is expected to be 33.846 rev/min.

Given.

t=1.3 min, Θ=22 rev, [tex]ω_{i}[/tex]=0

We know, Θ= [tex]ω_{i}[/tex]t+[tex]\frac{1}{2}[/tex][tex]\alpha[/tex][tex]t^{2}[/tex]

22=0+[tex]\frac{1}{2}[/tex][tex]\alpha[/tex][tex]1.3^{2}[/tex]

[tex]\alpha[/tex]=26.035 rev/[tex]min^{2}[/tex]

[tex]ω_{f} =ω_{i}+\alpha t[/tex]=0+26.035*1.3=33.846 rev/min

An object's rate of change in angular position or orientation over time is depicted by its angular velocity, rotational velocity, or both ( or ), also known as the angular frequency vector (i.e. how quickly an object rotates or revolves relative to a point or axis). The direction of the pseudovector is normal to the instantaneous plane of rotation or angular displacement, and its magnitude denotes the angular speed, or the rate at which the item rotates or revolves. It is customary to use the right-hand rule to specify the direction of angular motion. A general definition of angular velocity is "angle per unit time" (angle replacing distance from linear velocity with time in common). Radians per second is how angles are measured in the SI.

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The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Mass of the astronaut, [tex]m_a[/tex] = 126 kg

Speed he acquires, [tex]v_{a}[/tex]  = 2.70 m/s

Mass of the space capsule, [tex]m_{c}[/tex] = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

[tex]P_f = m_av_a + m_cv_c[/tex]

[tex]P_I[/tex] = 0

[tex]m_av_a + m_cv_c = 0[/tex]

[tex]v_c =\frac{- m_a v_a}{m_c}}\\\\[/tex]

   [tex]= \frac{126* 2.70}{1800}[/tex]

   [tex]= - 0.189[/tex] m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

    = 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

  = 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = [tex]\frac{1}{2} m v^2[/tex]

     [tex]= \frac{1}{2}[/tex] × 126 × [tex](2.70) ^2[/tex]

     = 459.27 J

The Kinetic Energy of the capsule;

K.E = [tex]\frac{1}{2} m v^2[/tex]

     = [tex]\frac{1}{2}[/tex]×1800×[tex](0.189) ^2[/tex]

     = 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

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Answer:

100rad because it angular velocity

Answer: 56.44°

Explanation:

Given:

Converting to SI Units (m/s):

= (1.2 mach)(340 ms^-1 / 1 Mach)

u  = 408 m/s

Find:

We have, sinθ = speed of sound / speed of object

               sinθ = v / u

                   θ = sin^-1 (v / u)  

                      = sin^-1 (340 / 408)

                   θ = 56.44°

The height of the water level in the water tank is 0.011 m.

t = (2× 1.72)/9.8 = 0.35 s

Velocity along x direction= x × t

= 1.33 × 0.35

= 0.46 m/s

As per Bernoulli's theorem, 1/2 × ρ × v² = ρ × g× h

=> h = v²/2g = 0.46²/(2×9.8)

= 0.011 m

Thus, we can conclude that the height of water level is 0.011m.

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Answer:

Approximately [tex]2.03 \times 10^{7}\; {\rm m}[/tex].

Explanation:

Assume that the radius of this orbit is [tex]r[/tex].

Let [tex]m[/tex] denote the mass of this satellite and let [tex]M[/tex] denote the mass of the Earth. At a distance of [tex]r[/tex] from the center of the earth, the magnitude of the gravitational attraction on this satellite would be [tex]G\, m\, M / (r^{2})[/tex].

The question implies that the gravitational pull from the earth is the only significant force on this satellite. Hence, the net force on this satellite would be also [tex]G\, m\, M / (r^{2})[/tex].

The acceleration of this satellite would thus be [tex]a = (\text{net force}) / (\text{mass}) = G\, M / (r^{2})[/tex].

Let [tex]\omega[/tex] denote the angular velocity of this satellite. Since this satellite in in a circular motion, the acceleration on this satellite would need to satisfy [tex]a = \omega^{2} \, r[/tex].

In other words:

[tex]\begin{aligned} \frac{G\, M}{r^{2}} = a = \omega^{2} \, r \end{aligned}[/tex].

[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3}\end{aligned}[/tex].

The question asks for a rotation of [tex]3\times (2\, \pi) = 6\, \pi\; {\text{rad}}[/tex] within a day, which is [tex]24 \times 3600\; {\rm s}[/tex]. The angular velocity of this satellite should be:

[tex]\begin{aligned}\omega = \frac{6\, \pi}{24 \times 3600\; {\rm s}} \end{aligned}[/tex].

Substitute this value into the expression for [tex]r[/tex] and evaluate:

[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3} \\ &= \left(\frac{(6.67 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}}) \times (5.97 \times 10^{24}\; {\rm kg})}{((6\, \pi) / (24 \times 3600\; {\rm s}))^{2}}\right)^{1/3} \\ &\approx 2.03 \times 10^{7}\; {\rm m}\end{aligned}[/tex].

(Note that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].)

The height of the light house is 17.86m.

To find the answer, we have to know about the simple pendulum.

                       [tex]T=2\pi \sqrt{\frac{l}{g} } \\[/tex]

where, g is acceleration due to gravity, and l is the length of the pendulum.

              [tex]h=(\frac{T}{2\pi }) ^2g=(\frac{8.48}{2*3.14}) ^2*9.8=17.86m[/tex]

Thus, we can conclude that, the height of the light house is 17.86m.

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Answer:

s^ -1    ( or    1/sec)

Explanation:

Velocity is given in units of displacement / sec

like feet /sec   or   m/sec    

so b would have units of   s^-1

(or perhaps a more general term would be   time^-1)

Answer:

453 gm

Explanation:

Immersed objects are buoyed up by force equal to mass of displaced liquid

400 + 53 = 453 gm  in air

The correct answer is 5828.675 J.

Given combined mass 4kg and mass of bullet 150gm=0.150kg.

Total mass= 4+0.150=4.150kg

Velocity=53 m/s

Kinetic energy = [tex]\frac{1}{2} *m*v^{2}[/tex] =0.5*4.150*[tex]53^{2}[/tex] =5828.675 J

Kinetic energy is a type of power that an item or particle possesses as a result of motion. When an item undergoes work—the transfer of energy—by being subjected to a net force, it accelerates and consequently obtains kinetic energy. A moving object or particle's kinetic energy, which depends on both mass and speed, is one of its characteristics. Any combination of motions, including translation (or travel along a path from one location to another), rotation about an axis, and vibration, may be used as the type of motion.

A body's translational kinetic energy is equal to  [tex]\frac{1}{2} *m*v^{2}[/tex] , or one-half of the product of its mass, m, and square of its velocity, v.

a trolley and a sandbag have a combined mass of 4 kg. a bullet with a mass of 150 gram is fired towards the trolley and is lodged in the sand bag immediately after the Collision, on the trolley and sandbag in combination with the bullet, move backwards at 53 metre per second calculate the kinetic energy of the trolley , with the sandbag and bullet ,directly after the Collision ​

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The energy needed to move an electron in a hydrogenatome from the ground state (n=1) to n=3 will be 1.93 *10^-18J and 12.09 eV.

The following can be deduced:

Energy of electron in hydrogen atom is

En = -13.6 /n2 eV

where n is principal quantum number of orbit.

Energy of electron in first orbit = E1 = -13.6 / 12 = - 13.6eV

Energy of electron in third orbit = E3 = -13.6 /32 = -1.51 eV

Energy required to move an electron fromfirst to thirdorbit ΔE = E3- E1

ΔE = -1.51 - ( 13.6) = 12.09 eV

Energy in Joule = 12.09 *l/× 1.6 × 10^-19 = 1.93 × 10^-18 J.

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Complete question:

How much energy is needed to move an electron in a hydrogenatome from the ground state (n=1) to n=3? Give theanswer (a) in joules and (b) in eV.

(a) The electric field strength at a point 1.00 cm to the left of the middle is  2.0 x 10⁷ N/C.

(b) The magnitude of the force is 94.4 N and direction of the force on it towards the left.

The electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;

E = kq/r²

E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²

E1 = 1.35 x 10⁸ N/C

E2 =  -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²

E2 = - 1.35 x 10⁸ N/C

E3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²

E3 = -2.0 x 10⁷ N/C

E = E1 + E2 + E3

E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - (-2.0 x 10⁷ N/C)

E = +2.0 x 10⁷ N/C

F = Eq

F = 2.0 x 10⁷ x -4.72 x 10⁻⁶

F = -94.4 N

Thus, the direction of the force will be towards the left.

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A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest then

(a) The speed of the 0.350- kg puck after the collision - 2.40 m/s

(b) The direction of the velocity of the 0.350- kg puck after the collision is towards West.

c) The speed of the 0.950- kg puck after the collision is 2.82 m/s

d) The direction of the velocity of the 0.950- kg puck after the collision is towards East.

Mass of ice puck, m₁ = 0.350 kg

Mass of another puck, m₂ = 0.950 kg

Velocity of ice puck, v₁ = 5.22 m/s

Velocity of another puck, v₂ = 0 m/s

[tex]v^{'}_1= ?[/tex]

[tex]v^{'}_2= ?[/tex]

[tex]m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{2 }[/tex]

[tex]v_{1} - v_{2} = - ( v^{'}_1 - v^{'}_2 )\\v_{1} - v_{2} = - v^{'}_1 + v^{'}_2\\v^{'}_2 = v^{'}_1 + v_{1}\\\\\\m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{1 } + m_{2} v_{1 }[/tex]

[tex](0.35)(5.22) + 0 = 0.350 v^{'} _{1} + 0.950 v^{'}_{1 }+ (0.950)(5.22)\\1.827 = 0.350v^{'} _{1} + 0.950v^{'}_{1 } + 4.959\\1.827 - 4.959 = 1.3 v^{'} _{1}\\- 3.132 = 1.3 v^{'} _{1}\\v^{'} _{1} = -2.40 m/s\\\\\\[/tex]

Therefore, the speed of the 0.350- kg puck after the collision is - 2.40 m/s.

[tex]v^{'}_2 = v^{'} + v_1[/tex]

    [tex]= -2.40+5.22[/tex]

    [tex]= 2.82 m/s[/tex]

Therefore, the speed of the 0.950- kg puck after the collision is 2.82 m/s.

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The object that ends up with less electrons has a positive charge; option A.

Electrons are negatively charged particles which form a part of the three fundamental particles in an atom.

Electrons can easily be removed from the vicinity of atoms making the atom to become positively charged. On the other hand, when atoms gain extra electrons, they become negatively-charged.

On way of charging objects is by rubbing them against other objects in order to gain extra electrons or to lose electrons.

When two objects are rubbed together, the object that ends up with less electrons has a positive charge, whereas the object that ends up with more electrons has a negative charge.

In conclusion, objects can become charged when they are rubbed against each other, and they can either become negatively or positively-charged by gain or loss of electrons.

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The object indicated here that ends up with fewer electrons has an unknown charge (not enough info, Option B).

The atomic charge makes reference to the overall charge of an atom, which can be evidenced by knowing its net charge (positive or negative charge).

The net charge of a molecule and/or atom is obtained by calculating the amount of electrically negative electrons (e-) and the number of positive protons.

Neutrons are another type of subatomic particle that have not net charge (neither positive nor negative).

In conclusion, the object indicated here that ends up with fewer electrons has an unknown charge (not enough info, Option B).

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change in velocity is 3m/s,the force exterted by bat on ball is 4.8N and the acceleration is 30m/s².

1) Speed is scalar quantity but velocity is vector quantity.

2) Speed is distance followed by an individual with respect to  time and velocity is displacement with respect to time .

1) final velocity- intial velocity

 25-22= 3m/s

2) The force is

F= ma

 . 0.16×30= 4.8N

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Based on the calculations, the average velocity is equal to 360 m/s and the percent difference is equal to 4.72%.

An average velocity can be defined as the total distance covered by a physical object divided by the total time taken.

An average is also referred to as mean and it can be defined as a ratio of the sum of the total number in a data set to the frequency of the data set.

Mathematically, the average velocity for this data set would be calculated by using this formula:

Average = [F(v)]/n

Vavg = [v₁ + v₂ + v₃ + v₄ + v₅)/5

Since the values of the average velocity from the table are missing, we would assume the following values for the purpose of an explanation:

Substituting the parameters into the formula, we have:

Vavg = [300 + 450 + 500 + 250 + 300)/5

Vavg = 1800/5

Vavg = 360 m/s.

Next, we would calculate the percent difference by using this formula:

[tex]Percent \;difference = \frac{[V_{avg}\;-\;V_{sound}]}{V_{sound}} \times 100[/tex]

Percent difference = [360 - 343]/360 × 100

Percent difference = 17/360 × 100

Percent difference = 0.0472 × 100

Percent difference = 4.72%.

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The frictional force between the crate and the ramp is determined as 35.2 N.

Apply the principle of conservation energy to calculate the change in the energy of the crate.

Change in energy of the crate = energy loss to friction

P.Ei - K.Ef = E

mgh - ¹/₂mv² = E

where;

(30)(9.8)(2) - (0.5)(30)(2²) = E

528 J = E

E = Fd

where;

F = E/d

F = (528)/(15)

F = 35.2 N

Thus, the frictional force between the crate and the ramp is determined as 35.2 N.

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The maximum force constant of the spring Kmax is 2337.9 N/m.

The force constant or spring constant is defined as the force required to stretch or compress a spring such that the displacement in the spring is 1 meter.

Force constant is denoted by K and its unit is N/m.

Where;

K = spring constant

x = displacement

The work done by the spring is given below as follows:

Work done = Fx/2

Kinetic Energy = mv²/2

Force on an inclined plane = mgsinθ

Total force, F = mgsinθ + frictional force

F = 1390 * sin 22° + 515

F = 1035.7 N

Work done = change in KE

Fx/2 = mv²/2

Fx = mv²

m  = 1390/9.81 = 141.692

Solving for x;

x = mv²/F

x = 141.692 * 1.8²/1035.7

x = 0.443 m

The maximum force constant of the spring Kmax = 1035.7/0.443

Kmax = 2337.9 N/m

In conclusion, the maximum force constant of the spring  is the ratio of the total force and displacement.

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Note that the complete question is given below:

You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 2.0 m/s at the top of a ramp that slopes downward at an angle 21.0 ∘. The ramp will exert a 533 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp. Calculate the maximum force constant of the spring Kmax that can be used in order to meet the design criteria

Answer:

Gravity exist between two objects that have mass