The figure illustrates flow through a pipe with diameters of 1.0 mm and 2.0 mm and with different elevations. Px is the pressure in the pipe, and Vx is the speed of a non-viscous incompressible fluid at locations x = Q,R,S,T, or U. Options are: Greater than, Less than, Equal to

PU is ... PQ

VU is ... 2VT

PR is ... PU.

VR is ... VS

VQ is ... VU

PR is ... PS

From the calculation, the force constant is 192 N. Also, friction would decrease the extension.

We know that the force constant can be obtained by the use of the relation;

F = Ke

F = applied force

K = force constant

e = extension

We know from Hooks law that the force applied is directly proportional to the extension.

We can write;

F = mgcosθ

F = 43 Kg * 9.8 m/s^2 * sin31°

F = 217 N

K = 217 N/1.13 m

K = 192 N/m

If there is friction between the incline and the crate, it will stretch less because some work will be lost due to friction causing only some fraction of the elastic potential energy to be converted to kinetic energy.

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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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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:

[tex]\huge\underline{\underline{\boxed{{ \mathbb{SOLUTION:}}}}}[/tex]

[tex]\leadsto[/tex] Total internal reflection occurs when a ray of light is traveling through a medium and, when that medium changes, it does not refract into the second medium but instead it reflects back into the first medium.

By Snell's law, we know that

[tex]\longrightarrow \sf{n_1 sin \emptyset_1 = n_2 sin \emptyset_2}[/tex]

If this happens, then [tex]\sf{n_2 \angle n_1}[/tex], and the ray does not lose intensity due to refraction.

[tex]\huge\underline{\underline{\boxed{\mathbb{ANSWER:}}}}[/tex]

[tex]\large \bm{ A.}[/tex] A ray of light has the same intensity both entering and exiting a fiber optic cable.

The kinetic energy of the ejected electron (photoelectron) is equal to the energy of the photon minus the work function (E required to eject the photoelectron).

The  process of final kinetic energy of the electron upon reaching the anode compare to its initial potential energy immediately after it has been ejected -

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Answer:

average velocity = [tex]\bf 900 \space\ km/h[/tex]

Explanation:

We can find the average velocity using the following equation:

[tex]\boxed{average \space\ velocity = \frac{total \space\ distance \space\ travelled}{ \space\ time \space\ taken}}[/tex] .

In this case:

• total distance travelled = 450 km

• time taken = 30 min = 0.5 h

Substituting these values into the equation:

[tex]average \space\ velocity = \frac{450 \space\ km }{ 0.5 \space\ h}[/tex]

                          ⇒ [tex]\bf 900 \space\ km/h[/tex]

Answer:

D. Waves combine and their amplitude increases

Explanation:

amplitude is the height of the wave in a graph

constructive interference happens when two waves are lined up perfectly

example when two speakers playing same music while facing each other, music will appear louder

Destructive interference happens when the peaks match the valleys and they cancel perfectly

like with noise cancelling headphones

during constructive interference, waves collide in phase

in phase means add together making the height of the wave in a graph bigger (amplitude increases)

Example: When we see two speakers right next to each other, we can experience constructive interference when the distance from each speaker to the observer is the same

two waves destructively interfere, they must have the same amplitude in opposite directions

quizlet

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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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The answers are :

(a) 23π/90 radians ≈ 0.803 radians

(b) 17/2π revolutions ≈ 2.706 revolutions

(c) 79.5/60 rad/s = 13.25 rad/s

*The important formulas to remember*

     [tex]\boxed {Radian = Degree \times \frac{\pi}{180^{o}}}[/tex]

    2. Radians to Revolutions

     [tex]\boxed {Revolution = \frac{Radian}{2\pi}}[/tex]

    3. rpm to rad/s

     [tex]\boxed {rad/s = \frac{rpm}{60}}[/tex]

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].)

Answer:

One needs to know the initial speed of the bus to find distance traveled.

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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Answer:

by using a magnetic field

Use a magnet to attract the iron fillings from the sand!

Hope this helps!

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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Answer: The answer is A.

Explanation:

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

Answer:

the light will reflect parallel to the principal axis

The rate energy is consumed this statement describes power.

Hence, Option C is correct answer.

Power is related to energy by that it is the rate at which energy is transferred.

It is a measure of the rate at which work is done.

According to the definition of power, Power is the amount of energy transferred or converted or consumed per unit time.

SI unit of power is watt.

By definition, 1 watt is equal to one joule of work done per second. So if P represents power in watts, E is the change in energy (number of joules) and t is the time taken in seconds then:

P=[tex]\frac{E}{t}[/tex]= [tex]\frac{1 Joule}{1 Second}[/tex]= 1 Watt.

Thus from the above conclusion we can say that, The rate energy is consumed describes power.

Hence, Option C is correct answer

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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].

Answer:

A: Q = m*Lv

Explanation:

The equation formula for the enthalpy of vaporization is;

Q = m*ΔHv

Where;

q = heat energy

m = mass

ΔHv = heat of vaporization

The statement that best explains the type of chemical reaction represented by Maya's picture is that it is neither a synthesis reaction nor a decomposition reaction because two reactants form two products. That is option B.

A chemical reaction is the combination of two elements to yield a new product through the formation of bonds.

A chemical reaction is said to be a synthesis reaction when when two different atoms or molecules interact to form a different molecule or compound.

A chemical reaction is said to be a decomposition reaction when one reactant breaks down into two or more products.

Therefore, from the picture, the chemical reaction is neither a synthesis reaction nor a decomposition reaction because two reactants form two products.

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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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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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The gravitational acceleration on the board of the Deepsea Challenger when it is in the Mariana Trench, 11 km below the surface of the Pacific Ocean is very large due to the high pressure. That is option B

The gravitational acceleration is the acceleration of an object that is under free fall. The acceleration due to gravity is 9.81 m/s2. above the sea level.

One of the factors that affect gravitational acceleration is depth. This is because as the depth increases so will pressure increase.

Increase in pressure leads to increase in gravitational acceleration.

Therefore, the gravitational acceleration on the board of the Deepsea Challenger when it is in the Mariana Trench, 11 km below the surface of the Pacific Ocean is very large due to the high pressure.

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Answer:

Sulfur (Has six valence electrons). It has maximum valency due to belonging to VI groups of the Periodic Table.

Explanation:

The electrons found in an element's outermost atomic shell are known as valence electrons.

Sulfur, which has an atomic number of 16, has an electrical configuration of 2, 8, 6, meaning it has six electrons in its outermost shell. As a result, its valence electrons will also be six.

However, in its natural condition, sulfur exists as the S8 molecule, which has the classic chair structure where each sulfur atom is covalently connected to two other sulfur atoms. In that sense, there will be 8 valence electrons.

Consequently, the answer will be 6 if you're asking about the "sulphur atom," but 8 if you're talking about sulfur in general.

Thank you ,

Eddie

Aster's final position from her initial position is 63 m approximately. She will head north west direction to return to her initial position

Displacement is the distance travelled in a specific direction. It is a vector quantity.

Given that a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the direction 20° south of east, and finally walks 28 m in the direction 30° west of north.

a) Let P be the Aster's final position from her initial position?

We can use bearing by using Cosine formula to solve this question.

P² = 70² + 82² - 2 × 70 × 82 cos 73

P² = 4900 + 6724 - 11480 cos 73

P² = 11624 - 3356.43

P² = 8267.57

P = √8267.57

P = 90.9 m

P = 90.9 - 28

P = 62.9 m

We can get the angle by using Sine rule

82/ sin Ф = 90.9 / sin 73

sin Ф = 0.8627

Ф = [tex]Sin^{-1}[/tex] (0.8627)

Ф = 59.6°

Ф = 60°

b)  She will head north west direction to return to her initial position

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Answer:

The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)

Explanation:

Let [tex]u[/tex] denote the initial velocity of the vehicle ([tex]20\; \text{mph}[/tex] or [tex]60\; \text{mph}[/tex]) and let [tex]v[/tex] denote the velocity of the vehicle after braking ([tex]0\; \text{mph}[/tex]). Let [tex]x[/tex] denote the braking distance.

Assume that the acceleration during braking are both constantly [tex]a[/tex] in both scenarios. The SUVAT equations would apply. In particular:

[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}[/tex].

Since [tex]v = 0[/tex] (the vehicle has completely stopped), the equation becomes [tex]x = (-u^{2}) / (2\, a)[/tex].

Assuming that [tex]a[/tex] (braking acceleration) stays the same, the braking distance [tex]x[/tex] would be proportional to [tex]u^{2}[/tex], the square of the initial velocity.

Hence, increasing the initial speed from [tex]20\; \text{mph}[/tex] to [tex]60\; \text{mph}[/tex] would increase the braking distance by a factor of [tex]3^{2} = 9[/tex].

Answer:

9 times

Explanation:

Answer:

B

Explanation:

friction opposes motion

Friction typically Slows down objects

While this is almost true for a wide range of low speeds, as speed increases and air friction is reckoned with, it has been found that friction depends not only on speed, but also on speed squared and sometimes on higher powers of friction. speed.

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

The magnitude of the force is 94.4 N and direction of the force on it towards the right.

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

E = kq/r²

Electric field due to first charge

E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²

E1 = 1.35 x 10⁸ N/C

Electric field due to second charge

E2 =  -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²

E2 = - 1.35 x 10⁸ N/C

Electric field due to third charge

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 right.

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(a) The position of the coin placed next to the convex side of a thin spherical glass shell is 3.68 cm.

(b) The size of the image is 1.8.

The position of the coin placed next to the convex side of a thin spherical glass shell is calculated as follows;

1/d = 1/f - 1/d'

where;

Focal length = r/2 = 17 cm / 2 = 8.5 cm

1/d = 1/8.5 - (-1/6.5)

1/d = 1/8.5 + 1/6.5

1/d = 0.1176 + 0.1538

1/d = 0.2714

d = 1/0.2714

d = 3.68 cm

Magnification, M = d'/d

M = (6.5)/(3.68)

M = 1.8

Thus, the position of the coin placed next to the convex side of a thin spherical glass shell is 3.68 cm.

The size of the image is 1.8.

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