Water at 298 K discharges from a nozzle and travels horizontally hitting a flat, vertical wall. The nozzle diameter is 12 mm and the water exits the nozzle with a flat velocity profile at a velocity of 6.0 m/s. Use macroscopic balance for momentum to compute the force [N] on the wall neglecting gravitational and frictional effects

To solve this problem it is necessary to apply the kinematic equations of Energy for which the rotation of a circular body is described as

[tex]KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex]

Where,

m = Mass of the Vall

v = Velocity

I = Moment of inertia abouts its centre of mass

[tex]\omega =[/tex] Angular speed

Basically the two sums of energies is the consideration of translational and rotational kinetic energy.

a. so that it was also rotating?

The ball is rotating means that it has some angular speed:

[tex]KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex]

[tex]1000J = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex]

When there is a little angular energy (and not linear energy to travel faster), translational energy will be greater than the 1000J applied.

[tex]1000J > \frac{1}{2}mv^2[/tex]

The ball will not go faster.

c. so that it wasn't rotating?

For the case where the angular velocity does not rotate it is zero therefore

[tex]KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex]

[tex]1000J = \frac{1}{2}mv^2+\frac{1}{2}I(0)^2[/tex]

[tex]1000J = \frac{1}{2}mv^2[/tex]

All energy is transoformed into translational energy so it is possible to go faster. This option is CORRECT.

b. It makes no difference.

Although the order presented is different, I left this last option because as we can see with the previous two parts if there is an affectation regarding angular movement, therefore it is not correct.

The rotational movement of the ball doesn't contribute to its linear speed. If the same amount of energy is used, a ball, whether rotating or not, should travel at the same speed, ignoring air resistance.

The rotational movement of the ball in this case would not alter the speed of the ball's travel when projected. One reason is that only the kinetic energy associated with linear motion (or translational motion) would affect the speed. Rotational energy doesn't contribute to the ball's translational or linear speed. So, regardless of whether the ball is rotating or not if the same amount of energy (1000 J in this case) is used, the ball would travel at the same speed. Therefore, the correct option would be "It makes no difference."

However, it's worth noting that in the real world, where we cannot ignore air resistance, a spinning ball might travel a different path due to the Magnus effect.

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Answer:

C. less than 950 N.

Explanation:

Given that

Force in north direction F₁ = 500 N

Force in the northwest F₂ = 450 N

Lets take resultant force R

The angle between force = θ

θ = 45°

The resultant force R

[tex]R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}[/tex]

[tex]R=\sqrt{500^2+450^2+2\times 450\times 500\times cos\theta}[/tex]

R= 877.89 N

Therefore resultant force is less than 950 N.

C. less than 950 N

Note- When these two force will act in the same direction then the resultant force will be 950 N.

The total force on the stump is less than the sum of the individual forces because the forces are vectors and not aligned in the same direction. The correct answer is (c) less than 950 N.

When you pull on a stump with two ropes at different angles, the total force is not simply the sum of the individual forces because the forces are vectors. To find the total force, you must consider both the magnitude and direction of each force. In this case, you pull with a force of 500 N to the north and your friend pulls with 450 N to the northwest. Since these forces are not in the same direction, you need to use vector addition to find the resultant force. Without calculating the exact value, we can already state that the resultant force will be less than 950 N because part of the forces cancel each other out due to the angle at which they are applied. Therefore, the correct answer is (c) less than 950 N.

Answer:

Fundamental frequency= 174.5 hz

Explanation:

We know

fundamental frequency=[tex]\frac{velocity}{2 *length}[/tex]

velocity =[tex]\sqrt{\frac{tension}{mass per unit length} }[/tex]

mass per unit length=[tex]\frac{3.5}{1000*1.22}[/tex]=0.00427[tex]\frac{kg}{m}[/tex]

Now calculating velocity v=[tex]\sqrt{\frac{255}{0.00427} }[/tex]

                                           =244.3[tex]\frac{m}{sec}[/tex]

Distance between two nodes is 0.7 m.

Plugging these values into to calculate frequency

f = [tex]\frac{244.3}{2 *0.7}[/tex] =174.5 hz

Answer

given,

mass of an object = 5 Kg

mass of the other object = 5.98 x 10²⁴ Kg

radius of earth = 6371 Km = 6.371 x 10⁶ m

comparison between gravitational force between two object and weight of the object.

gravitational force between two objects

   [tex]F = \dfrac{Gm_1m_2}{r^2}[/tex]

   [tex]F = \dfrac{6.67 \times 10^{-11}\times 5 \times 5.98 \times 10^{24}}{(6.371 \times 10^6)^2}[/tex]

      F = 49.134 N

weight of the object on the earth surface

  F = m g

  F = 5 x 9.81

  F = 49.05 N

From the above two calculation we can conclude that gravitational force is approximately equal to weight of the body.

Answer:

C. The object will slow down, momentarily stopping, then pick up speed moving to the left.

Explanation:

Given that

Speed of the object = 10 m/s ( right direction)

Acceleration ,a=  - 2 m/s

A negative sign shows that acceleration is in the opposite direction to the speed.

We know that if speed and acceleration is in the opposite direction then the final speed of the objects becomes zero. for a moment and it will increase in the opposite direction.

Therefore the answer is C

The correct statement is c. The object will slow down, momentarily stop, then pick up speed moving to the left. This is due to the object's negative acceleration, which means it is slowing down.

In this physics problem, an object is moving to the right with an initial velocity of 10 m/s and an acceleration of -2.0 m/s2. Negative acceleration, also known as deceleration, doesn't mean the object is moving in a negative direction, rather it is slowing down. The correct statement is c. The object will slow down, momentarily stopping, then pick up speed moving to the left. This is because as time progresses, the negative acceleration reduces the speed of the object until it comes to a stop, after which it starts moving in the opposite direction, towards the left.

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Answer:

τext =(M*R*g)*(m₂ - m₁) / (m₁ + m₂)

The wheel is rotating clockwise. Its vector direction is - k  which is perpendicular to the plane of the wheel.

Explanation:

Given info

m₁ = mass of block 1

m₂ = mass of block 2

m₁ < m₂

M = Mass of the wheel

R = Radius of the wheel

I = Moment of inertia of the wheel = M*R²

We assume that the wheel is rotating clockwise since m₁ < m₂

In order to get an expression for the net external torque acting on the system we can apply the following equation

τext = I*α

where α is the angular acceleration of the wheel which can be found as follows

at = R*α   ⇒    α = at / R

then we have to compute the acceleration of the system (at), using the Newton's 2nd Law

Block 1:

∑Fy = m₁*at  (↑+)  ⇒    T - m₁*g = m₁*at    (I)

Block 2:

∑Fy = m₂*at  (↓+)  ⇒    - T + m₂*g = m₂*at    (II)

If  we apply  

(I) + (II)  ⇒        at = g*(m₂ - m₁) / (m₁ + m₂)

Now, we get α:

α = at / R = (g*(m₂ - m₁) / (m₁ + m₂)) / R

⇒     α = (g / R)*(m₂ - m₁) / (m₁ + m₂)

Finally

τext = I*α = (M*R²)*((g / R)*(m₂ - m₁) / (m₁ + m₂))

⇒   τext =(M*R*g)*(m₂ - m₁) / (m₁ + m₂)

The wheel is rotating clockwise. Its vector direction is - k  which is perpendicular to the plane of the wheel.

Answer:

Answer:u=66.67 m/s

Explanation:

Given

mass of meteor m=2.5 gm\approx 2.5\times 10^{-3} kg

velocity of meteor v=40km/s \approx 40000 m/s

Kinetic Energy of Meteor

K.E.=\frac{mv^2}{2}

K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}

K.E.=2\times 10^6 J

Kinetic Energy of Car

=\frac{1}{2}\times Mu^2

=\frac{1}{2}\times 900\times u^2

\frac{1}{2}\times 900\times u^2=2\times 10^6  

900\times u^2=4\times 10^6

u^2=\frac{4}{9}\times 10^4

u=\frac{2}{3}\times 10^2

u=66.67 m/s

Answer:

Half as large.

Explanation:

Using Newton's law of universal gravitation, if the mass of the planet is M and of the Moons 1 and 2 is m, them the force exerted by the planet on them will be:

[tex]F_1=\frac{GMm}{r}[/tex]

[tex]F_2=\frac{GMm}{2r}[/tex]

Which clearly shows that the force that the planet exerts on the Moon 2 is half  the force it exerts on the Moon 1.

The gravitational force exerted by the planet on Moon 2, which orbits twice as far from the planet as Moon 1, is one-fourth as large as the force exerted on Moon 1 due to the inverse-square law of gravity.

This question is about the gravitational force that a planet exerts on its moons. Each moon has an identical mass but they differ in their distances from the planet - the radius of their orbits. One moon orbits at a radius of r, while the second moon orbits at a radius of 2r.

The gravitational force that a planet exerts on an object is inversely proportional to the square of the distance between the center of the planet and the object. So, if you double the distance, the gravitational force becomes one-fourth as large. Therefore, the magnitude of the gravitational force exerted by the planet on Moon 2, which is twice as far from the planet as Moon 1, is one-fourth as large as the gravitational force exerted by the planet on Moon 1.

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Answer:

Initial speed of car = 25 m/s north direction

correct option is A. 25 m/s

Explanation:

given data

Final velocity of both = 28 m/s at 27 degrees north of east

to find out

how fast was the car initially traveling

solution

we consider both vehicle have same mass

mass of car  m1  = mass of truck m2

so we will apply here conservation of momentum  law

m1 × u1 j + m2 × u2 i  = (m1 + m2) × ( v cosθ i + v sinθ j )      ...............1

here m is mass and u is initial velocity and v is final velocity

m × u1 j + m × u2 i  = (2m) × ( 28 × cos27 i + 28 × sin27 j )  

u2 i + u1 j = 2 ( 24.94 i + 12.71 j)

u2 i + u1 j = 49.88 i + 25.42 j

so compare both side we get

Initial speed of car u1 = 25.42 m/s north direction

To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description.

From the definition we know that the speed of a body can be described as a function of gravity and height

[tex]V = \sqrt{2gh}[/tex]

[tex]V = \sqrt{2*9.8*0.15}[/tex]

[tex]V = 1.714m/s[/tex]

Then applying the kinematic equation of displacement, the height can be written as

[tex]H = \frac{1}{2}gt^2[/tex]

Re-arrange to find t,

[tex]t = \sqrt{2\frac{h}{g}}[/tex]

[tex]t = \sqrt{2\frac{0.5}{9.8}}[/tex]

[tex]t = 0.3194s[/tex]

Thus the calculation of the displacement would be subject to

[tex]x = vt[/tex]

[tex]x =1.714*0.3194[/tex]

[tex]x = 0.547m[/tex]

Therefore the required distance must be 0.547m

To solve the exercise it is necessary to apply the equations necessary to apply Newton's second law and the concept related to frictional force.

An angle of 30 degrees is formed on the vertical at an applied force of 2.3N

In this way the frictional force, opposite to the movement will be given by

[tex]f_k = \mu_k N[/tex]

where,

[tex]\mu_k =[/tex] Kinetic friction constant

N = Normal Force (Mass*gravity)

The friction force is completely vertical and opposes the rising force of 2.3 N. The Normal force acts perpendicular to the surface (vertical) therefore corresponds to the horizontal component of the applied force.

The ascending force would be given by

[tex]F_v = 2.3N*Cos30 = 1.99N[/tex]

As the block is moving upward, the friction force acts downward, also its weight acts downward. We can have

[tex]2.3N+f_k = 1.99N[/tex]

[tex]f_k = 0.31N[/tex]

Considering the horizontal force the normal force on the block must be balanced by the horizontal component of pishing foce

[tex]N = 2.3sin30[/tex]

[tex]N = 1.15N[/tex]

Then the frictional force

[tex]f_k = \mu_k N[/tex]

[tex]0.31N = \mu_k 1.15[/tex]

[tex]\mu_k = \frac{0.31}{1.15}[/tex]

[tex]\mu_k =0.26[/tex]

Therefore the coefficient of kinetic friction is 0.26

Answer:

Explanation:

a)

Using Ohms Law

[tex]R= \rho \frac{l}{\pi a^2 } \\V = IR = E l = I \rho \frac{l}{\pi a^2 } \\E = \frac{I \rho}{\pi a^2 } \\E = J \rho [/tex]

Where J is the current density [tex]J = \frac{I}{\pi a^2 } [/tex]

and the direction of E is the same as the direction of the current. Since J is uniform throughout the conductor [tex]E = \rhoJ[/tex] just inside at a radius a (and anywhere else).

b)

Since we have no changing electric fields we can use Ampere’s law in it’s simplest form without displacement current

[tex]\oint B .dl = B 2 \pi a = \mu_{o} I [/tex]

such that

[tex]B = \frac{\mu_{o} I}{2 \pi a } [/tex]

and by the right hand rule, since the current is going to the right, the magnetic field is circling around the conductor such that it’s pointing out of the page at the top and into the page at the bottom.

c)

The Poynting vector is given by

[tex]S = \frac{1}{ \mu_o} |E \textrm{x}B| = \frac{\rho I^2}{2 \pi^2 a^3} [/tex]

and by the right hand rule it’s always pointing in towards the center of the conductor.

d)

Note: directions of these three vectors are mentioned along with their magnitudes in above 3 parts a , b and c

a) The magnitude of the electric field vector E at a point just inside the wire at a distance a from the axis is given by E = I / (2*pi*a*ε0). b) The magnitude of the magnetic field vector B at the same point is given by B = (μ0 * I)/(2*pi*a). c) The magnitude of the Poynting vector S at the same point is given by S = (I^2 * μ0) / (4 * pi * a * ε0). d) The electric field vector E is radially outward, the magnetic field vector B is tangent to circles centered on the conductor, and the Poynting vector S is in the same direction as the cross product of E and B.

a) The magnitude of the electric field vector E at a point just inside the wire at a distance a from the axis can be determined using Gauss's Law. Since the conductor is cylindrical and the current is constant, the electric field is uniform and given by E = I / (2*pi*a*ε0), where I is the current and ε0 is the permittivity of free space.

b) According to Ampère's Law, the magnitude of the magnetic field vector B at the same point is given by B = (μ0 * I) / (2 * pi * a), where μ0 is the permeability of free space.

c) The magnitude of the Poynting vector S at the same point is given by S = E x B = (I2 * μ0) / (4 * pi * a * ε0).

d) The direction of the electric field vector E is radially outward, away from the axis of the conductor. The magnetic field vector B is tangent to concentric circles centered on the conductor. The Poynting vector S is in the same direction as the cross product of E and B.

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Using conservation of momentum and the work-energy principle, the initial speed of the bullet can be calculated. The collision is inelastic because kinetic energy is not conserved. The impulse on the block is the change in momentum, calculated from the initial combined momentum to zero when it stops.

Initial Speed of the Bullet

To find the initial speed of the bullet, we use the principle of conservation of momentum for the collision and the work-energy principle for the block's movement after the collision.

For the conservation of momentum:

After the collision, the bullet and block move together, so the total final momentum is the combined mass times their velocity.

The normal force is equal to the weight of the block since the surface is horizontal.

Kinetic energy can then be equated to this work to find the velocity right after the collision.

Type of Collision

The initial momentum right after impact is the combined mass of the block and bullet times the velocity just after collision, which we obtained earlier.

Analyzing the result, we can discuss the magnitude of impulse and how it relates to the force of friction and the distance slid.

To solve this problem it is necessary to apply the concepts related to the kinetic energy expressed in terms of simple harmonic movement, as well as the concepts related to angular velocity and acceleration and linear acceleration and velocity.

By definition we know that the angular velocity of a body can be described as a function of mass and spring constant as

[tex]\omega = \sqrt{\frac{k}{m}}[/tex]

Where,

k = Spring constant

m = mass

From the given values the angular velocity would be

[tex]\omega = \sqrt{\frac{277}{0.4}}[/tex]

[tex]\omega = 26.31rad/s[/tex]

The kinetic energy on its part is expressed as

[tex]E = \frac{1}{2} m\omega^2A^2[/tex]

Where,

A = Amplitude

[tex]\omega[/tex] = Angular Velocity

[tex]m = Mass[/tex]

PART A) Replacing previously given values the energy in the system would be

[tex]E = \frac{1}{2} m\omega^2A^2[/tex]

[tex]E = \frac{1}{2} (0.4)(26.31)^2(3*10^{-2})^2[/tex]

[tex]E= 0.1245J[/tex]

PART B) Through the amplitude and angular velocity it is possible to know the linear velocity by means of the relation

[tex]v = A\omega[/tex]

[tex]v = (3*10^{-2})(26.31)[/tex]

[tex]v = 0.7893m/s[/tex]

PART C) Finally, the relationship between linear acceleration and angular velocity is subject to

[tex]a = A\omega^2[/tex]

[tex]a = (3*10^{-2})(26.31)^2[/tex]

[tex]a = 20.76m/s^2[/tex]

(a) The mechanical energy of the system is 1.245 J (b) The maximum speed of the oscillating mass is 0.6225 m/s (c) The magnitude of the maximum acceleration of the oscillating mass is 20.775m/s².

(a) The mechanical energy of a mass†“spring system in simple harmonic motion (SHM) is given by the equation:

[tex]\[ E = \frac{1}{2}kA^2 \][/tex]

where [tex]$k$[/tex] is the spring constant and [tex]$A$[/tex]is the amplitude of oscillation. Plugging in the given values:

[tex]\[ E = \frac{1}{2} \times 277 \, \text{N/m} \times (0.03 \, \text{m})^2 \][/tex]

[tex]\[ E = \frac{1}{2} \times 277 \times 0.0009 \][/tex]

[tex]\[ E = 1.245 \, \text{J} \][/tex]

(b) The maximum speed of the mass occurs at the equilibrium position (when the displacement is zero) and is given by:

[tex]\[ v_{max} = A\sqrt{\frac{k}{m}} \][/tex]

where[tex]$m$[/tex] is the mass of the object. Using the given values:

[tex]\[ v_{max} = 0.03 \, \text{m} \times \sqrt{\frac{277 \, \text{N/m}}{0.40 \, \text{kg}}} \][/tex]

[tex]\[ v_{max} = 0.03 \times \sqrt{692.5} \][/tex]

[tex]\[ v_{max} = 0.03 \times 27.0676 \][/tex]

[tex]\[ v_{max} = 0.6225 \, \text{m/s} \][/tex]

(c) The maximum acceleration of the mass occurs at the maximum displacement (amplitude) and is given by:

[tex]\[ a_{max} = \frac{F_{max}}{m} \][/tex]

where [tex]$F_{max}$[/tex]is the maximum force exerted by the spring, which is [tex]$kA$[/tex]. Thus:

[tex]\[ a_{max} = \frac{kA}{m} \][/tex]

Using the given values:

[tex]\[ a_{max} = \frac{277 \, \text{N/m} \ times 0.03 \, \text{m}}{0.40 \, \text{kg}} \][/tex]

[tex]\[ a_{max} = \frac{8.31}{0.40} \][/tex]

[tex]\[ a_{max} = 20.775 \, \text{m/s}^2 \][/tex]

Answer: 0.999959 c

Explanation:  

According to the special relativity theory, time is measured differently by two observers moving one relative another, according to the Lorentz Transform Equation, as follows:

t = t’ / t=t^'/√(1-(v)2/c2 )

where t= time for the moving observer (relative to the spacecraft, fixed on Earth) = 110 years.

t’= time for the observer at rest respect from spacecraft = 1 year

v= spacecraft constant speed

c= speed of light

Solving for v, with a six decimals precision as a multiple of c, we get:

v = 0.999959 c

Answer:

a)  ΔL/L = F / (E A),  b)   [tex]L_{f}[/tex] = L (1 + L F /(EA) )

Explanation:

Let's write the formula for Young's module

     E = P / (ΔL / L)

Let's rewrite the formula, to have the pressure alone

    P = E ΔL / L

The pressure is defined as

    P = F / A

Let's replace

   F / A = E ΔL / L

   F = E A ΔL / L

   ΔL / L = F / (E A)

b) To calculate the elongation we must have the variation of the length, so the length of the bar must be a fact. Let's clear

    ΔL = L [F / EA]

    [tex]L_{f}[/tex] -L = L (F / EA)

    [tex]L_{f}[/tex] = L + L (F / EA)

    [tex]L_{f}[/tex] = L (1 + L (F / EA))

Answer:

(a) Tangential velocity will be 38.648 m/sec

(b) Acceleration will be [tex]133.617m/sec^2[/tex]

Explanation:

We have given radius r = 11.2 m

Angular speed [tex]\omega =0.550rev/sec=0.550\times 2\pi =3.454rad/sec[/tex]

(a) We have to find the tangential velocity

We know that tangential velocity is given by  

[tex]v_t=\omega r=3.454\times 11.2=38.684m/sec[/tex]

(b) We know that acceleration is given by

[tex]a=\frac{v^2}{r}=\frac{38.684^2}{11.2}=133.617m/sec^2[/tex]

To solve this problem it is necessary to apply the Discharge of flow equations.

From the theory the flow rate is defined as

Q = AV

Where,

A =Area

V = Velocity

PART A) The question is telling us about the total fluid flow rate then

[tex]Q_T = Q_1+Q_2+Q_3[/tex]

[tex]Q_T = 27+19+12[/tex]

[tex]Q_T = 58L/min[/tex]

PART B) The radius would be given between another pipe with a flow rate of 27L / min.

For proportionality ratio we have to

[tex]\frac{Q_T}{Q'} = \frac{A_TV_T}{A'V'}[/tex]

[tex]\frac{V_T}{V'} = \frac{A_'Q_T}{A_TQ'}[/tex]

[tex]\frac{V_T}{V'} = \frac{(\pi r_T^2)Q_T}{(\pi r'^2)Q'}[/tex]

[tex]\frac{V_T}{V'} = \frac{1.2*^2 58}{1.8^2 27}[/tex]

[tex]\frac{V_T}{V'} = 0.9547[/tex]

Answer:

230 Volts

Explanation:

Emf of a solenoid is given by

[tex]E=-N\frac{d\phi}{dt}[/tex]

Where,

N = Number of turns

[tex]\phi[/tex] = Magnetic flux

t = Time taken

The negative sign denotes the direction of flow or polarity.

It can be seen that the induce emf is proportional to the number of turns.

So, if the initial voltage was 10 volts at a set number of turns then at 23 volts the emf will be

[tex]E\times N=23\times 10=230\ V[/tex]

230 Volts

The voltmeter would read 230 volts with a coil of 23 turns. This is calculated using Faraday's law of electromagnetic induction, tying in the principles of EMF produced per turn and the number of turns.

The subject of the question is Electromagnetic Induction in Physics. Based on Faraday's law of electromagnetic induction, the electromotive force (emf) induced in a circuit is directly proportional to the rate of change of magnetic flux through the loop. When you use a single wire, the voltmeter reads 10 volts. This means, that in the case of a single-turn coil, the change in flux induces an emf of 10 volts.

If the coil has more turns, then the total emf is equal to the emf produced per turn multiplied by the number of turns. In this case, if we use 23 turns instead of 1, the total emf becomes 10 volts * 23 turns = 230 volts. Thus, the voltmeter would read 230 volts.

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Answer:

The power in this flow is [tex]9.56\times10^{8}\ W[/tex]

Explanation:

Given that,

Distance = 221 m

Power output = 680 MW

Height =150 m

Average flow rate = 650 m³/s

Suppose we need to calculate the power in this flow in watt

We need to calculate the pressure

Using formula of pressure

[tex]Pressure=\rho g h[/tex]

Where, [tex]\rho[/tex]= density

h = height

g = acceleration due to gravity

Put the value into the formula

[tex]Pressure=1000\times9.8\times150[/tex]

[tex]Pressure=1470000\ Nm^2[/tex]

We need to calculate the power

Using formula of power

[tex]P=Pressure\times flow\ rate[/tex]

Put the value into the formula

[tex]P=1470000\times650[/tex]

[tex]P=9.56\times10^{8}\ W[/tex]

Hence, The power in this flow is [tex]9.56\times10^{8}\ W[/tex]

Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating  system = 6.74 J

We know that energy is given by

(a) [tex]Ke=\frac{1}{2}mv_{max}^2[/tex]

[tex]6.74=\frac{1}{2}\times 0.267\times v_{max}^2[/tex]

[tex]v_{max}=7.1m/sec[/tex]

(b) Now [tex]\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec[/tex]

We know that [tex]\omega =\sqrt{\frac{k}{m}}[/tex]

[tex]35.68=\sqrt{\frac{k}{0.267}}[/tex]

[tex]k=339.9N/m[/tex]

(c) We know that energy is given by

[tex]E=\frac{1}{2}KA^2[/tex]

[tex]6.74=\frac{1}{2}\times 339.9\times A^2[/tex]

[tex]A=19.91cm[/tex]

Answer:

vf₁  =  15.29 cm/s : to the right

vf₂ =  25.29 cm/s : to the right

Explanation:

Theory of collisions  

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:  

p=m*v  

where  

p:Linear momentum  

m: mass  

v:velocity  

There are 3 cases of collisions : elastic, inelastic and plastic.  

For the three cases the total linear momentum quantity is conserved:  

P₀ = Pf Formula (1)  

P₀ :Initial linear momentum quantity  

Pf : Final linear momentum quantity  

Data

m₁= 11 g : mass of  object₁

m₂= 24 g  : mass of  object₂

v₀₁ = 29 cm/s , to the right : initial velocity of m₁

v₀₂=  19 cm/s, to the right  i :initial velocity of m₂

Problem development

We appy the formula (1):

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂  

We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:

( 11)*( 29) + (24 )*(19) = ( 11)*vf₁ +(24)*vf₂

775 = ( 11)*vf₁ +(24)*vf₂ Equation (1)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

[tex]e = \frac{v_{f2} -v_{f1}}{v_{o1} -v_{o2}}[/tex]

1*(v₀₁ - v₀₂ )  = (vf₂ -vf₁)

(29 -  19 )  = (vf₂ -vf₁)

10 =  (vf₂ -vf₁)

vf₂ = 10 + vf₁  Equation (2)

We replace Equation (2) in the Equation (1)

775 = ( 11)*vf₁ +(24)*(10 + vf₁ )

775 = ( 11)*vf₁ +240+(24) vf₁

775 - 240= ( 35)*vf₁

535 = ( 35)*vf₁

vf₁  = 535 / 35

vf₁  =  15.29 cm/s : to the right : Final velocity of object₁  

We replace vf₁  =  15.29  cm/s in the Equation (2)

vf₂ = 10 + vf₁

vf₂ =10 +  15.29

vf₂ =  25.29 cm/s, to the right: Final velocity of object₂

Answer:

[tex]v_1=-2.616\ m/s[/tex]

Explanation:

Given that,

Mass of ball 1, [tex]m_1=0.25\ kg[/tex]

Initial speed of ball 1, [tex]u_1=5\ m/s[/tex]

Mass of ball 2, [tex]m_2=0.8\ kg[/tex]

Initial speed of ball 2, [tex]u_2=0[/tex] (at rest)

After the collision,

Final speed of ball 2, [tex]v_2=2.38\ m/s[/tex]

Let [tex]v_2[/tex] is the final speed of ball 1.

Initial momentum of the system is :

[tex]p_i=m_1u_1+m_2u_2[/tex]

[tex]p_i=0.25\times 5+0[/tex]

[tex]p_i=1.25\ m/s[/tex]

Final momentum of the system is :

[tex]p_f=m_1v_1+m_2v_2[/tex]

[tex]p_f=0.25\times v_1+0.8\times 2.38[/tex]

[tex]p_f=0.25 v_1+1.904[/tex]

According the law of conservation of linear momentum :

initial momentum = final momentum

[tex]1.25=0.25 v_1+1.904[/tex]

[tex]v_1=-2.616\ m/s[/tex]

So, the final velocity of ball 1 is (-2.616)m/s.

Answer:

mass of ball 1= 0.25 kg

initial speed of ball 1= 5. m/s

mass ball 2= 0.8 kg

initial speed ball 2= 0 ( it was at rest)

final speed of ball 2= 2.38m/s

the formula is:

m1 . v 1 + m2 . v2 = m1 . v1f+ m2 . v2f

0.25 . 5 + 0.8 . 0 = o.25 . x + 0.8 . 2.38

-2.6288 m/s

Explanation:

i got it right on the flvs

All the devices listed namely the Prism, Transmission Grating, and Reflection Grating, are applicable in separating different wavelengths emitted by a light source during an experiment. They operate on the principles of dispersion and diffraction.

To answer your question, all the mentioned devices - Prism, a Transmission Grating and a Reflection Grating can be used to separate the different wavelengths emitted by a light source in the experiment. Prisms and diffraction gratings operate on the principle of dispersion and diffraction respectively, which is the separation of light into its constituent colors (wavelengths). For example, when white light, which is made up of multiple wavelengths, is passed through a prism or a grating, it is separated into its individual wavelengths, producing a spectrum of colors.

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In the experiment, a prism, a transmission grating, or a reflection grating can be used to separate different wavelengths of light. They all operate on principles of dispersion, refraction, interference, and diffraction to achieve this. Hence, the correct answer is D. All of the above.

In the experiment described, the task of separating different wavelengths of light emitted by the source can be accomplished by any of the following: A. Prism, B. A transmission grating, or C. A reflection grating. All of these devices operate on the principle of dispersion and diffraction, respectively. They split light into different constituent wavelengths or colors, similar to how a rainbow is formed. The distinguishing factor is how they do this. A prism uses refraction to disperse light. Transmission and reflection gratings, on the other hand, use interference and diffraction to separate the light spectrally. Therefore, the answer is D. All of the above.

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Rowan's feet appear closer than usual when they are under 0.60 m of water due to refraction, making the apparent depth approximately 0.45 m.

When Rowan is looking at her feet under 0.60 m of water, they appear closer to her than usual due to the phenomenon of refraction. Refraction occurs because light travels at different speeds in different mediums. When light passes from water to air, it speeds up and changes direction. This bending of light causes objects under water to look closer to the surface than they really are.

For apparent depth (how far away her feet appear), we can use Snell's law of refraction, which states that n1 × sin(θ_1) = n2 × sin(θ_2), where n1 and n2 are the refractive indices of water and air, respectively, and θ_1 and θ_2 are the angles of incidence and refraction.

Knowing that the refractive index of water is approximately 1.33 and that of air is 1 (approximating a normal incidence angle where sin(θ_1) ≈ sin(θ_2)), we get an apparent depth d’ = d/n, where d is the real depth (0.60 m). Thus, the apparent depth of Rowan's feet would be 0.60 m / 1.33 ≈ 0.45 m.

Answer:

Explanation:

Atmospheric pressure = 7 x 10⁴ Pa

force on  a disk-shaped region 2.00 m in radius at the surface of the ocean due to atmosphere  = pressure x area

= 7 x 10⁴ x 3.14 x 2 x 2

= 87.92 x 10⁴ N

b )

weight, on this exoplanet, of a 10.0 m deep cylindrical column of methane with radius 2.00 m

Pressure x area

height x density x acceleration of gravity x π r²

= 10 x 415 x 6.2 x 3.14 x 2 x 2

=323168.8 N

c ) Pressure at a depth of 10 m

atmospheric pressure + pressure due to liquid column

= 7 x 10⁴ + 10 x 415 x 6.2 ( hρg)

= 7 x 10⁴ + 10 x 415 x 6.2

(7 + 2.57 )x 10⁴ Pa

9.57 x 10⁴ Pa

The Force exerted by the atmosphere on a disk-shaped area is 8.8 x 10^5 N. The weight of a 10 m cylindrical column of methane is 1.31 x 10^7 N. The pressure at a depth of 10 m in the methane ocean is 1.22 x 10^5 Pa.

To answer this question, we need to apply concepts of fluid pressure and buoyancy.

(a) The force exerted by the atmosphere on a disk-shaped region is given by the atmospheric pressure multiplied by the area of the disk. This is calculated as Force = Pressure x Area. The Area of the disk is given by πr², where r is the radius. Therefore, the force = (7.00 x 10^4 Pa) x π(2 m)² = 8.8 x 10^5 N.

(b) The weight of the methane column can be found by using the formula for the weight of a liquid column which is Weight = density x gravity x height x Area. Plugging in the values from the question, it comes to (415 kg/m³) x (6.20 m/s²) x (10 m) x π(2 m)² = 1.31 x 10^7 N.

(c) The pressure at a depth in the methane ocean can be calculated using the equation P = Po + pgd, where Po is atmospheric pressure, p is the density of the fluid, g is the acceleration due to gravity and d is the depth. Substituting the given values, P = 7.00 x 10^4 Pa + (415 kg/m³)(6.20 m/s²)(10 m) = 1.22 x 10^5 Pa.

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Answer:

0.25 kg m^2

Explanation:

mass of each , m = 500 g = 0.5 kg

distance, r = 50 cm = 0.5 m

Moment of inertia about the axis passing through one corner and perpendicular to the plane of triangle

I = mr^2 + mr^2

I = 2 mr^2

I = 2 x 0.5 x 0.5 x 0.5

I = 0.25 kgm^2

Answer:

θ= 5 radian

Explanation:

Given data:

Radius r = 0.70 m

Initial angular speed ω_i = 2rev/s

Time t = 5 s

Final angular speed ω_f =0

so we have angular displacement

[tex]\theta= \frac{\omega_f-\omega_i}{2}\times t[/tex]

putting values

[tex]\theta= \frac{0-2}{2}\times5[/tex] = 5 rad

The bicycle wheel rotated through 5 revolutions during the 5 seconds of braking, calculated using the initial angular velocity and time to find the angular deceleration and the total number of revolutions.

To find out through how many revolutions the bicycle wheel rotated during the 5.0 seconds of braking, we first need to know the angular deceleration. The initial angular speed is 2.0 rev/s, and the wheel comes to stop in 5.0 s. We can calculate the angular deceleration as follows:

Initial angular speed, ω0 = 2.0 rev/s

Final angular speed, ω = 0 rev/s (since the bicycle stops)

Time, t = 5.0 s

Angular acceleration, α = (ω - ω0) / t = (0 - 2.0) / 5.0 = -0.4 rev/s²

Now, we can find the total number of revolutions by integrating the angular speed over the time:

Total revolutions, θ = ω0 * t + (1/2) * α * t² = 2.0 * 5.0 + (1/2) * (-0.4) * (5.0)² = 10 - 5 = 5 revolutions

The wheel therefore rotates through a total of 5 revolutions during the breaking period.

To solve this problem we must rely on the concepts related to wavelength in terms of the speed of light and frequency.

Mathematically one of the ways to express the wavelength is,

[tex]\lambda = \frac{c}{f}[/tex]

Where,

c = Speed of light [tex](3*10^8m/s)[/tex]

f = Frequency

The value of the frequency given is

f = 45Hz

Therefore replacing we can find it,

[tex]\lambda = \frac{3*10^8}{45Hz}[/tex]

[tex]\lambda = 6.6667*10^6m[/tex]

Hence the wavelength of the very low frequency electromagnetic wave is [tex]6.6667*10^6m[/tex]

Final answer:

The wavelength of this very-low-frequency electromagnetic wave is approximately 6.67 x 10^6 meters.

Explanation:

The wavelength (λ) of an electromagnetic wave can be calculated using the formula:

λ = c/f

Where c is the speed of light, approximately 3 x 10^8 m/s, and f is the frequency of the wave. In this case, the frequency is 45 Hz. Substituting the values into the formula, we get:

λ = 3 x 10^8 m/s / 45 Hz = 6.67 x 10^6 m

Therefore, the wavelength of this very-low-frequency electromagnetic wave is approximately 6.67 x 10^6 meters.

To develop the problem it is necessary to apply the concepts related to Hooke's Law. For which it is understood that

[tex]F = kx[/tex]

Where,

k = Spring constant

x = Displacement of Spring

By the theorem of work we understand that it is the force realized when there is the path of an object therefore

[tex]dW =Fdx[/tex]

Our value of F is

[tex]dW = (kx)dx[/tex]

Integrating

[tex]W = k\int\limit_0^{x_1} xdx[/tex]

Therefore the expresion is

[tex]W = \frac{kx_1^2}{2}[/tex]