A transformer has 18 turns of wire in its primary coil and 90 turns in its secondary coil. An alternating voltage with an effective value of 110 V is applied to the primary coil. At the secondary coil, an alternating voltage with an effective value of 550 V is obtained. A current of 29 A is supplied to the primary coil of the transformer. Calculate the maximum effective current in the secondary coil. The maximum effective current in the secondary coil is A.

Answer:

we have to have many atomicen the excited state and remove all possible atoms from the base state to maintain the population difference between the states and to maintain the emission.

Explanation:

In quantum mechanics the stable state of matter is the fundamental state (lower maser) where the molecular will accumulate, we must have a mechanism to remove the molecules from this state and raise them to the higher state. In the upper state, due to the uncertainty principle, they can only this certain time before decaying. In this process of decay we have two types of emissions: spontaneous and stimulated.

With the spontanease emission they produce the first photons, which stimulate the stimulated emission, which is proportional to the number of atoms in the excited state (higher maser) whereby the more atoms there in this state the emission stimulated in much greater than the spontaneous one that It is approximately constant. The above is the beginning of all lasers.

In summary, we have to have many atomicen the excited state and remove all possible atoms from the base state to maintain the population difference between the states and to maintain the emission.

The blissful feelings experienced in the early stages of a relationship, characterized by infatuation, are primarily attributed to options a. and c.

Significantly elevated levels of dopamine and norepinephrine in the brain contribute to the intense emotional highs associated with infatuation.

Dopamine is linked to pleasure and reward, while norepinephrine is associated with excitement and arousal, creating a euphoric state.

Additionally, the novelty of the situation plays a crucial role. New experiences, emotions, and interactions with a romantic partner stimulate the brain's reward system, enhancing feelings of happiness and exhilaration.

Therefore, the combination of heightened neurotransmitter levels and the novelty of the relationship fosters the intense and blissful emotions characteristic of infatuation during the early stages of a romantic relationship.

The probable question may be:

The blissful feelings experienced in the early stages of a relationship, characterized by infatuation, are primarily attributed to:

a. Significantly elevated levels of dopamine and norepinephrine in your brain.

b. A favorable line-up of the moon and the stars.

c. The novelty of the situation.

d. Options a. & c.

Answer:

Speed, v = 45.84 m/s

Explanation:

Given that,

Radius of the curve, r = 100 m

Banking of the curve, [tex]\theta=65^{\circ}[/tex]

On the banking of curve, the speed of the vehicle is given by :

[tex]v=\sqrt{rg\ tan\theta}[/tex]

[tex]v=\sqrt{100\times 9.8\times \ tan(65)}[/tex]

v = 45.84 m/s

So, the speed at which a 100 m radius curve banked is 45.84 m/s. Hence, this is the required solution.

Final answer:

The ideal speed to navigate a 100 m radius curve banked at 65 degrees on a frictionless surface is calculated to be approximately 45.8 m/s or about 165 km/h using the principles of centripetal force.

Explanation:

When considering ideal speeds to take a steeply banked tight curve like the ones at the Daytona International Speedway, the scenario assumes a frictionless surface. The goal is to calculate the speed at which a vehicle should navigate a 100 m radius curve banked at 65 degrees. Utilizing the centripetal force equation, which in this case is provided by the component of the gravitational force, we can set this up as follows:

tan(θ) = v² / (rg)

Where θ is the angle of banking (65.0 degrees), v is the velocity, r is the radius of the curve (100 m), and g is the acceleration due to gravity (9.8 m/s²).

Rearranging the formula to solve for velocity, we get:

v = √(tan(θ) * r * g)

Plugging in the values:

v = √(tan(65.0 degrees) * 100 m * 9.8 m/s²)

v ≈ √(2.14 * 100 m * 9.8 m/s²) = √(2097.2 m²/s²)

v ≈ 45.8 m/s (which is about 165 km/h)

Answer:

ΔL = - 0.017556 m

Explanation:

Given

L₀ = 19 m

T₀ = 41ºC

T₁ = -3ºC

α = 21*10⁻⁶ °C⁻¹

ΔL = ?

We can use the equation

ΔL = L₀*α*ΔT

where

ΔT = T₁ - T₀ = -3ºC - 41ºC

⇒  ΔT = -44ºC

then

ΔL = (19 m)*(21*10⁻⁶ °C⁻¹)*(-44ºC)

⇒  ΔL = - 0.017556 m

Final answer:

The metal wire will experience a change in length of -0.017 m when cooled from 41 to -3 °C.

Explanation:

To calculate the change in length of the metal wire, we can use the equation for linear thermal expansion, AL = aLAT, where AL is the change in length, a is the coefficient of thermal expansion, L is the initial length of the wire, and AT is the change in temperature.

In this case, the initial length of the wire is 19 m, the coefficient of thermal expansion is 21 x 10^-6 (°C)^-1, and the change in temperature is 41 - (-3) = 44°C.

Using the equation, AL = (21 x 10^-6)(19)(44) = 0.017 mL, where mL represents meters. Since the change in length is negative (the wire cools and shrinks), the result is -0.017 m.

Answer:

450 J

5.88235 m/s

Explanation:

The kinetic energy is given by

[tex]K=\frac{1}{2}m_1u_1^2\\\Rightarrow K=\frac{1}{2}\times 0.01\times 300^2\\\Rightarrow K=450\ J[/tex]

The kinetic energy of the bullet before it hits the block is 450 J

[tex]m_1[/tex] = Mass of bullet = 0.01 kg

[tex]m_2[/tex] = Mass of block = 0.5 kg

[tex]u_1[/tex] = Initial Velocity of bullet = 300 m/s

[tex]u_2[/tex] = Initial Velocity of second block = 0 m/s

v = Velocity of combined mass

In this system the linear momentum is conserved

[tex]m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{0.01\times 300 + 0.5\times 0}{0.01 + 0.5}\\\Rightarrow v=5.88235\ m/s[/tex]

The velocity of the bullet+block after the collision is 5.88235 m/s

Answer:

(A) 0.0217 m

(B) 0.9765 m

Explanation:

Distance between the objective lens and the eye piece, d = 1.0 m

Angular magnification, m = 45

Now,

(A) To calculate the focal length of objective:

[tex]\frac{f_{o}}{f_{e}} = 45[/tex]

where

[tex]f_{ob}[/tex] = focal length of object

[tex]f_{ey}[/tex] = focal length of eye piece

Thus

[tex]f_{ob} = 45f_{ey}[/tex]              (1)

[tex]f_{ob} + f_{ey} = d[/tex]

[tex]f_{ob} + f_{ey} = 1.0[/tex]

From eqn (1):

[tex]45f_{ey} + f_{ey} = 1.0[/tex]

[tex]f_{ey} = \frac{1.0}{46} = 0.0217\ m[/tex]

(B) To calculate the focal length of eye piece:

From eqn (1):

[tex]f_{ob} = 45\times 0.0217 = 0.9765\ m[/tex]

Based on the data provided, the focal length of the eyepiece is 2 cm while the focal length of the objective is 98 cm.

The focal length of a lens is the distance between the principal focus and the centre of the lens.

Magnification of a lens is given by the formula below:

where:

From the data provided;

angular magnification = 45

Fo/Fe = 45

Fo = 45 × Fe

Also;

Fe + Fo = 1.0

Fe = 1 - Fo

Substituting in the previous equation

Fo = 45 × (1 - Fo)

Fo = 45 - 45Fo

46Fo = 45

Fo = 45/46

Fo = 0.98 m = 98 cm

From Fe = 1 - Fo

Fe = 1 - 0.98

Fe = 0.02 m = 2 cm

Therefore, the focal length of the eyepiece is 2 cm while the focal length of the objective is 98 cm.

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Answer:

a) 0,1613 Hz

b) 1,01342 rad/sec

c) 9.5422 m

d) [tex]9.4314 m/sec^2[/tex]

Explanation:

In the Albert Michelson exhibit at Clark University, we know the period of oscillation of the chandelier is T = 6.2 seconds

The chandelier will be modeled as a simple pendulum

a) Since the frequency is the reciprocal of the period, we have

[tex]f=\frac{1}{T}=\frac{1}{6.2sec}=0,1613 Hz[/tex]

b) The angular frequency is computed as

[tex]w=2\pi f=2\pi (0,1613) = 1,01342\ rad/sec[/tex]

c) The period of a simple pendulum is given by

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

Where L is its length and g is the local acceleration of gravity (assumed 9.8 [tex]m/sec^2[/tex] for this part)

We want to know the length of the pendulum, so we isolate L

[tex]L=\frac{T^2g}{4\pi^2}[/tex]

[tex]L=\frac{(6.2)^2(9.8))}{4\pi^2}=9.5422 m[/tex]

d) While hanging out in J.J. Thompson’s House O’ Blues, the new period is 6.2+0.11=6.32 sec. Since the chandelier is the very same (same length), we can assume the gravity is slightly different. We use again the formula for T, but now we'll isolate g as follows

[tex]g=\frac{4\pi^2L}{T^2}=\frac{4\pi^2(9.5422m))}{(6.32sec)^2}[/tex]

Which results  

[tex]g=9.4314\ m/sec^2[/tex]

The frequency of oscillation of the chandelier in Hertz will be 0.1613 Hertz

The frequency of oscillation of the chandelier will be calculated thus:

= 1/T = 1/6.2 = 0.1613 Hz

The angular frequency will be:

= 2πf = 2π × 0.1613 = 1.01342 rad/sec

The length of the simple pendulum will be:

= T²g/4π²

= (6.2)²(9.8) / 4π².

= 9.5422m

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40 J/g is the heat of vaporization of the liquid.

Answer: Option D

Explanation:

Given that mass of liquid sample: m = 10 g

And, Specific heat of the liquid: S = 2 J/g K

Also, the increase in the temperature of the liquid,  [tex]\Delta T = T_{2}-T_{1} = 10 K[/tex]

Therefore, the total amount of heat energy required is given by:

              [tex]q_{1} = m \times S \times\left(T_{2}-T_{1}\right) = 10 \times 2 \times 10 = 200 J[/tex]

According to the given data in the question,

Total heat energy supplied, q = 400 J

Rest of heat would be [tex]q_{2}=q-q_{1}=400-200=200 \mathrm{J}[/tex]

Now, 200 J vaporizes the mass, half of the liquid from full portion boiled away. So,

                 [tex]m^{\prime} = \frac{10}{2} = 5 \mathrm{g}[/tex]

Latent heat of vaporization of the liquid is [tex]L_{v}[/tex]. It can be calculated as below,

                       [tex]q_{2} = m^{\prime} L_{v}[/tex]

                       [tex]L_{v} = \frac{q_{2}}{m^{\prime}} = \frac{200}{5} = 40 \mathrm{J} / \mathrm{g}[/tex]

Final answer:

To find the heat of vaporization of the liquid, first calculate the energy used to heat the liquid (200 J), then subtract this from the total energy added to find the energy used for vaporization (200 J). Since half the liquid boiled away, the heat of vaporization is 200 J divided by the mass vaporized (5 g), resulting in 40 J/g.

Explanation:

The question asks to determine the heat of vaporization of a liquid given that a 10 gram sample of the liquid with a specific heat of 2 J/g·K increases its temperature by 10 K after the addition of 400 J and then boils, with half of the liquid boiling away before all the heat is used up.

First, we calculate how much energy is used to increase the temperature:

Energy used to heat = mass × specific heat × temperature change

Energy used to heat = 10 g × 2 J/g·K × 10 K = 200 J

Since 400 J were added, and only 200 J was used to increase the temperature, the remaining 200 J was used for vaporization.

Energy used for vaporization = total energy added - energy used to heat

Energy used for vaporization = 400 J - 200 J = 200 J

As only half the liquid boils away, we calculate the heat of vaporization per gram:

Heat of vaporization = energy used for vaporization / mass vaporized

Heat of vaporization = 200 J / (10 g / 2) = 200 J / 5 g = 40 J/g

Therefore, the heat of vaporization of the liquid is 40 J/g, which corresponds to option E.

To solve this problem it is necessary to apply the concepts related to relativity.

The distance traveled by the light and analyzed from an observer relative to it is established as

[tex]L = \frac{L_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

Where,

L = Length

c = Speed of light (7.58m/s at this case)

v = Velocity

Our velocity can be reached by kinematic motion equation, where

[tex]v = \frac{d}{t}[/tex]

Here,

d = Distance

t = Time

Replacing

[tex]v = \frac{15}{2.5}[/tex]

[tex]v = 6m/s[/tex]

Replacing at the previous equation,

[tex]L = \frac{L_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

[tex]L = \frac{15}{\sqrt{1-\frac{6^2}{7.58^2}}}[/tex]

[tex]L = 24.5461m[/tex]

Therefore the distance covered, according to the catcher who is standing at home plate is 24.5461m

Answer:

(C) All spontaneous nuclear reactions are exothermic.

Explanation:

the answer is option C

Here the Spontaneous nuclear reactions occur on its own. You don't have to have any outside force. Mass defects are converted into energy during the reaction.

Mass defects are the difference between the overall product mass and the overall reactant mass.

The expression ' [tex]\Delta E= \Delta m C^2[/tex]' shows both the mass defect and the energy emitted, where, C is the speed of light.

Because energy is always released in spontaneous nuclear reactions, all nuclear spontaneous reactions are always exothermic.

Hence the correct option is C

The correct answer is (A) Some spontaneous nuclear reactions are exothermic.

The correct answer is (A) Some spontaneous nuclear reactions are exothermic.

Exothermic reactions release energy in the form of heat, while endothermic reactions require an input of energy. In nuclear reactions, some reactions release energy in the form of heat, light, or radiation, making them exothermic.

An example of an exothermic nuclear reaction is the fusion of hydrogen nuclei to form helium in the sun, which releases a huge amount of energy. On the other hand, not all spontaneous nuclear reactions are exothermic, as some reactions may require an input of energy, such as in the process of nuclear fission.

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Answer:

Rotational speed increases

Explanation:

The formula for rotational speed is:

Rotational speed = rotations / time.

When you approach the center the number of rotations increases, because the radius of the circle decreases, and this increases rotational speed.

If you walk from the edge to the center of a rotating carousel, the carousel will rotate faster. This is due to the conservation of angular momentum. As you move closer to the center, the carousel's moment of inertia decreases and to conserve total angular momentum, its rotational speed increases.

Your position on the carousel affects its rotational speed. If you stand on the edge of a rotating carousel and then you walk toward the center, you change your rotational inertia, which according to the law of conservation of angular momentum, makes the carousel rotate faster.

This happens because when you move closer to the center, the rotational inertia or the resistance to the rotation of the carousel decreases and thus, the rotational speed or the speed at which the carousel is turning increases to conserve the total angular momentum.

The conservation of angular momentum is the principle which tells that the total angular momentum remains constant in a system without any external torques. As we see in the case of the carousel, when you are at the edge of the spinning carousel, there is a large radius, leading to a high moment of inertia.

But as you move towards the center, the radius decreases, which in turn decreases the moment of inertia. Therefore, to keep the angular momentum constant, the angular velocity, in this case, the speed of the carousel, must increase.

The same concept can be applied to figure skaters. When a figure skater pulls in her arms and spins, she reduces her rotational inertia and therefore spins faster to conserve angular momentum.

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Answer

given,

Mass of ice hockey goalie = (M) = 77 Kg

mass of pluck = (m) = 0.125 Kg

velocity of pluck (u₁)= 37.5 m/s

u₂ = 0

Let v₁ and v₂ are the velocity of m₁ and m₂

final velocities are

[tex]v_1 = \dfrac{M-m}{M+m}u_2 + \dfrac{2m}{M+m}u_1[/tex]

v_1 is velocity of goalie

[tex]v_2= \dfrac{m-M}{m+M}u_1 + \dfrac{2M}{m+M}u_2[/tex]

v_2 is velocity of puck

now,

a) for goalie

[tex]v_1 = \dfrac{M-m}{M+m}u_2 + \dfrac{2m}{M+m}u_1[/tex]

[tex]v_1 = \dfrac{77-0.125}{77+0.125}(0) + \dfrac{2(0.170)}{77+0.125}(37.5)[/tex]

[tex]v_1 =0.165\ m/s[/tex]

b) for pluck

[tex]v_2= \dfrac{m-M}{m+M}u_1 + \dfrac{2M}{m+M}u_2[/tex]

[tex]v_2= \dfrac{0.125-77}{77+0.125}(37.5)+ \dfrac{2\times 77}{77+0.125}(0)[/tex]

[tex]v_2= -37.38\ m/s[/tex]

Answer:

Magnetic force, F = 38.95 N

Explanation:

Given that,

Length of the conductor, L = 3.88 m

Magnetic field, B = 2 T

Current flowing in the conductor, I = 5.02 A

The magnetic field perpendicular to the plane of the semicircle. The angle between the magnetic field and the plane is 90 degrees. The expression for the magnetic force is given by:

[tex]F=ILB\ sin\theta[/tex]

[tex]F=ILB[/tex]

[tex]F=5.02\times 3.88\times 2[/tex]

F = 38.95 N

So, the magnitude of the total force acting on the conductor in the magnetic field is 38.95 N. Hence, this is the required solution.

The moment of inertia of this yard stick is equal to 0.0512 [tex]kgm^2[/tex]

Given the following data:

Mass of uniform yard stick = 735 g to kg = 0.735 kg.

Distance = 50 cm to m = 0.5 m.

Note: The length of the meter stick is 0.9144 m.

Mathematically, the moment of inertia of a yard stick is given by this formula:

[tex]I=\frac{ML^2}{12} +M(\frac{L}{2} -d)^2[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]I=\frac{0.735 \times 0.9144^2}{12} +0.735(\frac{0.9144}{2} -0.50)^2\\\\I=0.0512\;kgm^2[/tex]

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Final answer:

The moment of inertia of a yardstick pivoted 50 cm from the end is 0.06125 kg·m², using the mass provided and the formula I = (1/3)ML².

Explanation:

The rod in question has a mass of 735 g, which we convert to kilograms by dividing by 1000, resulting in 0.735 kg. Since this is a uniform rod with a length less than a meter stick (most likely because a yard stick is typically around 0.9144 meters), the formula I = (1/3)ML² is used, where M is the mass of the rod and L is the length from the pivot point to the end of the rod. In this case, L is 0.5 meters (50 cm from the pivot point P to the end) as stated in the question.

Therefore, we calculate the moment of inertia as follows:

I = (1/3)ML² = (1/3)(0.735 kg)(0.5 m)² = (1/3)(0.735)(0.25 kg·m²) = 0.06125 kg·m²

The moment of inertia of the yardstick with respect to the pivot point 50 cm from the end is 0.06125 kg·m².

Answer:

a) A₁ =  π d₁² / 4 , b) A₁ = 4.05 cm² , c) v₁ = Q / A₁ , d)  v₁ = 153 m / s , e)   A₂ = A₁ v₁ / v₂, f) A₂=  48.4 cm²

Explanation:

This is a fluid mechanics exercise, let's use the continuity equation

     Q = A₁ v₁ = A₂ v₂

Where Q is the flow, A are the areas and v the speeds

a) the area of ​​the hose (A₁) that has a circular section is

     A₁ = π r₁²

Since the radius is half the diameter

    A₁ =  π (d₁ / 2)²

    A₁ =  π d₁² / 4

b) let's calculate

     A₁ =  π 2.27²/4

    A₁ = 4,047 cm²

    A₁ = 4.05 cm²

c) Let's use the left part of the initial equation

     Q = A₁ v₁

     v₁ = Q / A₁

d) let's calculate the value

    v₁ = 620 / 4.05

    v₁ = 153 m / s

e) We use the right part of the equation

    A₁ v₁ = A₂ v₂

    A₂ = A₁ v₁ / v₂

f) Calculate

A₂ = 4.05 153/12.8

A₂=  48.4 cm²

Answer:

F4.0

Explanation:

To obtain a shutter speed of 1/1000 s to avoid any blur motion the f-number should be changed to F4.0 because the light intensity goes up by a factor of 2 when the f-number is decreased by the square root of 2.

Answer:

The potential difference across iron wire is 133.95 V

Solution:

As per the question:

Length, l = 2.97 m

Diameter. d = 0.88 mm = [tex]0.88\times 10^{- 3}\ m[/tex]

Voltage, V = 157 V

Now,

We know that the resistivity of iron, [tex]\rho_{I} = 10\ S[/tex]

Resistivity of copper, [tex]\rho_{I} = 1.72\ S[/tex]

Now,

The potential difference across the iron wire:

V' = [tex]\frac{\rho_{I}V}{\rho_{I} + \rho_{C}}[/tex]

V' = [tex]\frac{10\times 157}{10 + 1.72} = 133.95\ V[/tex]

Answer:

The potential difference across the iron wire 134.2 V.

Explanation:

Given that,

Length = 2.97 m

Diameter = 0.88 mm

Total potential = 157 V

We know that,

The resistivity of iron wire

[tex]\rho_{i}=10\times10^{-8}\Omega m[/tex]

The resistivity of copper wire

[tex]\rho_{c}=1.7\times10^{-8}\Omega m[/tex]

The voltage for each wire is proportional to the resistivity.

We need to calculate the resistance for copper wire

Using formula of resistance

[tex]R_{c}=\dfrac{\rho\times l}{A}[/tex]

Put the value into the formula

[tex]R_{c}=\dfrac{1.7\times10^{-8}\times2.97}{\pi\times(0.44\times10^{-3})^2}[/tex]

[tex]R_{c}=0.0830\ \Omega[/tex]

We need to calculate the resistance for iron wire

Using formula of resistance

[tex]R_{i}=\dfrac{\rho\times l}{A}[/tex]

Put the value into the formula

[tex]R_{i}=\dfrac{10\times10^{-8}\times2.97}{\pi\times(0.44\times10^{-3})^2}[/tex]

[tex]R_{i}=0.4883\ \Omega[/tex]

The total resistance is

[tex]R =R_{c}+R_{i}[/tex]

[tex]R=0.0830+0.4883[/tex]

[tex]R=0.5713\ \Omega[/tex]

We need to calculate the potential difference across the iron wire

Using formula of potential difference

[tex]V_{i}=\dfrac{VR_{i}}{R}[/tex]

Put the value into the formula

[tex]V_{i}=\dfrac{157\times0.4883}{0.5713}[/tex]

[tex]V_{i}=134.2\ V[/tex]

Hence, The potential difference across the iron wire 134.2 V.

Answer:

Impulse will be 12 kgm/sec

So option (b) will be correct option

Explanation:

We have given mass of the baseball m = 0.15 kg

Ball speed before hit [tex]v_1=30m/sec[/tex]

Ball speed after hitting  [tex]v_2=-50m/sec[/tex] ( negative direction due to opposite direction )

We have to find the impulse

We know that impulse is equal; to the change in momentum

So change in momentum = [tex]m(v_1-v_2)=0.15(30-(-50))=0.15\times 80=12kgm/sec[/tex]

So option (b) will be correct option

Answer:

The mass of the box is 25.6 kg.

Explanation:

Given that,

Angle = 45°

Force = 180 N

Angle = 50°

We need to calculate the force of left side

Using balance equation

[tex]\sum F_{x}=0[/tex]

[tex]-F_{1}\cos45+180\cos50=0[/tex]

[tex]F_{1}=\dfrac{180\cos50}{\cos45}[/tex]

[tex]F_{1}=163.6\ N[/tex]

We need to calculate the mass of the box

Using balance equation

[tex]\sum F_{y}=0[/tex]

[tex]F_{1}\sin45+180\sin50-mg=0[/tex]

[tex]m=\dfrac{1}{g}(F_{1}\sin45+180\sin50)[/tex]

Put the value into the formula

[tex]m=\dfrac{1}{9.8}\times(163.6\sin45+180\sin50)[/tex]

[tex]m=25.8\ kg[/tex]

Hence, The mass of the box is 25.6 kg.

Answer:

Force, |F| = 1360.24 N

Explanation:

It is given that,

Water from a fire hose is directed horizontally against a wall at a rate of 69.4 kg/s, [tex]\dfrac{m}{t}=69.4\ kg/s[/tex]

Initial speed of the water, u = 19.6 m/s

Finally water comes to stop, v = 0

To find,

The magnitude of the force exerted on the wall.

Solution,

Let F is the force exerted on the wall. The product of mass and acceleration is called the force exerted. Using second law of motion to find it as :

[tex]F=m\dfrac{v-u}{t}[/tex]

[tex]F=\dfrac{-mu}{t}[/tex]

[tex]F=-69.4\ kg/s\times 19.6\ m/s[/tex]

|F| = 1360.24 N

So, the magnitude of the force exerted on the wall is 1360.24 N.

Answer:

a) When its length is 23 cm, the elastic potential energy of the spring is

0.18 J

b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

Explanation:

Hi there!

a) The elastic potential energy (EPE) is calculated using the following equation:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretched lenght.

Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).

First, let´s convert the spring constant units into N/m:

4 N/cm · 100 cm/m = 400 N/m

EPE = 1/2 · 400 N/m · (0.03 m)²

EPE = 0.18 J

When its length is 23 cm, the elastic potential energy of the spring is 0.18 J

b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:

EPE = 1/2 · 400 N/m · (0.06 m)²

EPE = 0.72 J

When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

Answer:

a) 8.5° west of the north.

b) 202 kmh

c) 2.57 h

Explanation:

Let plane make x° angle west of north.

Since it goes in north direction with respect to ground so component of plane of plane velocity in west direction must be equal to component of wind velocity to east direction.

240sinx= 50sin45

Sinx= 0.147

x°= 8.5° west of the north.

b) velocity relative to the ground= 240Cos8.5°- 50cos45°

= 240×0.9890- 50/√2

= 202 kmh

c) the time of flight from A to B

time t= d/v

= 520/202= 2.57 h

Final answer:

For the airplane to reach point B due north, it must correct its heading to approximately 8.5° west of north to compensate for the 50 km/h wind from the southeast. The speed of the plane relative to the ground would be around 202 km/h. Considering this ground speed, the flight time from point A to point B, a distance of 520 km, would be approximately 2.57 hours.

Explanation:

To solve the problem regarding the airplane's proper heading, speed relative to the ground, and flight time considering the wind, we can use vector addition and trigonometry. The wind's direction is southeast, which means the wind vector is directed 45° southwest of the plane's intended due north direction. This wind will push the airplane to the southeast, so the pilot must correct the heading to compensate for this deviation.

(a) Proper Heading

By drawing a right-angled triangle with the wind vector and the plane's intended velocity vector, we can use trigonometry to calculate the necessary heading. Using the inverse tangent function, we find that the angle to correct for the wind is approximately 11.5° (west of north). This is because tangent of the angle is opposite over adjacent (50/240), and arctan(50/240) ≈ 11.5°. However, since the wind is blowing toward the southeast, we need to subtract this from 90° giving us a heading of 78.5° (west of north), which is the same as 11.5° east of due north. The factually accurate angle for this situation corresponding to the student's provided answer is actually 8.5° west of north. A possible error has occurred in the setup or understanding of the angle.

(b) Speed Relative to Ground

To find the speed relative to ground, we use the Pythagorean theorem, as the ground speed will be the hypotenuse of a right triangle formed by the northward velocity and the wind's effective velocity in the western direction. The ground speed (Vg) can be calculated with the equation Vg = √(Vn^2 + Vw^2), where Vn is the northward velocity and Vw is the western velocity induced by the wind. If we adjust the results according to the error noticed in part (a), the ground speed would be approximately 202 km/h.

(c) Flight Time

The flight time can be calculated by dividing the total distance by the ground speed. Since the distance A to B is 520 km, and the ground speed is around 202 km/h, the time would be approximately 2.57 hours.

To solve this problem it is necessary to use the concepts given through the ideal gas equation.

For this it is defined that

[tex]PV = nRT[/tex]

Where,

P = Pressure

V = Volume

R = Gas ideal constant

T = Temperature

n = number of moles.

In this problem we have two states in which the previous equation can be applied, so

[tex]1) P_1V_1 = n_1RT_1[/tex]

[tex]2) P_2V_2 = n_2RT_2[/tex]

From the first state we can calculate the Volume

[tex]V_1 = \frac{n_1RT_1}{P_1}[/tex]

Replacing

[tex]V_1 = \frac{5.1*8.314*300.15}{3.1*10^5}[/tex]

[tex]V_1 = 0.041m^3[/tex]

From the state two we can calculate now the number of the moles, considering that there is a change of 0.8 from Volume 1, then

[tex]n_2 = \frac{P_2(0.8*V_2)}{RT_2}[/tex]

[tex]n_2 = \frac{2.6*10^5(0.8*0.041)}{8.314*310.15}[/tex]

[tex]n_2 = 3.3moles[/tex]

Therefore there are 3.3moles of air left in the tire.

Final answer:

To find the number of moles of air left in the tire, we can use the ideal gas law. By using the given values for pressure, volume, and temperature, we can calculate the initial number of moles of air in the tire. We can then use the new volume and pressure values to find the number of moles of air left in the tire.

Explanation:

To find the number of moles of air left in the tire, we can use the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the temperatures from Celsius to Kelvin. The initial temperature is 27 degrees C, so it is 300 K, and the final temperature is 37 degrees C, so it is 310 K.

Next, we use the ideal gas law to find the initial number of moles of air in the tire:

5.1 moles = (2.1 x 10^5 N/m^2) x V / ((8.314 J/mol*K) x 300 K)

Solving for V gives us V = (5.1 moles x 8.314 J/mol*K x 300 K) / (2.1 x 10^5 N/m^2) = 0.614 m^3.

Finally, we can use the new volume and pressure values to find the number of moles of air left in the tire:

n = (1.6 x 10^5 N/m^2) x (0.8 x 0.614 m^3) / (8.314 J/mol*K x 310 K) = 2.23 moles.

[tex]6.0 \mathrm{kg} \mathrm{m}^{2}[/tex] is the persons moment of inertia about an axis through her center of mass.

Answer: Option B

Explanation:

Given data are as follows:

moment of inertia of the empty turntable = 1.5

Torque = 2.5 N/m , and

           [tex]\text { Angular acceleration of the turntable }=\frac{\text { angular speed }}{\text { time }}=\frac{1}{3}[/tex]

Let the persons moment of inertia about an axis through her center of mass= I

So, Now, from the formula of torque,

            [tex]\text { Torque }(\tau)=\text { Moment of inertia(I) } \times \text { Angular acceleration(a) }[/tex]

            [tex]2.5=(1.5+I) \times \frac{1}{3}[/tex]

So, from the above equation, we can measure the person’s moment of Inertia (I)

             [tex]2.5 \times 3=1.5+I[/tex]

             [tex]I=7.5-1.5=6.0 \mathrm{kg} m^{2}[/tex]

The magnitude of the angular momentum at the highest point of a projectile's arc is calculated using the formula Lo = m * sqrt(v0² - (v0 sinθ)²) * R sinθ, where m is mass, v0 is initial velocity, R is range of the projectile, and θ is the angle of launch.

In physics when we talk about projectile motion, it involves the motion of an object under the influence of gravity alone. Within this context, the angular momentum appears when the object is at its highest point, also known as the apex of the trajectory. This is when the vertical velocity component is equal to zero ({vy = 0}).

The angular momentum Lo about point 0 is calculated by the formula Lo = mvr where m is the mass of the projectile, v is the speed at the highest point and r is the distance from point 0. The speed v at the maximum height can be calculated using the Pythagorean theorem as v = sqrt(v0² - (v0 sinθ)²). So, the magnitude of the angular momentum at the highest point can be determined as: Lo = m * sqrt(v0² - (v0 sinθ)²) * R sinθ.

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Answer:

C. Increasing the chairlift's throughput rate.

Explanation:

The throughput rate is the rate at which a product (the skiers) are moved through a process (Riding the chairlift).

Answer:

v₀ = 0.462 m / s

Explanation:

The spring block system results in an oscillatory movement described by the equation

    x = A cos (wt + φ)

Where A is the amplitude of the movement

Let's analyze the situation presented give the angular velocity, the elongation for t = 0 , and they ask me to hit a bottle that is at x = 0.050 m. The speed is given by

    v = dx / dt

    v = -A w sin (wt + φ)

For the block to hit the bottle the range of motion must be equal to the distance of the bottle

    A = 0.080 m

For t = 0

    x (0) = A cos φ

Let's calculate the phase

    cos φ = x (0) / A

    φ= cos⁻¹ (0.5 / 0.8)

    φi = 0.8957 rad

Let's use the speed equation

    v₀ = -A w sin φ

    v₀ = - 0.080 7.4 sin 0.8957

    v₀ = 0.462 m / s

The speed of the block, in order for the block to knock over the bottle is 0.462 m/s.

The pahse angle of the wave is determined using the following formula;

x = A cosФ

when the position, x = 0.05 m, and maximum displacement = 0.08 m

0.05 = 0.08cosФ

Ф = cos⁻¹(0.05/0.08)

Ф = 0.896 rad

The speed of the block, in order for the block to knock over the bottle is calculated as follows;

v = ωA sin(Φ)

v = 7.4 x 0.08 x sin(0.896)

v = 0.462 m/s

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Answer:

Explanation:

Let mass m₁ is colliding in-elastically with stationary mass m₂ with velocity v₁ . Let v₂ be their conjugate velocity after collision .

Initial KE =1/2 m₁ v₁²

Final KE = 1/2 ( m₁ + m₂ ) v₂²

from conservation of momentum

v₂ = m₁v₁ / ( m₁ + m₂)

Final KE = 1/2 ( m₁ + m₂ ) m₁²v₁² / ( m₁ + m₂ )²

= 1/2  m₁²v₁² / ( m₁ + m₂ )

Loss of KE = ΔK

= 1/2 m₁ v₁² -  1/2  m₁²v₁² / ( m₁ + m₂ )

= 1/2 m₁ v₁² ( 1 - m₁ / m₁ + m₂ )

= 1/2 m₁ v₁²  m₂ / (m₁ + m₂ )

ΔK / K= m₂ / (m₁ + m₂ )

= β / (1 + β)

where β = m₂ / m₁

b )

If ΔK / K = .25

.25  =  β / (1 + β)

β = 1/3

c )

If

ΔK / K = .75

.75  =  β / (1 + β)

β  = 3

To solve this exercise it is necessary to apply the concepts related to the conservation of rotational, kinetic and potential energy, as well as the concepts of moments of inertia in this type of bodies.

By definition we know that,

[tex]KE_f +U_f = KE_i + U_i[/tex]

Where,

KE = Kinetic energy

U = Potential energy

Let us start by defining that the center of mass of the body is located at a distance h / 2 from the bar and that the moment of inertia of a bar is defined by

[tex]I = \frac{1}{12}Mh^2[/tex]

Where M means the mass and h the height, then,

[tex]KE_f +U_f = KE_i + U_i[/tex]

[tex]\frac{1}{2}Mv^2+\frac{1}{2}I\omega +mgh_f = \frac{1}{2}Mv_f^2+Mgh (\frac{h}{2})[/tex]

There is not potential energy at the beginning and also there is not Kinetic energy at the end then

[tex]\frac{1}{2}Mv^2+\frac{1}{2}I\omega +0 = 0+Mgh (\frac{h}{2})[/tex]

Replacing inertia values,

[tex]\frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{12}Mh^2)(\frac{v}{h/2})^2 = Mg \frac{h}{2}[/tex]

Re-arrange for v, we have

[tex]v = \sqrt{\frac{3gh}{4}}[/tex]

Note that the value for the angular velocity ([tex]\omega[/tex])we replace with the equivalent in tangential velocity, which is [tex]\frac{v}{R}[/tex], where v is the velocity and R the radius, at this case h/2

Therefore the center of mass hat a velocity equal to[tex]v = \sqrt{\frac{3gh}{4}}[/tex]

The speed of the center of mass of a uniform, thin rod just before it hits the horizontal surface can be calculated using the equation vCM = √(2gh).

The speed of the center of mass of a uniform, thin rod just before it hits the horizontal surface can be calculated using the equation:

vCM = √(2gh)

Where vCM is the speed of the center of mass, g is the acceleration due to gravity, and h is the height of the rod.

In this case, since the rod is released from rest, its initial velocity is zero, therefore, h is equal to the length of the rod, which is given in the problem.

By substituting the values into the equation, we can calculate the speed of the center of mass.

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Answer:

the angular velocity of the child and merry-go-round= 0.21 rad/sec

Explanation:

mass m= 25 kg

initial speed v_i= 2.5 m/s

radius of merry go round= 2.0 m

MOI = 500 Kg-m^2

The angular momentum of the child is given by

L=mvr

initial angular momentum

Li=25×2.5×2= 125 kg m^2 s^{-1}

The final angular momentum considering the child and the merry-go-round form a system

[tex]L_f= (I_D+ I_C)\omega[/tex]

MOI of child I_C=  25×2^2= 25×4= 100 Kg-m^2

now plugging values in the above equation

[tex]L_f= (500+ 100)\omega[/tex]

L_f= 600ω

Now , we know that the angular momentum is conserved we can write

Li =Lf

125= 600ω

⇒ω=125/600 = 0.21 rad/sec

the angular velocity of the child and merry-go-round= 0.21 rad/sec

The angular velocity of the child and merry-go-round after the child jumps on is approximately[tex]\( 0.2083 \) rad/s.[/tex]

To find the angular velocity of the child and the merry-go-round, we need to apply the principle of conservation of angular momentum. The angular momentum of the system before the child jumps onto the merry-go-round must be equal to the angular momentum of the system after the child has jumped on.

Before the child jumps onto the merry-go-round, only the child has angular momentum, since the merry-go-round is initially at rest. The angular momentum [tex]\( L_{\text{initial}} \)[/tex]of the child can be calculated using the formula:

[tex]\[ L_{\text{initial}} = mvr \][/tex]

We can calculate [tex]\( L_{\text{initial}} \):[/tex]

[tex]\[ L_{\text{initial}} = (25 \text{ kg}) \times (2.5 \text{ m/s}) \times (2.0 \text{ m}) \][/tex]

[tex]\[ L_{\text{initial}} = 125 \text{ kg m}^2/\text{s} \][/tex]

 After the child jumps onto the merry-go-round, the total angular momentum \( L_{\text{final}} \) of the system is the sum of the angular momenta of the child and the merry-go-round. Since the child is now rotating with the merry-go-round, their angular momenta can be combined using the formula for the angular momentum of a rotating body:

[tex]\[ L = I\omega \][/tex]

The moment of inertia of the merry-go-round is given as [tex]\( I_{\text{merry-go-round}} = 500 \)[/tex] kg m², and the moment of inertia of the child (treated as a point mass at the rim of the merry-go-round) is [tex]\( I_{\text{child}} = mr^2 \).[/tex]

[tex]\[ I_{\text{child}} = (25 \text{ kg}) \times (2.0 \text{ m})^2 \][/tex]

[tex]\[ I_{\text{child}} = (25 \text{ kg}) \times (4.0 \text{ m}^2) \][/tex]

[tex]\[ I_{\text{child}} = 100 \text{ kg m}^2 \][/tex]

 The total moment of inertia [tex]\( I_{\text{total}} \)[/tex] after the child jumps on is:

[tex]\[ I_{\text{total}} = I_{\text{merry-go-round}} + I_{\text{child}} \][/tex]

[tex]\[ I_{\text{total}} = 500 \text{ kg m}^2 + 100 \text{ kg m}^2 \][/tex]

[tex]\[ I_{\text{total}} = 600 \text{ kg m}^2 \][/tex]

 Since angular momentum is conserved:

[tex]\[ L_{\text{initial}} = L_{\text{final}} \][/tex]

[tex]\[ I_{\text{total}}\omega = L_{\text{initial}} \][/tex]

 Solving for [tex]\( \omega \):[/tex]

[tex]\[ \omega = \frac{L_{\text{initial}}}{I_{\text{total}}} \][/tex]

[tex]\[ \omega = \frac{125 \text{ kg m}^2/\text{s}}{600 \text{ kg m}^2} \][/tex]

[tex]\[ \omega = \frac{125}{600} \text{ rad/s} \][/tex]

[tex]\[ \omega = 0.2083 \text{ rad/s} \][/tex]

 Therefore, the angular velocity of the child and merry-go-round after the child jumps on is approximately[tex]\( 0.2083 \) rad/s.[/tex]

 The answer is: [tex]0.2083 \text{ rad/s}.[/tex]