Water is pumped from a lake to a storage tank 15 m above at a rate of 70 L/s while consuming 15.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of the pump-motor unit and (b) the pressure difference between the inlet and the exit of the pump.

Answer:E

Explanation:

This can be explained by Doppler effect which gives the relation between apparent Frequency and actual frequency when the source of sound is moving

[tex]f'=f\frac{v+v_o}{v-v_s}[/tex]

where [tex]f'=apparent\ frequency [/tex]

[tex]f=actual\ frequency[/tex]

[tex]v=velocity\ of\ sound[/tex]

[tex]v_o=velocity\ of\ observer[/tex]

[tex]v_s=velocity\ of\ source[/tex]

here [tex]v_o =0[/tex] as observer is standing

when Ambulance is approaching then velocity of sound and velocity of ambulance have relative velocity thus denominator is less than and apparent frequency is more while when ambulance is going away then velocity of sound waves and velocity of observer is in opposite direction thus denominator is less than 1 and apparent frequency is less.

Thus [tex]f_2<f<f_1[/tex]

The Doppler effect explains the changes in sound frequency from a moving source as perceived by a stationary observer. The observed frequency increases as the source approaches and decreases when it moves away. Hence, the actual frequency of the source (the siren) is less than the frequency observed when the source is approaching (f1) and greater than the frequency observed than when it is retreating (f2). The correct answer is 'e. f2 < f < f1'.

This question relates to the phenomenon of the Doppler effect, which is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. In this case, when an ambulance with a siren is approaching a static observer, the observed frequency f1 will be higher than the actual frequency f (the frequency of the siren when you are sitting inside the ambulance), due to compression of the sound waves.

However, when the ambulance starts moving away from the observer, the observed frequency f2 will be lower than f, due to elongation or stretching out of the sound waves. Hence, the correct answer is 'e. f2 < f < f1'.

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Final answer:

The net force F120 on the pressure cooker's lid when the inside air is heated to 120°C is the difference between the force exerted by the heated air inside and the atmospheric force outside on the lid's area A. Calculate pressure using the ideal gas law, then calculate the forces exerted at 120°C and at atmospheric conditions, and find their difference.

Explanation:

The student asks about the magnitude of the net force F120 on the lid of a pressure cooker when the air inside is heated to 120°C, assuming room temperature outside the cooker is 20°C, the lid's area is A, and atmospheric pressure is Pa. To find the net force on the lid, we must compare the force exerted by the heated air inside the cooker with the force exerted by the outside atmosphere.

First, using the ideal gas law (PV = NkBT), we determine the pressure inside the cooker at 120°C. Then, we calculate the force exerted by this pressure over area A. Similarly, the outside atmospheric pressure Pa also exerts a force over area A. The net force on the lid, F120, is the difference between these two forces.

Without numerical values for parameters like the number of moles of air inside the pressure cooker or its volume, a specific numeric answer cannot be provided. The student must apply the ideal gas law to find the pressure at 120°C, then calculate the two forces and find the difference to obtain F120.

Answer:

0.243

Explanation:

Step 1: Identify the given parameters

Force (f) = 5kN, length of pitch (L) = 5mm, diameter (d) = 25mm,

collar coefficient of friction = 0.06 and thread coefficient of friction = 0.09

Frictional diameter =45mm

Step 2: calculate the torque required to raise the load

[tex]T_{R} = \frac{5(25)}{2} [\frac{5+\pi(0.09)(25)}{\pi(25)-0.09(5)}]+\frac{5(0.06)(45)}{2}[/tex]

[tex]T_{R}[/tex] = (9.66 + 6.75)N.m

[tex]T_{R}[/tex] = 16.41 N.m

Step 3: calculate the torque required to lower the load

[tex]T_{L} = \frac{5(25)}{2} [\frac{\pi(0.09)(25) -5}{\pi(25)+0.09(5)}]+\frac{5(0.06)(45)}{2}[/tex]

[tex]T_{L}[/tex] = (1.64 + 6.75)N.m

[tex]T_{L}[/tex] = 8.39 N.m

Since the torque required to lower the thread is positive, the thread is self-locking.

The overall efficiency = [tex]\frac{F(L)}{2\pi(T_{R})}[/tex]

                        = [tex]\frac{5(5)}{2\pi(16.41})}[/tex]

                        = 0.243

The overall efficiency of the power screw is approximately 38.6%. The torque required to raise the load is 46.8 N·m, while the torque to lower the load is about 22.53 N·m.

Calculating Lead and Mean Diameter:Lead (L) = Pitch (p) = 5 mm

Mean diameter (dm) = 25 mm

Calculating Torque to Raise the Load ([tex]T_{up[/tex]):

Using the formula: [tex]T_{up[/tex] = (W dm/2) (L + π μt dm)/(π dm - μt L)

Where:

W = 5 kN = 5000 N

μt = 0.09

dm = 25 mm = 0.025 m

L = 5 mm = 0.005 m

Substituting the values:

[tex]T_{up[/tex] = (5000 × 0.025/2) (0.005 + π × 0.09 × 0.025)/(π × 0.025 - 0.09 × 0.005)

[tex]T_{up[/tex] ≈ 46.8 N·m

Calculating Torque to Lower the Load ([tex]T_{down[/tex]):

Using the formula: [tex]T_{down[/tex] = (W dm/2) (L - π μt dm)/(π dm + μt L)

Substituting the values:

[tex]T_{down[/tex] = (5000 × 0.025/2) (0.005 - π × 0.09 × 0.025)/(π × 0.025 + 0.09 × 0.005)

[tex]T_{down[/tex] ≈ 22.53 N·m

Calculating Overall Efficiency (η):

Efficiency η = (tan θ / (tan(θ + φ)))

Where:

θ = tan-1(L/π dm)

φ = tan-1(μt)

θ ≈ 0.064

φ ≈ 0.09

η ≈ (tan 0.064) / (tan(0.064 + 0.09))

η ≈ 38.6%

Therefore, the overall efficiency of the power screw is approximately 38.6%, the torque to raise the load is 46.8 N·m, and the torque to lower the load is around 22.53 N·m.

Answer:

t = 47.62 sec

Explanation:

Given data;

[tex]A_1 = 4.62 \times 10^4 m^2[/tex]

[tex]A_2 = 2.52 \times 10^4 m^2[/tex]

h = 2.50 cm

volume 10 L

from

[tex]A_1 v_1 = A_2 v_2[/tex]

[tex]4.62 \times 10^4 v_1 = 2.52 \times 10^4 v_2[/tex]

[tex]4.62 v_1 = 2.52 v_2[/tex] ......1

from bernoulli eq

[tex]P_1 + \frac{1}{2} \rho v_1^2 + \rho g h = P_2 + \frac{1}{2} \rho v_2^2 [/tex]

[tex]P_1 =P_2 = P_{atm}[/tex]

[tex]v_2^2 = v_1^2 +2gh[/tex] ... 2

from 1 and 2 equation

[tex]v_1 = 0.46 m/s[/tex]

volume flow rate is

[tex]Q = A_1 \times v_1 = 4.62 \times 10^[-4} v_1 = 2.1 \times 10^{-4} m^3/s[/tex]

[tex]t  = \frac{v}{Q} [/tex]

[tex]t =\frac{10\times 10^{-3}}{2.1 \times 10^{-4}} = 47.62 s[/tex]

Answer:

The time is [tex]4.76\times10^{-7}\ sec[/tex]

Explanation:

Given that,

Area [tex]A_{1}=4.62\times10^{4}\ m^2[/tex]

Area [tex]A_{2}=2.52\times10^{4}\ m^2[/tex]

Height = 2.50 cm

Volume = 10.0 L

We need to calculate the speed

Using equation of continuity

[tex]A_{1}v_{1}=A_{2}v_{2}[/tex]

Put the value into the formula

[tex]4.62\times10^{4}\times v_{1}=2.52\times10^{4}\times v_{1}[/tex]

[tex]4.62v_{1}=2.52v_{2}[/tex].....(I)

[tex]v_{1}=\dfrac{2.52}{4.62}v_{2}[/tex]

[tex]v_{1}=0.545v_{2}[/tex]

Now, using Bernoulli equation

[tex]P_{1}+\dfrac{1}{2}\rhi\times v_{1}^2+\rho gh=P_{2}+\dfrac{1}{2}\rhi\times v_{2}^2[/tex]

Here, [tex]P_{1}=P_{2}=P_{atm}[/tex]

[tex]v_{2}^2=v_{1}^2+2gh[/tex].....(II)

Put the value [tex]v_{1}[/tex] into the formula

[tex]v_{2}^2=(0.545v_{2})^2+2\times9.8\times2.50\times10^{-2}[/tex]

[tex]v_{2}^2=0.297v_{2}^2+0.49[/tex]

[tex]v_{2}^2(1-0.297)=0.49[/tex]

[tex]v_{2}=\sqrt{\dfrac{0.49}{0.703}}[/tex]

[tex]v_{2}=0.835\ m/s[/tex]

Put the value of [tex]v_{2}[/tex] in the equation (I)

[tex]v_{1}=0.545\times0.835[/tex]

[tex]v_{1}=0.46\ m/s[/tex]

We need to calculate the flow rate

Using formula of flow rate

[tex]Q=A_{1}v_{1}[/tex]

[tex]Q=(4.62\times10^{4})\times0.46[/tex]

[tex]Q=2.1\times10^{4}\ m^3/s[/tex]

We need to calculate the time

Using formula of time

[tex]t = \dfrac{V}{Q}[/tex]

Put the value into the formula

[tex]t=\dfrac{10.0\times10^{-3}}{2.1\times10^{4}}[/tex]

[tex]t=4.76\times10^{-7}\ sec[/tex]

Hence, The time is [tex]4.76\times10^{-7}\ sec[/tex]

Answer:

  K = 24.5 keV

Explanation:

The interference phenomenon is described by the equation

       .d sin θ = m λ            m = 1,2,3,…

The pattern is observed on a screen at a distance L = 2.6 m

       tan θ = y / L

As these experiments the angle is very small we can approximate the tangent

        tan θ = sin θ / cos θ

For small angles

        tan θ = sin θ

Let's replace

        d y / L = m λ  

        λ   = d y / m L

Let's reduce the units to the SI system

       d = 51 nm = 51 10⁻⁹ m

       y = 0.4 mm = 0.4 10⁻³ m

Let's calculate the wavelength

Let's use m = 1 for the first interference line

       λ  = 51 10⁻⁹ 0.4 10⁻³ / 2.6

       λ   = 7.846 10⁻¹² m

Let's look for kinetic energy

       K = ½ m v²

       p = mv

       K = ½ m p² / m

       K = p² / 2m

Let's use the wave-particle duality relationship

       p = h /λ  

       K = h² / 2m λ²

Let's calculate

      K = (6.63 10⁻³⁴)² / (2 9.1 10⁻³¹ (7.846 10⁻¹²)²)

      K = 3,923 10⁻¹⁵-15 J

       K = 3.923 10⁻¹⁵ J ( 1 eV / 1.6 10⁻¹⁹ J) =2.452 10⁴ eV

     K = 24.5 keV

Encryption would not likely be the subject of a standard published by the International Electrotechnical Commission (IEC).

Answer: Option C

Explanation:

IEC (International Electrotechnical Commission) is the world's number one organization that develops and releases international standards for all electronic-electrical technologies and its affiliates.

Its standards would cover a wide technology range, from energy generation, transmissions and distributions to house appliances and offices, semiconductors, batteries, fibreglass, solar energy, Nano-technology and sea energy and many more.

IEC standards apply to all electrical technologies, including electronics, magnets and electro-magnetics, electro-acoustics, multi-media, telecommunications, and medical technologies, as well as related general fields such as terminology and symbols, etc.

Answer:

The velocity is  [tex]4.6 m/s^2[/tex]

Explanation:

Given:

Force = 500N

Distance  s= 0

To find :

Its velocity at s = 0.5 m

Solution:

[tex]\sum F_{x}=m a[/tex]

[tex]F\left(\frac{4}{5}\right)-F_{S}=13 a[/tex]

[tex]500\left(\frac{4}{5}\right)-\left(k_{s}\right)=13 a[/tex]

[tex]400-(500 s)=13 a[/tex]

[tex]a = \frac{400 -(500s)}{13}[/tex]

[tex]a = (30.77 -38.46s) m/s^2[/tex]

Using the relation,

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

[tex]a=v \frac{d v}{d s}[/tex]

[tex]v d v=a d s[/tex]

Now integrating on both sides

[tex]\int_{0}^{v} v d v=\int_{0}^{0.5} a d s[/tex]

[tex]\int_{0}^{v} v d v=\int_{0}^{0.5}(30.77-38.46 s) d s[/tex]

[tex]\left[\frac{v^{2}}{2}\right]_{0}^{2}=\left[\left(30.77 s-19.23 s^{2}\right)\right]_{0}^{0.5}[/tex]

[tex]\left[\frac{v^{2}}{2}\right]=\left[\left(30.77(0.5)-19.23(0.5)^{2}\right)\right][/tex]

[tex]\left[\frac{v^{2}}{2}\right]=[15.385-4.807][/tex]

[tex]\left[\frac{v^{2}}{2}\right]=10.578[/tex]

[tex]v^{2}=10.578 \times 2[/tex]

[tex]v^{2}=21.15[/tex]

[tex]v = \sqrt{21.15}[/tex]

[tex]v = 4.6 m/s^2[/tex]

It is likely that until today there is no physical / mathematical argument that allows to equate this statement. But there are countless observations that have led to that conclusion.

From these observations two types of highlights can be summarized: Magnetic fields and Gravity Force.

Electromagnetic Field: If there is a negative or positive charge, these charges should be observed through the electric and magnetic fields by means of microwaves. However, there is currently no observation that supports any electromagnetic charge in space.

Force of gravity: The only force capable of maintaining the attraction of the planets, without affecting (at certain scales), is the force of gravity. An electromagnetic force often exceeds that of gravity, but has not been observed at the moment.

Final answer:

Mathematical and observational evidence from electromagnetism and cosmology, such as the Maxwell-Einstein theoretical framework and charge quantization, supports the assertion that the universe is electrically neutral on large scales.

Explanation:

To support the assertion that the universe is electrically neutral on large scales, one can turn to the principles of electromagnetism and cosmology melded in the framework of the general theory of relativity. A key aspect of this inquiry relates to the near uniform distribution of matter and the overall balance of positive and negative charges in the cosmos.

On quantum scales, charge quantization ensures that matter is made up of atoms with equal numbers of protons and electrons, contributing to the electrical neutrality on a smaller scale. This neutrality scales up to larger structures in the universe. Observational evidence also points towards the large-scale neutrality of the universe; otherwise, we would observe huge electrostatic forces that would significantly alter the structure of the cosmos.

Answer:

I think d is the answer haha

Galaxy collisions were more common in the past as C. the universe was smaller, hence galaxies were closer together.

Galaxy collisions were more common in the past than they are today primarily because galaxies were closer together in the past due to the smaller size of the universe. As the universe expanded, it carried galaxies along with it, increasing the distances between them and making collisions less frequent. When the universe was younger, galaxy interactions and collisions were frequent events that significantly influenced galaxy evolution, increased star-formation rates, and contributed to the prevalence of quasars during that time.

Observations indicate that when the universe was about 20% of its current age, interactions such as galaxy mergers happened most frequently, likely accounting for the active quasar populations observed from that epoch. These collisions often resulted in starburst galaxies, and the debris from these encounters could fuel the supermassive black holes at the centers of galaxies.

Answer:

given,

mass of the block = m₁

mass of the another block = 3 m₁

initial Amplitude, A = 6 cm

final amplitude = 6 cm

total mechanical energy = 12 J

total energy of the block spring

   [tex]E = \dfrac{1}{2}kA^2[/tex]

A is the amplitude and k is spring constant

initial energy is equal to 12 J

from the above expression we can say that

Energy of the given system depends up on the magnitude of spring constant and the amplitude.

so, energy of both the system will be same.

we know,

[tex]E = \dfrac{1}{2}mv^2[/tex]

[tex]12= \dfrac{1}{2}\times 3 m_1 v^2[/tex]

  [tex]v^2 = \dfrac{8}{m_1}[/tex]

  [tex]v = \sqrt{ \dfrac{8}{m_1}}[/tex]

Answer: a. three times as large as the initial value.

Explanation:

To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature (isothermal).  

The equation given by this law is:

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.

[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.

We are given:

[tex]P_1=p\\V_1=v\\P_2=?\\V_2=\frac{v}{3}[/tex]

Putting values in above equation, we get:

[tex]p\times v=P_2\times \frac{v}{3}\\\\P_2=3p[/tex]

Thus the resulting pressure will be three times as large as the initial value.

When an ideal gas is compressed isothermally to one-third of its initial volume, the resulting pressure will be three times as large as the initial value. This is derived from the ideal gas law which illustrates the inverse relationship between pressure and volume during an isothermal process.

The behavior of an ideal gas during an isothermal compression can be understood using the ideal gas law, which states that the Pressure times the Volume (PV) equals the number of gas moles (n), times the gas constant (R), times the temperature (T). Symbolically, this can be written as PV=nRT. In an isothermal process, the temperature (T) remains constant, meaning that pressure and volume are inversely proportional. If the volume of the gas is compressed to one third of its initial value (V = V_initial/3), the pressure would increase three times the initial value (P = 3*P_initial). Therefore, the answer to the question is (a) the resulting pressure will be three times as large as the initial value.

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Answer:

A. The time it takes the projectile to reach the top of its path is about 1 second.

Explanation:

Hi there!

The equation of the velocity of a projectile fired straight up is the following:

v = v0 + g · t

Where:

v = velocity of the projectile.

v0 = initial velocity.

g = acceleration due to gravity (≅ -9.8 m/s² considering the upward direction as positive)

t = time.

When the projectile reaches the top of its path, its velocity is zero, then, using the equation of velocity, we can solve it for the time:

v = v0 + g · t

0 = 10 m/s - 9.8 m/s² · t

t = -10 m/s / -9.8 m/s²

t = 1.0 s

The time it takes the projectile to reach the top of its path is about 1 second.

Final answer:

a. The largest angular magnification that Galileo could have obtained with the given lenses is 100x. b. The length of the telescope between the two lenses is 40 cm.

Explanation:

a. Largest angular magnification

The angular magnification of a telescope is equal to the product of the magnification produced by the objective lens and the magnification produced by the eyepiece lens. In this case, we need to find the largest possible angular magnification using the given lenses.

The magnification produced by a lens can be calculated using the formula: magnification = - / , where is the distance of the final image from the eyepiece and is the distance of the object from the objective lens.

Since the angular magnification is the product of the magnifications produced by the two lenses, we can find the largest possible angular magnification by maximizing the product of the two magnifications.

The largest possible angular magnification in this case is 100x.

b. Length of the telescope

The length of the telescope can be calculated by summing the focal lengths of the two lenses and the distances between them. In this case, the length of the telescope between the two lenses is the sum of the focal lengths of the two lenses, which can be calculated using the formula: length = + , where is the focal length of the objective lens and is the focal length of the eyepiece lens.

The length of the telescope between the two lenses is 40 cm.

Answer

given,

frequency of sound (f)= 120 Hz

speed of sound (v)= 343 m/s

a) speed of approaching = 3 % of speed of sound

                          = 0.03 x 343 = 10.3 m/s

     frequency you would hear

     [tex]f' = f\dfrac{v+v_0}{v}[/tex]

     [tex]f' = 120 \times \dfrac{343+10.3}{343}[/tex]

            f' = 123.60 Hz

b)  speed of going away = 3 % of speed of sound

                          = 0.03 x 343 = 10.3 m/s

     frequency you would hear

     [tex]f' = f\dfrac{v-v_0}{v}[/tex]

     [tex]f' = 120 \times \dfrac{343-10.3}{343}[/tex]

            f' = 116.39 Hz

Answer:

correct answer is b    Z --->Z+1

Explanation:

In the processes of radioactive decay there are three basic processes the emission of alpha particles and the emission of beta rays and the emission of ayos range

The emission of a beta ray implies the transformation of a neutral into a proton, which implies the increase of the atomic number in a unit

          Z   ---->  Z +1

the atomic mass does not change since the mass of the two particles is practically the same, to balance the reaction antineutrino must also be emitted

The daughter particle is in an execrated state and passes to its base state with the emission of a gamma ray that does not change its atomic number or its atomic mass.

Consequently, from the above the correct answer is b

Answer:

74 N

Explanation:

T = Tension in the string = 74 N

[tex]\rho[/tex] = Density of the steel = 7800 kg/m³

A = Area = [tex]\pi r^2[/tex]

r = Radius = [tex]8.5\times 10^{-4}\ m[/tex]

Linear density is given by

[tex]\mu=\rho A\\\Rightarrow \mu=7800\times \pi (8.5\times 10^{-4})^2\\\Rightarrow \mu=0.0177\ kg/m[/tex]

The linear density is 0.0177 kg/m

Velocity is given by

[tex]v=\sqrt{\dfrac{T}{\mu}}\\\Rightarrow v=\sqrt{\dfrac{74}{0.0177}}\\\Rightarrow v=64.65903\ m/s[/tex]

The velocity of the wave on the guitar string is 64.65903 m/s

Entropy is the energy that cannot be used to do useful work.

The correct answer is b. entropy. Entropy is a measure of the disorder or randomness in a system. It is a form of energy that cannot be converted into useful work. For example, when energy is transferred from one form to another, such as from chemical energy to thermal energy, some of the energy is lost as heat, which is a form of entropy.

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The weight of a 15kg dog on Earth is 147 Newtons. Weight is calculated by multiplying the object's mass by the acceleration due to gravity on Earth, which is approximately 9.8 m/s². The result is rounded to the tenths place if necessary, though not required in this calculation.

The weight of an object is the force due to gravity acting on that object's mass. On Earth, the weight of an object can be calculated by multiplying its mass by the acceleration due to gravity, which is approximately 9.8 m/s2. Therefore, the weight of a 15kg dog on Earth is calculated as 15 kg × 9.8 m/s2, which equals 147 Newtons (N).

When working with measurements and answers in physics, it's important to consider precision. Since the least precise measurement provided in the reference information is 13.7 kg, expressed to the 0.1 decimal place, any calculated weight should also be expressed to the tenths place. However, since our calculation already gives a whole number without a decimal component, further rounding is not required for this particular scenario.

Answer:

FN₁ = 1146.6N : Force exerted on the ladder by the floor , vertical and upward

FN₂ = 407.5 N : Force exerted on the ladder by the wall , horizontal and opposite to friction force between the floor and the ladder

Explanation:

The equilibrium equation are:

∑Fx=0

∑Fy=0

∑M = 0  

M = F*d  

Where:  

∑M : Algebraic sum of moments

M : moment  ( N*m)  

F : Force ( N)  

d :Perpendicular distance of the force to the point  ( meters )

Data

m =45 kg  : mass of the ladder

M =72 kg : mass of the fire fighter

g = 9.8 m/s²: acceleration due to gravity

L = 12 m : ladder length

h =  9.3 m: ladder height

L/3 = 12/3 = 4m Location of the center of mass of the ladder of the way up

L/2 = 12/2 = 6m Location of the center of mass of the  fire fighter

µ = 0 : coefficient of friction between the ladder and the wall

θ  : angle that makes  the  ladder  with the floor

sinθ = h/L = 9.3 m/12 m

θ =sin⁻¹( 9.3 / 12)

θ = 50.8°

Forces acting on the ladder

W₁ =m*g = 45 kg* 9.8 m/s² = 441 N: Weight of the ladder (vertical downward)

W₂ =M*g = 72 kg * 9.8 m/s² = 705.6 N : Weight of the fire fighter(vertical downward)  

FN₁ :Normal force that the floor exerts on the ladder (vertical upward)  (point A)

fs : friction force that the floor exerts on the ladder (horizontal and opposite the movement )(point A)  

FN₂ :  Normal Force that the wall exerts on the ladder ( horizontal and opposite to friction force between the floor and the ladder)

∑Fy=0

FN₁ -W₁ -W₂= 0

FN₁ = W₁ + W₂

FN₁ = 441N+ 705.6N

FN₁ = 1146.6N : Force exerted on the ladder by the wall (vertical and upward)

Calculation of the distances of the forces at the point A (contact point of the ladder on the floor)

d₁ = 4*cos 50.8° (m) = 2.53 m:  Distance from W₁ to the point A

d₂ =6*cos 50.8° (m)= 3.79 m  : Distance from W₂ to the point A

d₃ = 9.3 m : Distance from FN₂ to the point A

The equilibrium equation of the moments at the point A  (contact point of the ladder with the floor)  

∑MA = 0  

FN₂(d₃) - W₁( d₁) - W₂(d₂) = 0

FN₂(d₃) = W₁(d₁) + W₂(d₂)

FN₂(9.3) = (441 )(2.53) + (705.6)( 3.79 )

FN₂(9.3) = 1115.73 + 2674.2

FN₂ = (3790) / (9.3)  

FN₂ = 407.5 N : Force exerted on the ladder by the wall , horizontal and opposite to friction force between the floor and the ladder

Force excreted by floor on ladder is 1146.6 N (vertically upward) and force excreted by wall on ladder is 407.5 N (horizontally towards the normal force).

Force is the effect of pull or push due to which the object having a mass changes its velocity.

The force is of two types-

The length of the ladder is 12 meter and the mass of the ladder is 45 kg.Its upper end is a distance h of 9.3 and the mass of fire fighter is 72 kg.

The sine angle for the ladder can be given as,

[tex]\sin\theta=\dfrac{9.3}{12}\\\theta=50.8^o[/tex]

The forces acting on the system is normal force and force due to the weight of the two bodies.

The summation of the vertical forces is equal to the zero to keep it at rest. Therefore,

[tex]F_n-45\times9.8-72\times9.8\\F_n=1146.6 \rm N[/tex]

By the equilibrium of the momentum for the system,

[tex]F_{N2}\times(9.3)+45(9.8)\times(4\cos50.8)+72(9.8)\times(6\cos50.8)\\F_{N2}=407.5\rm N[/tex]

Thus, the force excreted by the floor on ladder is 1146.6 N (vertically upward) and the force excreted by the wall on ladder is 407.5 N (horizontally towards the normal force).

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The tensile strength and ductility cannot be directly calculated from the provided information, but the proportion of cold work (or percentage deformation) can be computed based on the diameter change.

Despite the detailed context provided, the question lacks sufficient data or formulas to directly compute the tensile strength nor the ductility of a copper rod from its diameter change due to cold working. Usually, the tensile strength and ductility mean the ultimate tensile strength (the maximum stress that a material can withstand while being stretched or pulled before failing or breaking) and the percent elongation after a material specimen has been pulled and rupture occurs. To get these values, experimenting with the material and measuring would be necessary.

However, we can calculate the percentage of cold work using the geometric deformation change. The percentage of cold work (%CW) based on change in diameter can be calculated with the formula: %CW = [(initial area - final area)/initial area] x 100%. Here, the area is that of the rod cross-section, which for a cylinder is pi*(d/2)². Thus you can substitute and calculate for %CW with the given diameters.

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Answer:

a)    t = 0.113 s , b)     W = 0.1756 N , c)   # λ = 49

Explanation:

a) Let's use the relationship

         v = λ f

Of the equation

         y = 8.55 10-3 cos (172 x + 2730 t)

When comparing this with the general equation

         y = A cos (kx - wt)

The wave number k = 172

       k = 2π /λ

       λ = 2π / 172

       λ = 0.03653 m

The angular velocity w = 2730

     w = 2π f

      f = w / 2π

     f = 2730 / 2π

     f = 434.49 Hz

The speed of the wave is

      v = 0.03653    434.49

      v = 15.87 m / s

The speed the wave on a string is constant, so

      v = d / t

      t = d / v

      t = 1.80 /15.87

      t = 0.113 s

b) The weight applied to the rope

      v = √ T /μ

The density

     μ = m / l

     μ = (0.0123 / 9.8) /1.80

     μ = 6.97 10-4 kg / m

The tension equal to the applied weight

    T = v² μ

    W = T = 15.87²   6.97 10⁻⁴

    W = 0.1756 N

c) let's use a rule of proportions

    # λ = 1.8 /0.03653

      # λ = 49

Answer:

5.5 × 10^14 Hz or s^-1

no orange light has less frequency so no photoelectric effect

Explanation:

hf = hf0 + K.E

HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s

f is frequency of incident photon and f0 is threshold frequency

hf0 = hf- k.E

6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20

6.63 × 10 ^-34 × f0 = 3.64158×10^-19

                           f0 = 3.64158×10^-19/ 6.63 × 10 ^-34

                           f0 = 5.4925 × 10^14

                            f0 =5.5 × 10^14 Hz or s^-1

frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light

In order to solve this problem it is necessary to apply the concepts related to Newton's second law and the respective representation of the Forces in their vector components.

The horizontal component of this force is given as

F_x = Fcos(6.7)

While the vertical component of this force would be

F_y = Fsin(6.7)

In the vertical component, the sum of Force indicates that:

[tex]\sum F_y= 0 [/tex]

The Normal Force would therefore be equivalent to the weight and vertical component of the applied force, therefore:

[tex]N = mg+Fsin(6.7)[/tex]

In the horizontal component we have that the Force of tension in its horizontal component is equivalent to the Force of friction:

[tex]\sum F_x = 0[/tex]

[tex]F_x = F_{friction}[/tex]

[tex]Fcos (6.7) = N\mu[/tex]

Using the previously found expression of the Normal Force and replacing it we have to,

[tex]Fcos(6.7)= \mu (mg+Fsin(6.7))[/tex]

Replacing,

[tex]Fcos(6.7)= (0.87) (mg+Fsin(6.7))[/tex]

[tex]Fcos(6.7) = (0.87)(mg) + (0.87)(Fsin(6.7))[/tex]

[tex]Fcos(6.7) -(0.87)(Fsin(6.7)) = 0.87 (mg)[/tex]

[tex]F(cos(6.7)-0.87sin(6.7)) = 0.87 (mg)[/tex]

[tex]F = \frac{0.87 (mg)}{(cos(6.7)-0.87sin(6.7))}[/tex]

[tex]F = \frac{0.87(128000*9.8)}{(cos(6.7)-0.87sin(6.7))}[/tex]

[tex]F = 1.95*10^6N[/tex]

Finally the acceleration would be by Newton's second law:

[tex]F = ma[/tex]

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{ 1.95*10^6}{128000}[/tex]

[tex]a = 15.234m/s^2[/tex]

Therefore the greatest acceleration the man can give the airplane is [tex]15.234m/s^2[/tex]

the greatest acceleration the man can give the airplane is approximately 7.38 m/s²

The greatest acceleration the man can give the airplane when pulling at an angle of 6.7° above the horizontal is given by the equation:

[tex]\[ a = \frac{\mu (m_m + m_a) g - m_a g \sin(\theta)}{m_m + m_a \cos(\theta)} \][/tex]

where:

- [tex]\( \mu \)[/tex] is the coefficient of static friction between the man's shoes and the runway,

- [tex]\( m_m \)[/tex] is the mass of the man,

- [tex]\( m_a \)[/tex] is the mass of the airplane,

- g is the acceleration due to gravity,

- [tex]\( \theta \)[/tex] is the angle at which the man pulls the cable.

Given:

[tex]- \( \mu = 0.87 \),[/tex]

[tex]- \( m_m = 76 \) kg,[/tex]

[tex]- \( m_a = 128000 \) kg,[/tex]

[tex]- \( g = 9.8 \) m/s²,[/tex]

[tex]- \( \theta = 6.7° \).[/tex]

First, we need to convert the angle from degrees to radians because the sine and cosine functions in trigonometry typically use radians:

[tex]\[ \theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180} \][/tex]

[tex]\[ \theta_{\text{radians}} = 6.7° \times \frac{\pi}{180} \][/tex]

Now, we can plug in the values into the equation:

[tex]\[ a = \frac{0.87 (76 \text{ kg} + 128000 \text{ kg}) \times 9.8 \text{ m/s}^2 - 128000 \text{ kg} \times 9.8 \text{ m/s}^2 \times \sin(6.7°)}{76 \text{ kg} + 128000 \text{ kg} \times \cos(6.7°)} \][/tex]

Calculating the sine and cosine of 6.7°:

[tex]\[ \sin(6.7)\approx 0.117 \][/tex]

[tex]\[ \cos(6.7) \approx 0.993 \][/tex]

Now, we substitute these values into the equation:

[tex]\[ a = \frac{0.87 (76 + 128000) \times 9.8 - 128000 \times 9.8 \times 0.117}{76 + 128000 \times 0.993} \][/tex]

Solving the numerator:

[tex]\[ (0.87 \times 128076 \times 9.8) - (128000 \times 9.8 \times 0.117) \][/tex]

[tex]\[ = (0.87 \times 128076 \times 9.8) - (128000 \times 1.1466) \][/tex]

[tex]\[ = 939608.16 \][/tex]

Solving the denominator:

[tex]\[ 76 + (128000 \times 0.993) \][/tex]

[tex]\[ = 127308 \][/tex]

Finally, we divide the numerator by the denominator to find the acceleration:

[tex]\[ a = \frac{939608.16}{127308} \][/tex]

[tex]\[ a \approx 7.38 \text{ m/s}^2 \][/tex]

The answer is: 7.38.

Answer:

distance cover is  = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

[tex]t_1 = 3.32 sec[/tex]

[tex]t_2 = 5.08 sec[/tex]

from equation of motion we know that

[tex]d_1 = vt_1 + \frac{1}{2} gt_1^2[/tex]

where d_1 is distance covered in time t1

so[tex] d_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2 [/tex]=

[tex]d_1 = 110.78 m[/tex]

[tex]d_2 = vt_2 + \frac{1}{2} gt_2^2[/tex]

where d_2 is distance covered in time t2

[tex]d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2[/tex]

[tex]d_2 = 213.31 m[/tex]

distance cover is  = 213.31 - 110.78 = 102.53 m

Answer:

Both equilibrium price and quantity will increase.

Explanation:

An increase in the income of video game players will surely lead to an upward shift in the supply and demand curve. This shift in the supply and demand curve would affect the equilibrium price and quantity positively leading to a correspondent increase in the equilibrium price and quantity.

If video game disks are a normal good, an increase in the income of video game players would likely lead to an increased demand for video games. This is because, with increased income, players now have more purchasing power ability to buy more video games. Other factors such as popularity, population, and price of substitutes can also influence this demand.

Your question relates to the concept of normal goods within the field of economics. According to economic theory, a normal good is a good for which demand increases as income increases. Therefore, if the income of video game players increases, according to economic principles, you can expect the demand for video games (assuming they are a normal good) to also increase.

This increase in demand in the market for video games originates from the increased purchasing power. That is, players now have the ability to buy more video games than before. Additionally, factors such as taste shift to greater popularity, the population likely to buy rises, and price of substitutes can further influence this demand.

However, it's essential to also consider that economic theory often simplifies real-world conditions. In reality, numerous variables could influence the market demand for video games beyond income changes.

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Answer:

Explanation:

Fission means breaking or splitting i.e. Nuclear fission involves breaking up of unstable nucleus into smaller pieces and simultaneously producing energy.  

Fusion means joining of different elements to form a unified whole. In Nuclear fusion smaller nuclei combined to form a new heavy nucleus associated with a large amount of energy release.

Thus nuclear Fusion and fission can release Energy.                                        

Answer:

The magnitude of the electric field at a point 4.0 cm from the center of the two surfaces is 4.10 N/C

Explanation:

Step 1: Data given

Charge of uniform density of first sphere= 40 pC/m² = 40.0 * 10^-12 C/m²

Radius of first sphere is 1.0 cm

Radius of second sphere = 3.0 cm

Charge of uniform density of second sphere= 60 pC/m² = 60.0 * 10^-12 C/m²

Step 2: Calculate the magnitude

Sphere surface area = 4πr²

Charge on inner sphere Qi = 40.0*10^-12C/m² * 4π(0.01m)² = 5.027*10^-14 C

NET charge on outer sphere Qo = 60.0*10^-12 * 4π(0.03m)² = 6.786*10^-13 C

Inner sphere induces a - 5.027*10^-14 C  charge (-Qi) on inside of the outer shell

This means there is a net zero charge within the outer shell.

For the outer shell to show a NET charge +6.786*10^-13C, it's must have a positivie charge {+6.786*10^-13C + (+5.027*10^-14C)} =  +7.2887*10^-13 C

Regarding the outer shell as a point charge (field at 0.04m is)

E = kQ /r²

E = (8.99^9)*(7.2887*10^-13 C) / (0.04)² .. .. ►E = 4.10 N/C

========

Using Gauss's law .. with a spherical Gaussian surface  at 0.04m enclosing a net charge +7.2887*10^-13 C

Flux EA = Q / εₒ

  ⇒ with εₒ  = 8.85 *10^-12

⇒ with Q = 7.2887*10^-13 C

E * 4π(0.04)² = 7.2887*10^-13 C / 8.85*10^-12  

E =  7.2887*10^-13 C / {8.85*10^-12 * 4π(0.04)²}

E = 4.10 N/C

The magnitude of the electric field at a point 4.0 cm from the center of the two surfaces is 4.10 N/C

A correct option is option (2).

Given,

The Surface area of a sphere [tex]A=4\pi r^2[/tex]

Find the charge on the inner sphere.

[tex]Q_i =40.0^{2} c/m^2\times4\pi (0.01m)^2\\=5.03^{-14 C}[/tex]

In this case, the Inner sphere induces a [tex]5.03^{-14 C}[/tex]charge [tex](-Q_i)[/tex] on the inside of the outer shell.

Therefore, there is a net-zero charge within the outer shell.

For the outer shell to show a net charge [tex]+6.79^{-13}C[/tex], it's must have a +ve charge.

[tex]{+6.79^{-13} C + (+5.03^{-14} C)} = +7.29^{-13} C[/tex]

Now, Use Gauss's law, with a spherical Gaussian surface at 0.03m enclosing a net charge[tex]+7.29^{-13} C[/tex]

[tex]Flux=E_A\\=\frac{Q}{\varepsilon _0 }[/tex]

Substitute numerical values we get,

[tex]E\times4\pi (0.04)^2=\frac{.29^{-13} }{8.85^{-12} } \\E=4.5 N/C[/tex]

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Answer:

option D.  

Explanation:

The correct answer is option D.                

Diffusion is the particle movement from high concentration to low concentration, such as air, water, etc.                      

Example: if you spray perfume spread throughout the room at one part of the fragrance scent.                                                        

The gas molecule thus spreads out in a gradient of concentration

Diffusion is equivalent to gas molecules spreading out in a concentration gradient. Option D.

Diffusion is the movement of particles (such as gas molecules, liquid molecules, or even ions) from an area of higher concentration to an area of lower concentration.

This process occurs spontaneously and is driven by the natural tendency of particles to move and distribute themselves evenly in a given medium.

Diffusion is essential in various biological, chemical, and physical processes and plays a crucial role in the movement of substances within and between cells, as well as in the mixing of gases and liquids in the environment.

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Answer:

a) Please, see the attached figure.

b) The horizontal component of the initial velocity is 8.5 m/s

The vertical component of the initial velocity is 6.0 m/s

c) The clown will return to a height of 1.0 m after 1.2 s of the launch.

d) The clown will land safely on the mattress, 10.2 m from the cannon. If we include air resistance in the calculation, he will surely not reach the mattress because, without air resistance, he lands just 20 cm from the closest edge of the mattress.

Explanation:

Hi!

a) Please, see the attached figure.

b) As shown in the figure, the initial velocity vector is the following:

v0 = (v0x, v0y)

Using trigonomety of right triangles:

cos angle = adjacent side / hypotenuse

In this case:

Adjacent side = v0x

hypotenuse = v0

(see figure)

Then:

cos 35° = v0x / v0

v0 · cos 35° = v0x

v0x = 10.4 m/s · cos 35°

v0x = 8.5 m/s

The horizontal component of the initial velocity is 8.5 m/s

We proceed in the same way to find the vertical component:

sin angle = opposite side / hypotenuse

sin 35° = v0y / v0 (see figure to notice that opposite side = v0y)

v0 · sin 35° = v0y

10.4 m/s · sin 35° = v0y

v0y = 6.0 m/s

The vertical component of the initial velocity is 6.0 m/s

c) The equation of the position vector of the clown at time t is the following:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

We have to find the time at which the vertical component of the position vector is 1 m. Let´s place the origin of the system of reference at the point where the cannon is located on the ground so that x0 = 0 and y0 = 1.0 m.

Using the equation of the vertical component of the position:

y = y0 + v0 · t · sin α + 1/2 · g · t²

1.0 m = 1.0 m + 10.4 m/s · t · sin 35° - 1/2 · 9.8 m/s² · t²

0 = 10.4 m/s · t · sin 35° - 4.9 m/s² · t²

0 = t (10.4 m/s · sin 35° - 4.9 m/s² · t) (t = 0, when t = 0 the clown is 1.0 m above the ground, just leaving the cannon).

0 = 10.4 m/s · sin 35° - 4.9 m/s² · t

-10.4 m/s · sin 35° / -4.9 m/s² = t

t = 1.2 s

The clown will return to a height of 1.0 m after 1.2 s of the launch.

d) Now, let´s calculate the horizontal traveled distance after 1.2 s using the equation of the horizontal component of the position vector:

x = x0 + v0 · t · cos α (x0 = 0)

x = 10.4 m/s · 1.2 s · cos 35°

x = 10.2 m

Since the mattress is located at 10 m from the cannon and it is 2.0 m long, the clown will land safely on the mattress. However, the clown almost miss the mattress (he lands just 20 cm from the closest edge), so, if we include air resistance in the calculation, he will surely not reach the mattress.

The question seems incomplete. I found a similar version with different details. For the purpose of helping you answering the question, I will use the details from the original questions added with the sub-questions from other question. You can change the details accordingly using the explanation given below:

At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the other side of the pivot point an adult pushes straight down on the teeter-totter with a force of 190 N. Determine the direction the teeter-totter will rotate if the adult applies the force at each of the following distances from the pivot. (Assume the teeter-totter is horizontal when the adult applies the force and that the child's weight applies a clockwise torque.)

(a) 3.0 m

(b) 2.5 m

(c) 1.5 m

Answer:

a) anticlockwise

b) anticlockwise

c) clockwise

Explanation:

To answer this question, we will need the knowledge of moment/torque

Torque, T = r*Fsinθ

r = radial distance from pivot to force, F

F = Force in the direction perpendicular to the distance

θ = angle between force line and r line

Note that the direction of movement will be affected by total torque on the system. Assuming both adult and the child is initially sitting on the same level, we'll have force due to the child weight and adult force going downward and perpendicular to their respective seat on teeter-totter.

Hence,

θ = 90 degrees ->  sinθ = 1

Therefore,

T = r*F(1)

T = r*F

RHS:

Torque of the child,

Tc = 1.8(21*9.8) = 370.44 Nm

LHS:

Torque of adult

Ta = d(190) =190d

with d as the distance of the adult from pivot

a) d = 3.0m

Ta = 190*3 = 570 Nm

Ta > Tc. Therefore the teeter-totter will rotate towards the adult side (LHS - going anti-clockwise)

b) d = 2.5m

Ta = 190*2.5 = 475 Nm

Ta > Tc. Therefore the teeter-totter will rotate towards the adult side (LHS - going anti-clockwise)

c) d = 1.5m

Ta = 190*1.5 = 285

Ta < Tc. Therefore the teeter-totter will rotate towards the child side (RHS - going clockwise)