Initially water is coming out of a hose at a velocity of 5 m/s. You place your thumb on the opening, reducing the area the water comes through to .001 $m^2$. You then find the velocity of the water to be 20 m/s. Calculate the radius of the hose.

Answer:

[tex]\dfrac{I}{I_0}=\dfrac{3}{8}[/tex]

Explanation:

Given that

Angle ,θ = 30°

From Malus law,Intensity given as

[tex]I=\dfrac{I_0}{2}cos^2\theta[/tex]

Io=Intensity of unpolarized light

I=Intensity of emerging light

Now by putting the value of angle

[tex]I=\dfrac{I_0}{2}cos^2\theta[/tex]

[tex]I=\dfrac{I_0}{2}cos^230^{\circ}[/tex]

We know that

[tex]cos30^{\circ}=\dfrac{\sqrt{3}}{2}[/tex]

[tex]I=\dfrac{I_0}{2}\times \dfrac{3}{4}[/tex]

[tex]\dfrac{I}{I_0}=\dfrac{3}{8}[/tex]

Therefore ratio will be [tex]\dfrac{3}{8}[/tex]

Answer:

Ratio of the intensity of emerging light[tex]$\frac{I}{I_{0}}=\frac{3}{8}$[/tex]

Explanation:

Given:

Angle,[tex]$\theta=30^{\circ}$[/tex]

Step 1:

According to Malus law,

Intensity,

[tex]$I=\frac{I_{0}}{2} \cos ^{2} \theta$[/tex]

[tex]l_{0} =[/tex]Intensity of unpolarized light

[tex]l=[/tex]Intensity of emerging light

Step 2:

Put the value of angle

[tex]$I=\frac{I_{0}}{2} \cos ^{2} \theta$[/tex]

[tex]$I=\frac{I_{0}}{2} \cos ^{2} 30^{\circ}$[/tex]

We know,

[tex]$\cos 30^{\circ}=\frac{\sqrt{3}}{2}$[/tex]

[tex]$I=\frac{I_{0}}{2} \times \frac{3}{4}$[/tex]

So, the intensity of the ratio

[tex]$\frac{I}{I_{0}}=\frac{3}{8}$[/tex]

Therefore, the ratio of the intensity of light is  [tex]$\frac{3}{8}$[/tex]

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Answer:

[tex]\omega_f = 0.602\ rad/s[/tex]

Explanation:

given,

radius of merry- go- round = 2.80 m

moment of inertia = I = 2400 kg⋅m²

child apply force tangentially = 20 N

for time = 25 s

angular speed after 25 speed = ?

initial angular speed of the merry go round = 0 rad/s

we know,

torque = I α.............(1)

α is angular acceleration

and also

τ = F.r........................(2)

computing equation (1) and (2)

F . r = I α

[tex]\alpha = \dfrac{\omega_f - \omega_i}{t}[/tex]

[tex]F . r =I \times \dfrac{\omega_f - \omega_i}{t}[/tex]

[tex]20 \times 2.89 =2400\times \dfrac{\omega_f -0}{25}[/tex]

[tex]\omega_f = 0.602\ rad/s[/tex]

the angular speed of merry-go-round after 25 second is equal to [tex]\omega_f = 0.602\ rad/s[/tex]

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:

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:

(a) 0.42 m

(b) 20.16 N/m

(c) - 0.42 m

(d) - 0.21 m

(e) 17.3 s

Solution:

As per the question:

Mass, m = 0.56 kg

x(t) = (0.42 m)cos[cos(6 rad/s)t]

Now,

The general eqn is:

[tex]x(t) = Acos\omega t[/tex]

where

A = Amplitude

[tex]\omega[/tex] = angular frequency

t = time

Now, on comparing the given eqn with the general eqn:

(a) The amplitude of oscillation:

A = 0.42 m

(b) Spring constant k is given by:

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

[tex]\omega^{2} = \frac{k}{m}[/tex]

Thus

[tex]k = m\omega^{2} = 0.56\times 6^{2} = 20.16\ N/m[/tex]

(c) Position after one half period:

[tex]x(t) = 0.42cos\pi = - 0.42\ m[/tex]

(d) After one third of the period:

[tex]x(t) = 0.42cos(\frac{2\pi}{3}) = - 0.21\ m[/tex]

(e) Time taken to get at x = - 0.10 m:

[tex]-0.10 = 0.42cos6t[/tex]

[tex]6t = co^{- 1} \frac{- 0.10}{0.42}[/tex]

t = 17.3 s

The amplitude of oscillation is 0.42 m. The force constant of the spring is approximately 0.03 N/m. The position of the mass after one half a period, one-third of a period, and at x = -0.10 m can be determined using the given equation x(t).

The amplitude of oscillation is the maximum displacement from the equilibrium position. In this case, the amplitude is given as 0.42 m.

The force constant (k) of the spring can be determined using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from equilibrium. In this case, the force constant is calculated by dividing the mass (0.56 kg) by the square of the angular frequency (6 rad/s) squared, which gives a force constant of approximately 0.03 N/m.

The position of the mass after it has been oscillating for one half a period can be found by substituting the value of time (T/2) into the equation x(t), which gives a position of approximately -0.42 m.

The position of the mass one-third of a period after it has been released can be determined by substituting the value of time (T/3) into the equation x(t), which gives a position of approximately 0.33 m.

The time it takes the mass to get to the position x = -0.10 m after it has been released can be found by rearranging the equation x(t) and solving for time. The time is approximately 0.14 seconds.

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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:

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.

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

[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:0.122 N/m

Explanation:

Given

Maximum Amplitude can be obtained when tapped after every second

i.e. Time Period [tex]T=1 s[/tex]

mass of spider [tex]m=3.1 gm[/tex]

and we know

[tex]T\cdot \omega =2\pi [/tex]

where [tex]\omega [/tex]=natural frequency of oscillation

T=time Period

[tex]\omega [/tex] is also given by [tex]\omega =\sqrt{\frac{k}{m}}[/tex]

[tex]1\cdot \sqrt{\frac{k}{m}}=2\pi [/tex]

[tex]1\cdot \sqrt{\frac{k}{3.1\times 10^{-3}}}=2\pi [/tex]

[tex]k=(2\pi )^2\times 3.1\times 10^{-3}[/tex]

[tex]k=0.122 N/m[/tex]

The answer is: [tex]k \approx 0.12 \text{ N/m}.[/tex]

To determine the spring constant of the silk thread, we need to consider the spider's motion as simple harmonic motion (SHM). The frequency of the motion is given by the tapping rate, which is once every second, so the frequency [tex]\( f \)[/tex]is 1 Hz. The mass [tex]\( m \)[/tex]of the spider is 3.1 g, which we need to convert to kilograms for consistency with standard units. Since 1 g = 0.001 kg, we have:

 [tex]\[ m = 3.1 \text{ g} = 3.1 \times 10^{-3} \text{ kg} \][/tex]

The period [tex]\( T \)[/tex] of the motion is the reciprocal of the frequency:

[tex]\[ T = \frac{1}{f} = \frac{1}{1 \text{ Hz}} = 1 \text{ s} \][/tex]

For SHM, the period [tex]\( T \)[/tex] is related to the mass[tex]\( m \)[/tex] and the spring constant [tex]\( k \)[/tex] by the following equation:

[tex]\[ T = 2\pi \sqrt{\frac{m}{k}} \][/tex]

Solving for [tex]\( k \),[/tex] we get:

 [tex]\[ k = \frac{4\pi^2 m}{T^2} \][/tex]

Substituting the known values for [tex]\( m \), \( T \)[/tex], and using[tex]\( \pi \approx 3.14159 \)[/tex], we have:

[tex]\[ k = \frac{4\pi^2 \times 3.1 \times 10^{-3} \text{ kg}}{(1 \text{ s})^2} \][/tex]

[tex]\[ k = 4\pi^2 \times 3.1 \times 10^{-3} \text{ N/m} \][/tex]

[tex]\[ k = 4 \times (3.14159)^2 \times 3.1 \times 10^{-3} \text{ N/m} \][/tex]

[tex]\[ k \approx 4 \times 9.8696 \times 3.1 \times 10^{-3} \text{ N/m} \][/tex]

[tex]\[ k \approx 120.47 \times 10^{-3} \text{ N/m} \][/tex]

[tex]\[ k \approx 0.12047 \text{ N/m} \][/tex]

Therefore, the spring constant[tex]\( k \)[/tex]of the silk thread is approximately[tex]\( 0.12047 \text{ N/m} \).[/tex]

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:

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:

[tex]P_{sol}=50.4\ mm.Hg[/tex]

Explanation:

According to given:

Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.

moles of water in the given quantity:

[tex]n_w=\frac{m_w}{M_w}[/tex]

[tex]n_w=\frac{313.5}{18}[/tex]

[tex]n_w=17.42 moles[/tex]

moles of glycerin in the given quantity:

[tex]n_g=\frac{m_g}{M_g}[/tex]

[tex]n_g=\frac{154}{92}[/tex]

[tex]n_g=1.674 moles[/tex]

Now the mole fraction of water:

[tex]X_w=\frac{n_w}{n_w+n_g}[/tex]

[tex]X_w=\frac{17.42}{17.42+1.674}[/tex]

[tex]X_w=0.912[/tex]

Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.

[tex]\therefore P_{sol}=X_w\times P_x[/tex]

[tex]\therefore P_{sol}=0.912\times 55.32[/tex]

[tex]P_{sol}=50.4\ mm.Hg[/tex]

Answer:

Explanation:

frequency, f = 6.7 MHz = 6.7 x 10^6 Hz

Time period is defined as the time taken by the oscillating body to complete one oscillation.

the formula for the time period is

T = 1/ f = 1 / (6.7 x 10^6) = 1.5 x 10^-7 second

The angular frequency is given  by

ω = 2 π f = 2 x 3.14 x 6.7 x 10^6

ω = 4.2 x 10^7 rad/s

The time for each oscillation is approximately 1.49 x 10^-7 seconds, and the angular frequency is approximately 4.21 x 10^7 rad/s.

To find the time for each oscillation, we can use the formula T = 1/f, where T represents the period and f represents the frequency. In this case, the frequency is given as 6.7 MHz, which is equivalent to 6.7 x 10^6 Hz. So, substituting the value of the frequency into the formula, we get: T = 1 / (6.7 x 10^6) = 1.49 x 10^-7 seconds. Therefore, each oscillation takes approximately 1.49 x 10^-7 seconds.

The angular frequency can be calculated using the formula ω = 2πf, where ω represents the angular frequency and f represents the frequency. Substituting the given value of the frequency into the formula, we get: ω = 2π x 6.7 x 10^6 = 4.21 x 10^7 rad/s. Therefore, the angular frequency is approximately 4.21 x 10^7 rad/s.

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 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:

Power = 9.75 ×10^8[tex]\frac{kgm^2}{s^3}[/tex]

Explanation:

For one second 650m^3 of water flows out down to 150m oh depth.

So, the energy at a height of 150m is transformed to kinetic energy.

for a second,

       650m^3 of water flows down ⇒ (1000kg/m^3 × 650m^3) = 6.5×10^5kg of warer flos down.

The total gravitational potential energy stored in water is

    = mass of water × height× gravity

    = 6.5 ×10^5 × 150 × 10 =  9.75 ×10^8[tex]\frac{kgm^2}{s^2}[/tex]

As it is transformed in a second it is also equal to Power.

Answer:

power = 1407.77 MW

Explanation:

The basic principle of a hydro electric station is the conversion of potential energy to electrical energy. Here, water is allowed to fall from a height which will increase its the kinetic energy. This high speed flowing water is used to rotate the shaft of a turbine which will in turn produce electrical energy.

So here,

power = rate of change of potential energy with respect to time

power = [tex]\frac{d(mgh)}{dt}[/tex]

where,

m = mass of water

g = acceleration due to gravity

h = height through which the water falls

here h and g are constants ( h is the total height through which the water falls and it doesn't change with time). Therefore we can take them out of differentiation.

thus,

power = gh[tex]\frac{d(m)}{dt}[/tex]

now,

m = ρV

where,

ρ = density

V = volume

substituting this in the above equation we get,

power = gh[tex]\frac{d(ρV)}{dt}[/tex]

again ρ is a constant. Thus,

power = ρgh[tex]\frac{d(V)}{dt}[/tex]

Given that,

h = 221 m

[tex]\frac{d(V)}{dt}[/tex] = 650 [tex]m^{3}[/tex]/s

g = 9.8 m/[tex]s^{2}[/tex]

ρ = 1000 kg/[tex]m^{3}[/tex]

substituting these values in the above equation

power =  1000 x 9.8 x 221 x 650

power = 1407.77 MW

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.

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.

Answer:

e. Both the acceleration and net force on the car point inward.

Explanation:

If no net force acts on the car, the car must drive in a straight line, at constant speed.

As the acceleration is defined as the rate of change of the velocity vector, this means that it can produce either a change in the magnitude of the velocity (the speed) or in the direction.

In order to the car can follow a circular trajectory, it must be subjected to an acceleration, that must go inward, trying to take the car towards the center of the circle.

The net force that causes this acceleration, aims inward, and is called the centripetal force.

It is not a different type of force, it can be a friction force, a tension force, a normal force, etc., as needed.

The net force on a car driving around a circular track at steady speed is zero, resulting in zero acceleration. Hence, the correct option is c.

The net force on the car is zero. When a car drives at a steady speed around a circular track, the centripetal force required to keep it moving in a circle balances out the outward inertia. Therefore, there is no net force on the car. This leads to zero acceleration (choice c).

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:

option A.

Explanation:

The correct answer is option A.

Two rocks are off a bridge first rock is fallen 4  when the second rock is dropped.

Both the rock is dropped under the effect of acceleration due to gravity so, the rate of change of the velocity for both the rock particle will be the same.

Hence, the first rock will reach ground earlier than the second rock because the rate of change of both the rock is at the same rate.

At the time when the two rocks should continue to fall so their velocities should be increased.

The correct answer is option A.

When two rocks are off a bridge so here first rock is fallen 4 at the time when the second rock should be fallen. Both the rock should be fallen under the impact of acceleration because of gravity due to which the rate of change of the velocity for both the rock particle should be similar.

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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) 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:

Galaxy 1:

z = 0.0056

Galaxy 2:

z = 0.014

Galaxy 3:

z = 0.040

Explanation:

Spectral lines will be shifted to the blue part of the spectrum¹ if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect). The source in this particular case is represented for each of the galaxies of interest.

Hence, the redshift represents this shift of the spectral lines to red part in the spectrum of a galaxy or any object which is moving away. That is a direct confirmation of how the universe is in an expanding accelerated motion.

The redshift can be defined in analytic way by means of the Doppler velocity:

[tex]v = c\frac{\Delta \lambda}{\lambda_{0}}[/tex]  (1)

Where [tex]\Delta \lambda[/tex] is the wavelength shift, [tex]\lambda_{0}[/tex] is the wavelength at rest, v is the velocity of the source and c is the speed of light.

[tex]v = c(\frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}})[/tex]

[tex]\frac{v}{c} = \frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}}[/tex]  

[tex]z = \frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}}[/tex]  (2)

Where z is the redshift.

For the case of Galaxy 1:

Where [tex]\lambda_{measured} = 660.0 nm[/tex] and [tex]\lambda_{0} = 656.3 nm[/tex].

[tex]z = \frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}}[/tex]

[tex]z = (\frac{660.0 nm - 656.3 nm}{656.3 nm})[/tex]

[tex]z = 0.0056[/tex]

For the case of Galaxy 2:

Where [tex]\lambda_{measured} = 665.8 nm[/tex] and [tex]\lambda_{0} = 656.3 nm[/tex].

[tex]z = \frac{665.8 nm - 656.3 nm}{656.3 nm}[/tex]

[tex]z = 0.014[/tex]

For the case of Galaxy 3:

Where [tex]\lambda_{measured} = 682.7 nm[/tex] and [tex]\lambda_{0} = 656.3 nm[/tex].

[tex]z = \frac{682.7 nm - 656.3 nm}{656.3 nm}[/tex]

[tex]z = 0.040[/tex]

Key terms:

¹Spectrum: Decomposition of light in its characteristic colors (wavelengths).

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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