A small boat coasts at constant speed under a bridge. A heavy sack of sand is dropped from the bridge onto the boat. The speed of the boat

a. does not change.
b. increases.
c. Without knowing the mass of the boat and the sand, we can't tell.
d. decreases

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:

[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]

If a person pulls ten feet of the rope tied to a vehicle through a pulley, the vehicle will move the same distance, which is ten feet assuming ideal conditions with no slack or stretch in the system.

The question is asking about how far a vehicle moves when a rope attached to it and run through a pulley is pulled a certain distance. This scenario can be understood using the principles of distance and leverage as applied in physics. When you pull a rope through a pulley, the movement or distance covered by the object attached to the rope is equivalent to the length of the rope pulled.

In this case, if a person pulls ten feet of the rope, the vehicle will also move ten feet. It's important to note that this is applicable under ideal conditions where the rope and pulley do not stretch or bend and the there is no slack in the system.

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Answer:

Explanation:

Let the two inclined planes having angle of inclinations α and β.

The acceleration along the inclined plane acting on the body is gSinα and gSinβ.

If α > β

So, g Sinα > g Sinβ

So, more the inclination of the plane more be the acceleration of body and hence the time taken is less.

So, the body kept on the inclined whose inclination is more reaches at the bottom first.

The mass sitting on the plane inclined to a greater angle to the horizontal will reach the bottom first because it will experience greater acceleration.

The net horizontal force on the masses is calculated as follows;

[tex]\Sigma F_x = 0\\\\mgsin(\theta) = ma\\\\gsin(\theta) = a\ \ \ or \ \ \ \\\\gsin(\beta ) = a[/tex]

Assuming the value of the angles of inclination of the plane is the following;

a = gsin(60) = 0.866g

a = gsin(30) = 0.5g

Thus, we can conclude that the mass sitting on the plane inclined to a greater angle to the horizontal will reach the bottom first because it will experience greater acceleration.

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Answer:

45500 J

Explanation:

Pressure, P =350 kPa = 350 x 1000 Pa

V1 = 0.02 m^3

V2 = 0.15 m^3

Work done by the piston

W = Pressure x increase in volume

W = P x (V2 - V1)

W = 350 x 1000 (0.15 - 0.02)

W = 45500 J

Thus, the work done is 45500 J.

Final answer:

The piston performs 45500 Joules of work on the gas when the volume of the gas in the piston-cylinder apparatus increases from 0.02 m³ to 0.15 m³ at a constant pressure of 350 kPa.

Explanation:

To determine the work done by the piston on the gas when the cylinder volume increases from 0.02 m³ to 0.15 m³ at a constant pressure of 350 kPa, we can use the formula for work done during a quasi-static process in thermodynamics, which is W = PΔV, where W is the work done by the gas, P is the constant pressure, and ΔV is the change in volume.

Given:
P = 350 kPa = 350000 Pa (since 1 kPa = 1000 Pa)
ΔV = 0.15 m³ - 0.02 m³ = 0.13 m³

We can now calculate the work done:

W = PΔV
W = 350000 Pa × 0.13 m³
W = 45500 J

Therefore, the piston performs 45500 Joules of work on the gas when the volume increases to 0.15 m³.

Answer:

600N

Explanation:

If the ball bounces off upward with the same speed, the velocities before and after are equal and in opposite directions. That means the floor must have caused a momentum impact to change its direction.

By the law of momentum conservation:

[tex]mv + \Delta P_f = mV[/tex]

where m = 3 kg is the ball mass. v = -12 m/s is the velocity right before the impact. V = 12m/s is the velocity right after the impact. ΔP_f is the momentum caused by the floor.

[tex]\Delta P_f = m(V-v) = 3(12 - (-12)) = 3*24 = 72kgm/s[/tex]

Then the average force exerted during the 0.12s impact is

[tex]F = \frac{\Delta P_f}{\Delta t} = \frac{72}{0.12} = 600N[/tex]

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: 3. The puck B remains at the point of collision.

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved.

The initial momentum is due only to puck B, as Puck A is at rest.

The final momentum is given by the sum of the momenta of both pucks, so we can write the following equation:

mA*viA = (mA * vfA) + (mB * vfB)

As mA = mB = m, we can simplify the former equation as follows:

viA = vfA + vfB   (1)

Now, we also know, that the collision was an elastic collision, so total kinetic energy must be conserved too:

½ m viA²= ½ m vfA² + ½ m vfB²

Simplifying on both sides, we finally have:

viA² = vfA² + vfB²   (2)

Now, if we square both sides of (1), we get:

viA² = (vfA + vfB)²= vfA² + 2* vfA * vfB  +vfB² (3)

As the right side in (2) and (3) must be equal each other (as the left sides do), the only choice is that either vfA or vfB, be zero.

As we are told that puck A (initially stationary) after the collision, moves, the only possible choice is that puck B remain at rest in the point of collision, after the collision, exchanging his speed with puck A.

In an elastic collision where the total energy of the system is conserved, Puck B rebounds. This is because kinetic energy, along with momentum, is conserved in such a collision, causing the moving puck to rebound.

Considering this scenario from the lens of physics, the principle of conservation of momentum and kinetic energy comes into play, which means the total momentum and kinetic energy before and after the collision should remain constant. This is consistent with the properties of an elastic collision. Both pucks have the same mass and initially, Puck B is in motion while Puck A is stationary. Hence, all of the initial energy and momentum are with Puck B.

Upon collision, the energy is transferred, causing Puck A to move in the same direction that Puck B was initially moving. The correct statement is that if the collision is elastic and the total energy of the system is conserved, 'Puck B rebounds'. This rebounding comes from the conservation of kinetic energy principle which holds in an elastic collision. That is, the total initial kinetic energy equals the total final kinetic energy.

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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

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:

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

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:

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.

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:

a) 4458K b) 5048K, c) 6166K, d) 5573K

Explanation:

The temperature of the stars and many very hot objects can be estimated using the Wien displacement law

    [tex]\lambda_{max}[/tex] T = 2,898 10⁻³ [m K]

    T = 2,898 10⁻³ / [tex]\lambda_{max}[/tex]

a) indicate that the wavelength is

    Lam = 650 nm (1 m / 109 nm) = 650 10⁻⁹ m

    Lam = 6.50 10⁻⁷ m

    T = 2,898 10⁻³ / 6.50 10⁻⁷

    T = 4,458 10³ K

    T = 4458K

b) lam = 570 nm = 5.70 10⁻⁷ m

    T = 2,898 10⁻³ / 5.70 10⁻⁷

    T = 5084K

c) lam = 470 nm = 4.70 10⁻⁷ m

    T = 2,898 10⁻³ / 4.7 10⁻⁷

    T = 6166K

d) lam = 520 nm = 5.20 10⁻⁷ m

    T = 2,898 10⁻³ / 5.20 10⁻⁷

    T = 5573K

A ) the surface temperature of red star is about 4500 K

B ) the surface temperature of yellow star is about 5100 K

C ) the surface temperature of blue star is about 6200 K

D ) the surface temperature of white star is about 5600 K

[tex]\texttt{ }[/tex]

Let's recall the Wien's Displacement Law as follows:

[tex]\boxed {\lambda_{max}\ T = 2.898 \times 10^{-3} \texttt{ m.K}}[/tex]

where:

λ_max = the wavelength of the maximum radiation energy ( m )

T = surface temperature of the star ( K )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

wavelength of red light = λ_r = 650 nm = 650 × 10⁻⁹ m

wavelength of yellow light =  λ_y = 570 nm = 570 × 10⁻⁹ m

wavelength of blue light =  λ_b = 470 nm = 470 × 10⁻⁹ m

wavelength of white light =  λ_w = 520 nm = 520 × 10⁻⁹ m

Asked:

A ) the surface temperature of red star = T_r = ?

B ) the surface temperature of yellow star = T_y = ?

C ) the surface temperature of blue star = T_b = ?

D ) the surface temperature of white star = T_w = ?

Solution:

[tex]T_r = ( 2.898 \times 10^{-3} ) \div \lambda_r[/tex]

[tex]T_r = ( 2.898 \times 10^{-3} ) \div ( 650 \times 10^{-9} )[/tex]

[tex]\boxed {T_r \approx 4500 \texttt{ K} }[/tex]

[tex]\texttt{ }[/tex]

[tex]T_y = ( 2.898 \times 10^{-3} ) \div \lambda_y[/tex]

[tex]T_y = ( 2.898 \times 10^{-3} ) \div ( 570 \times 10^{-9} )[/tex]

[tex]\boxed {T_y \approx 5100 \texttt{ K} }[/tex]

[tex]\texttt{ }[/tex]

[tex]T_b = ( 2.898 \times 10^{-3} ) \div \lambda_b[/tex]

[tex]T_b = ( 2.898 \times 10^{-3} ) \div ( 470 \times 10^{-9} )[/tex]

[tex]\boxed {T_b \approx 6200 \texttt{ K} }[/tex]

[tex]\texttt{ }[/tex]

[tex]T_w = ( 2.898 \times 10^{-3} ) \div \lambda_w[/tex]

[tex]T_w = ( 2.898 \times 10^{-3} ) \div ( 520 \times 10^{-9} )[/tex]

[tex]\boxed {T_w \approx 5600 \texttt{ K} }[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Mathematics

Chapter: Energy

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:

e. 160.0 mm

Explanation:

Precision is defined as the accuracy in measurement that the measuring instrument measure any given specimen. The instrument having the lowest least count will be more precise than the instrument with higher least count.

Also precision of the reading is given by the number of the significant figures in the given sample of reading.

Option (e) that is 160.0 has a total number of significant figure of 4, which is the largest total significant figures among the other options.

Also the measurement reads 160.0 mm which measure to the lowest possible measurement of the given pen. Thus this reading is more precise.

Hence the correct option is --- Option (e) 160.0 mm

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]

[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]

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:

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:

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:

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:

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: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:

The Q₂ is 0.318 μC

Explanation:

The charge flows is the same on both, then:

[tex]V_{1} =\frac{kQ_{1} }{r_{1} } \\V_{2} =\frac{kQ_{2} }{r_{2} } \\Q_{2} =Q-Q_{1} \\\frac{kQ_{1} }{r_{1} } =\frac{k*(Q-Q_{1}) }{r_{2} } \\Q_{1} =\frac{\frac{Q}{r_{2} } }{(1/r_{1})+(1/r_{2}) }[/tex]

But:

[tex]\frac{1}{r_{1} } +\frac{1}{r_{2} }=\frac{1}{0.38} +\frac{1}{0.28} =6.2[/tex]

Q = = 0.75 μC

Replacing:

[tex]Q_{1} =\frac{\frac{0.75}{0.28} }{6.2} =0.432\mu C[/tex]

The Q₂ is equal:

Q₂ = 0.75 - 0.432 = 0.318 μC

The second conducting sphere, once connected and then disconnected from the first, ends up with the same charge as the first sphere, 0.75 µC or 0.75 x 10^-6 C.

When two conducting spheres are connected by a thin conducting wire, the charge distributes itself evenly across both spheres, assuming both spheres are identical in size. In the case of spheres with different radii, the amount of charge on each sphere once they are separated again will still be proportional to the original amount. Since the two spheres in the question are described to be separated by a large distance relative to their size, the thin wire connecting them would create an equipotential surface, allowing the charge to redistribute.

The total charge Q is conserved in the system, thus the charge on the second sphere Q2 after disconnecting the wire would be the same as the initial total charge Q since it was not charged before. Therefore, the total charge on sphere two, Q2, will also be 0.75 μC or 0.75 x 10^{-6} C.

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Answer:

r = 0.31 m

Explanation:

Given that,

Mass of the sculpture, m = 191 kg

The sculpture's moment of inertia with respect to the pivot is, [tex]I=17.2\ kg-m^2[/tex]

Frequency of oscillation, f = 0.925 Hz

Let r is the distance of the the pivot from the sculpture's center of mass. The frequency of oscillation is given by :

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{mgr}{I}}[/tex]

[tex]r=\dfrac{4\pi^2f^2I}{mg}[/tex]

[tex]r=\dfrac{4\pi^2\times 0.925^2\times 17.2}{191\times 9.8}[/tex]

r = 0.31 m

So, the pivot is 0.31 meters from the sculpture's center of mass. Hence, this is the required solution.

Answer:

Cost to leave this circuit connected for 24 hours is $ 3.12.

Explanation:

We know that,

[tex]\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \pi \mathrm{fc}}[/tex]

f = frequency (60 Hz)

c= capacitor (10 µF = [tex]10^-6[/tex])  

[tex]\mathrm{X}_{\mathrm{c}}=\text { Capacitive reactance }[/tex]

Substitute the given values

[tex]\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \times 3.14 \times 10 \times 10^{-6} \times 60}[/tex]

[tex]\mathrm{X}_{\mathrm{c}}=\frac{1}{3.768 \times 10^{-3}}[/tex]

[tex]\mathrm{x}_{\mathrm{c}}=265.39 \Omega[/tex]

Given that, R = 200 Ω

[tex]X^{2}=R^{2}+X c^{2}[/tex]

[tex]X^{2}=200^{2}+265.39^{2}[/tex]

[tex]X^{2}=40000+70431.85[/tex]

[tex]X^{2}=110431.825[/tex]

[tex]x=\sqrt{110431.825}[/tex]

X = 332.31 Ω

[tex]\text { Current }(I)=\frac{V}{R}[/tex]

[tex]\text { Current }(I)=\frac{120}{332.31}[/tex]

Current (I) = 0.361 amps

“Real power” is only consumed in the resistor,  

[tex]\mathrm{I}^{2} \mathrm{R}=0.361^{2} \times 200[/tex]

[tex]\mathrm{I}^{2} \mathrm{R}=0.1303 \times 200[/tex]

[tex]\mathrm{I}^{2} \mathrm{R}=26.06 \mathrm{Watts} \sim 26 \mathrm{watts}[/tex]

In one hour 26 watt hours are used.

Energy used in 54 hours = 26 × 24 = 624 watt hours

E = 0.624 kilowatt hours

Cost = (5)(0.624) = 3.12  

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:

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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