A.

Calculate the specific weight, density and specific gravity of one litre of a liquid, which weighs 7N.

Select one:

1. 7000 N/m3, 713.5 kg/m3, 0.7135

2. 700 N/m3, 71.35 kg/m3, 0.07135

3. 70 N/m3, 7.135 kg/m3, 0.007135

4. None of the above.

B.

The multiplying factor for converting one stoke into m2/s is

Select one:

1. 102

2. 104

3. 10-2

4. 10-4

Answer:

1.5 m/s²

Explanation:

m = mass of the clarinet case = 3.450 kg

F = upward force on the clarinet through the air = 38.92 N

W = weight of the clarinet case in down direction

weight of the clarinet case is given as

W = mg

W = (3.450) (9.8)

W = 33.81 N

a = acceleration of the case

Force equation for the motion of the case is given as

F - W = ma

38.92 - 33.81 = (3.450) a

a = 1.5 m/s²

Answer:

the lowest frequency is [tex]7.5\times 10^{14} Hz[/tex]

Explanation:

In the question it is given that wavelength(L) in the range of 200μm to 400μm.

let ν be frequency of wave v velocity = 3\times 10^8

velocity v= Lν

therefore ν=[tex]\frac{v}{L}[/tex]

frequency ν be lopwest when L will be heighest

ν(lowest)=[tex]\frac{3\times 10^8}{400\times 10^-9}[/tex]

ν=[tex]7.5\times 10^{14} Hz[/tex]

Answer:

9.4 N

Explanation:

m = mass of the book = 1.83 kg

[tex]\mu _{s}[/tex] = Coefficient of static friction = 0.522

[tex]\mu _{k}[/tex] = Coefficient of kinetic friction = 0.283

[tex]f_{s}[/tex] = Static frictional force

F = force needed to make the book move

force needed to make the book move is same as the magnitude of maximum static frictional force applied by the desk on the book

Static frictional force is given as

[tex]f_{s}[/tex] = [tex]\mu _{s}[/tex] mg

Hence, the force need to move the book is given as

F = [tex]f_{s}[/tex] = [tex]\mu _{s}[/tex] mg

F = [tex]\mu _{s}[/tex] mg

F = (0.522) (1.83 x 9.8)

F = 9.4 N

Explanation:

It is given that,

Mass of object, m = 2 kg

Initial velocity, u = 3 m/s (east)

Final velocity, v = - 7 m/s (west)

The Impulse can be calculated using the change in momentum of an object i.e.

J = m(v-u)

[tex]J=2\ kg(-7\ m/s-3\ m/s)[/tex]

J = -20 kg-m/s

So, the Impulse of this object is 20 kg-m/s but the direction is opposite. Hence, statement (1) is correct i.e. Student #1: "The impulse is equal to the change in momentum, which is (2 kg)(3 m/s + 7 m/s) = 20 kg m/s."

The correct student to agree with is Student #2. The impulse applied to an object is equal to the change in its momentum.

The change in momentum is calculated by multiplying the mass of the object by the change in its velocity.

First, let's calculate the change in velocity.

[tex]\[ \Delta v = v_f - v_i = (-7 \text{ m/s}) - (3 \text{ m/s}) = -10 \text{ m/s} \][/tex]

The negative sign indicates that the velocity has changed in the opposite direction. The magnitude of the change in velocity is 10 m/s.

Now, we multiply the mass of the object by the change in velocity to find the change in momentum:

[tex]\[ \Delta p = m \cdot \Delta v = (2 \text{ kg}) \cdot (-10 \text{ m/s}) = -20 \text{ kg m/s} \][/tex]

The impulse, which is equal to the change in momentum, is therefore:

[tex]\[ \text{Impulse} = \Delta p = -20 \text{ kg m/s} \][/tex]

The negative sign indicates that the impulse is in the opposite direction to the initial velocity. The magnitude of the impulse is 20 kg m/s.

Student #1 incorrectly adds the initial and final velocities without considering their directions. Student #3 incorrectly divides the change in momentum by the time interval, which is not necessary since impulse is the change in momentum, not the rate of change of momentum. Student #4 incorrectly states that we cannot find the impulse without knowing the force; however, we can find the impulse using the change in momentum, as we have done.

Given an AC source with a known voltage amplitude and frequency, and knowing the current amplitude, we can calculate the capacitance of a capacitor connected to this source using the formula for capacitive reactance and Ohm's law. The calculated value for the capacitance of the capacitor in this scenario is approximately 0.17 microfarads (µF).

The subject of this question is the physics concept of capacitive reactance, which has to do with a capacitor's ability to oppose change in voltage. The question is asking for the capacitance of a capacitor in an arrangement where it is connected across an AC source with a known voltage amplitude and frequency. The current amplitude is also provided.

To calculate the capacitance (C) of the capacitor, we should start with the formula for capacitive reactance (Xc): Xc = 1/(2πfC), where Xc is the capacitive reactance, f is the frequency of the AC voltage source, and C is the capacitance. However, in this case, we need to use Ohm's law (I=V/R or here I=V/Xc), manipulate it for Xc and then substitute it into our reactance formula to solve for C. Doing the math will yield the required capacitance.

Let's do the calculations: First, calculate the reactance Xc from the given amplitude of the voltage (V), of the current (I), Xc=V/I. Hence, Xc= 59.0V/5.05A ≈ 11.68 Ω. Now we substitute Xc into the equation for capacitive reactance Xc =1/(2πfC), manipulate for C to get C = 1/(2πfXc), & replace f & Xc with the given values, we get, C = 1/(2π*77 Hz*11.68 Ω) ≈ 0.176 x 10^-6 F. Thus, the capacitance of the capacitor is approximately 0.17 microfarads (µF).

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The capacitance of the capacitor connected to an alternating current source with a voltage amplitude of 59.0 V, a frequency of 77.0 Hz and a current amplitude of 5.05 A, can be calculated using the formula for capacitive reactance and Ohm's law. The calculated capacitance is approximately 0.217 μF.

The formula for calculating the capacitance (C) of a capacitor in an Alternating Current (AC) circuit is given by the equation Xc = 1/(2πfC), where Xc refers to the capacitive reactance, f is the frequency of the AC source in hertz, and C is the capacitance in farads. In your question, it's stated that the current amplitude is 5.05 A and the voltage amplitude is 59.0 V. The capacitive reactance, Xc, can then be calculated using Ohm's law (Xc = Voltage / Current), which in this case would give us a value of around 11.68 Ohm. Substituting this into the given formula, and equally rearranging for C, gives us a capacitance value of approximately 0.217 x 10⁻⁶ F or 0.217 μF. Hence, the capacitance of the capacitor is approximately 0.217 μF.

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Answer:

The work done is 360 J.

Explanation:

Given that,

Mass = 50 kg

Distance =3 m

We need to calculate the work done

The work done is equal to the product of force and displacement.

Using formula of work done

[tex]W = F\cdot d[/tex]

[tex]W = Fd\cos\theta[/tex]

Where, F = force

D = distance

θ = Angle between force and displacement

Put the value into the formula

[tex]W=120\times3\cos0^{\circ}[/tex]

[tex]W=360\ J[/tex]

Hence, The work done is 360 J.

Answer:

The contraction in the rod is 71 mm.

Explanation:

Given that,

original length L'= 2.99 m

Speed [tex]v= 6.49\times10^{7}\ m/s[/tex]

We need to calculate the length

Using expression for length contraction

[tex]L'=\gamma L[/tex]

[tex]L=\dfrac{L'}{\gamma}[/tex]

Where,

[tex]\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}[/tex]

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

Where, v = speed of observer

c = speed of the light

Put the value into the formula

[tex]L=\sqrt{1-\dfrac{(6.49\times10^{7})^2}{(3\times10^{8})^2}}\times2.99[/tex]

[tex]L=2.919\ m[/tex]

The expression for the contraction in the rod

[tex]d =L'-L[/tex]

[tex]d=2.99-2.919 [/tex]

[tex]d=0.071[/tex]

[tex]d= 71\ mm[/tex]

Hence, The contraction in the rod is 71 mm.

Answer:

The speed when it strikes the ground below is V= 157.64 m/s < -56.92º .

Explanation:

V= 150m/s

α= 55º

hi= 120m

Vy= V*sinα

Vy= 122.87 m/s

Vx= V * cos α

Vx= 86.03 m/s

h= hi + Vy * t - g*t²/2

clearing t we get the total flying time of the projectile:

t(total fly)= 26.01 sec

0= Vy - g*t

clearing t we get the maximum height time:

t(max height)= 12.53 sec

to get the fall time:

t(fall)= t(total fly) - t(max height)

t(fall)= 13.48 sec

Vy'= g* t(fall)

Vy'= 132.1 m/s

V'= √(Vx² +Vy'²)

V'= 157.64 m/s

α'= tg⁻¹ (Vy'/Vx)

α'= -56.92º

By using the conservation of energy principle, we find the speed of the projectile when it hits the ground to be roughly 165 m/s (rounded to 3 significant figures).

Given that a projectile is fired upward at an angle of 55° from the top of a 120 m cliff at a speed of 150 m/s, it is asked what its speed will be when it strikes the ground. To answer such a question, we employ the principle of conservation of energy, which states that the total mechanical energy (kinetic energy + potential energy) of an isolated system remains constant if non-conservative forces like air resistance are negligible.

Now, the energy of the projectile at the top of the cliff is equal to its kinetic energy (as it's launched) and its potential energy (due to its height). When the projectile hits the ground, all its potential energy will be converted into kinetic energy as the body has descended from the height, which the body will possess as its speed.

Let's calculate:

Initial Energy = Final Energy

1/2 * mass * (initial speed)² + mass * g * height = 1/2 * mass * (final speed)²

Solving for final speed, -(initial speed)² - 2gh = - (final speed)²

finalSpeed = sqrt((initial speed)² + 2*g*h)

Given that the initial speed = 150m/s, g (acceleration due to gravity) = 9.81m/s², and the height = 120m, the final speed would amount to 164.833 m/s. Therefore, when rounding to 3 significant figures, the speed when it hits the ground will be 165 m/s.

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Answer:

10.2 m

Explanation:

A = 70 mm^2 = 70 x 10^-6 m^2

F = 7 N

Pressure sustained by ear drum

P = F / A = 7 / (70 x 10^-6) = 10^5 N/m^2

Let the depth of water is h

Pressure due to the depth of water should be same as the pressure sustained by the ear drum

P = h x d x g

where, d be the density of water.

10^5 = h x 1000 x 9.8

h = 10.2 m

The discomfort and pressure on the eardrums when diving underwater is due to increased pressure caused by depth. The maximum depth a person can dive without rupturing their eardrum can be calculated by dividing the maximum pressure the eardrum can sustain by the pressure increase per meter. In this case, the maximum depth is about 10 meters.

The discomfort and pressure on your eardrums when diving underwater is due to the increased pressure that comes with depth. The pressure exerted on your eardrums underwater is a result of the weight of the water above you and the air above the water's surface. To calculate the maximum depth you can dive without rupturing your eardrum, you need to determine the pressure that the eardrum can sustain and then divide it by the pressure difference caused by the depth.

The maximum force the eardrum can sustain is 7 N, and the area of the eardrum is 7x10-5 m2. The maximum pressure the eardrum can sustain is therefore 7 N / (7x10-5 m2) = 1x105 Pa.

The pressure underwater increases by 1 atmosphere for every 33 feet (10 meters) of saltwater depth. Using this information, you can calculate the maximum depth:

Maximum Depth = Maximum Pressure / Pressure Increase per Meter = 1x105 Pa / (1x105 Pa/m) = 10 meters

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Answer:

velocity of red ball is either 0 or 1 m/s and then the velocity of green ball is 10 m/s or 8 m/s.

Explanation:

mass of red ball, m1 = 0.05 kg u 1 = 10 m/s

mass of green ball, m2 = 0.1 kg, u2 = 0

Let the velocity of red ball and green ball after the collision is v1 and v2, respectively.

By use of conservation of momentum

m1 u1 + m2 u2 = m1 v1 + m2 v2

0.05 x 10 + 0 = 0.05 x v1 + 0.1 x v2

0.5 = 0.05 (v1 + 2 v2)

v1 + 2v2 = 10 ...... (1)

Now use conservation of kinetic energy

1/2 m1 x u1^2 + 1/2 x m2 x u2^2 = 1/2 m1 v1^2 + 1/2 m2 x v2^2

0.05 x 10 x 10 + 0 = 0.05 x v1^2 + 01. x v2^2

5 = 0.05(v1^2 + 2v2^2)

10 = v1^2 + 2 v2^2 .....(2)

Put teh value of v1 from equation (1) in equation (2)

10 = 10 - 2 v2 + 2 v2^2

0 = v2^2 - v2

v2 =0 , 1 m/s

So, v1 = 10 - 2 x 0 = 10  m/s

or v1 = 10 - 2 x 1 = 8 m/s

Thus, the velocity of red ball is either 0 or 1 m/s and then the velocity of green ball is 10 m/s or 8 m/s.

Explanation:

The motion of satellite is an example of uniform circular motion. In this type of motion, the velocity of object varies at each and every point but its speed is constant. The satellite in a circular orbit travel at a constant speed.

The gravitational force of earth is balanced by the centripetal force such that,

[tex]F_g=F_c[/tex]

A satellite travel at a constant speed because there is no force acting in the direction of motion of satellite i.e. there is no resistive force that opposite the motion of the satellite.

Answer:

24 m/s

Explanation:

Let's say that under very icy conditions, there is no friction.

Draw a free body diagram.  There are 2 forces acting on the van.  Gravity straight down, and normal force perpendicular to the surface.

Sum of the forces in the radial (+x) direction:

∑F = ma

N sin θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ − mg = 0

Solve for N in the second equation:

N cos θ = mg

N = mg / cos θ

Substitute into the first equation:

(mg / cos θ) sin θ = m v² / r

mg tan θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

Given g = 9.8 m/s², r = 130 m, and θ = 24.8°:

v = √(9.8 m/s² × 130 m × tan 24.8°)

v = 24.3 m/s

Rounded to two significant figures, the maximum velocity is 24 m/s (approximately 54 mph).

Answer:

24 m/s

Explanation:

The maximum speed that a minivan can have and still follow the curve safely under very icy conditions is 24 m/s.

g = 9.8 m/s², r = 130 m, and θ = 24.8°:

v = √(9.8 m/s² × 130 m × tan 24.8°)

v = 24.3 m/s

Answer:

No answer is correct.

Explanation:

given data:

Q1 = 400 gpm

P1 = 28 psi

Q2 = ?

P2 = 30 psi

Change in pressure in a pipe is given as

[tex]\Delta P = \frac{32\mu vl}{D^{2}}[/tex]

where v is velocity and it is given as [tex]v = \frac{Q}{A}[/tex]

[tex]\Delta P = \frac{32\mu Ql}{AD^{2}}[/tex]

Therefore, change in pressure is directly proportional to flow

thus we have

[tex]\frac{P_{1}}{Q_{1}}=\frac{P_{2}}{Q_{2}}[/tex]

[tex]Q_{2}=\frac{P_{2}}{P_{1}}*Q_{1}[/tex]

[tex]Q_{2}=\frac{30}{28}*400 = 428.57 gpm[/tex]

[tex]Q_{2} =428.57 gpm[/tex]

no answer is correct

Answer:

428.5 gmp

Explanation:

Given that,

A piping system is operating at quantity = 400 gpm

Pressure = 28 psi

Increased pressure = 30 psi

We need to calculate the resultant flow rate

We know that,

The flow rate is directly proportional to pressure

[tex]Q\propto P[/tex]

Therefore,

[tex]\dfrac{Q_1}{Q_2}=\dfrac{P_1}{P_2}[/tex]

where,

[tex]Q_1\rightarrow 400\text{ gpm}[/tex]

[tex]Q_2\rightarrow x\text{ gpm}[/tex]

[tex]P_1\rightarrow 28\text{ psi}[/tex]

[tex]P_2\rightarrow 30\text{ psi}[/tex]

By substituting into formula

[tex]\dfrac{400}{x}=\dfrac{28}{30}[/tex]

[tex]x=\dfrac{12000}{28}\approx 428.5\text{ gpm}[/tex]

Hence, The resultant flow rate will be 428.5 gmp

Answer:

(a): The frequency of the waves is f= 0.16 Hz

Explanation:

T/4= 1.5 s

T= 6 sec

f= 1/T

f= 0.16 Hz (a)

(a) The frequency of the waves is ¹/₆ Hz ≈ 0.167 Hz

(b) The speed of the waves is 5¹/₃ m/s ≈ 5.33 m/s

[tex]\texttt{ }[/tex]

Let's recall the speed of wave and intensity of wave formula as follows:

[tex]\large {\boxed {v = \lambda f}}[/tex]

f = frequency of wave ( Hz )

v = speed of wave ( m/s )

λ = wavelength ( m )

[tex]\texttt{ }[/tex]

[tex]\large {\boxed {I = 2 \pi^2 A^2 f^2 \rho v}}[/tex]

I = intensity of wave ( W/m² )

A = amplitude of wave ( m )

f = frequeny of wave ( Hz )

ρ = density of medium ( kg/m³ )

v = speed of wave ( m/s )

Let's tackle the problem!

[tex]\texttt{ }[/tex]

Given:

time taken = t = 1.50 seconds

distance covered = d = 8.00 m

Asked:

(a) frequency of the waves = ?

(b) speed of the waves = ?

Solution:

[tex]t = \frac{1}{4}T[/tex]

[tex]1.50 = \frac{1}{4}T[/tex]

[tex]T = 4 \times 1.50[/tex]

[tex]T = 6 \texttt{ seconds}[/tex]

[tex]\texttt{ }[/tex]

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

[tex]f = \frac{1}{6} \texttt{ Hz}[/tex]

[tex]\texttt{ }[/tex]

[tex]d = \frac{1}{4}\lambda[/tex]

[tex]8.00 = \frac{1}{4}\lambda[/tex]

[tex]\lambda = 8.00 \times 4[/tex]

[tex]\lambda = 32 \texttt{ m}[/tex]

[tex]\texttt{ }[/tex]

[tex]v = \lambda f[/tex]

[tex]v = 32 \times \frac{1}{6}[/tex]

[tex]v = 5\frac{1}{3} \texttt{ m/s}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: College

Subject: Physics

Chapter: Sound Waves

Answer:

(a) 1.35 x 10^6 Joule

(b) 321.45 Kcal

Explanation:

Energy consumed in 1 hour = 270 kJ

So, Energy consumed in 5 hour = 270 x 5 = 1350 kJ

(a) Energy in joules = 1350 x 1000 J = 1.35 x 10^6 J

(b) 4.2 Joule = 1 calorie

So, 1.35 x 10^6 Joule = 1.35 x 10^6 / 4.2 = 0.32 x 10^6 Calorie

                                                                =  0.32 x 1000 Kcal = 321.45 Kcal

The electrical energy consumed by the lightbulb is 1..35 x 10⁶J or 322.65 kcal.

The electrical energy consumed by the lightbulb is the product of power and time of energy consumption.

E = Pt

1 h ------------------- 270 kJ

5 h -------------------- ?

= 5 x 270 kJ

= 1,350 kJ

= 1.35 x 10⁶ J.

1 kJ ---------------- 0.239 kcal

1,350 kJ ------------ ?

= 1,350 x 0.239

= 322.65 kcal.

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Answer:

The no of revolutions is 2.032 revolution.

Explanation:

Given that,

Moment of inertia = 0.85 Kgm²

Radius = 170 mm

Force = 32 N

Time =  2s

We need to calculate the angular acceleration

Using formula of torque

[tex]\tau=I\times\alpha[/tex]

[tex]\alpha=\dfrac{\tau}{I}[/tex]

[tex]\alpha=\dfrac{F\times r}{I}[/tex]

Where, F = force

r = radius

I = moment of inertia

Put the value into the formula

[tex]\alpha=\dfrac{32\times170\times10^{-3}}{0.85}[/tex]

[tex]\alpha=6.4\ m/s^2[/tex]

We need to calculate the rotational speed

Using equation of angular motion

[tex]\omega_{f}=\omega_{i}+\alpha t[/tex]

[tex]\omega_{f}=6.4\times2[/tex]

[tex]\omega=12.8\ rad/s[/tex]

We need to calculate the angular position

Using equation of angular motion

[tex]\theta=\omega_{i}+\dfrac{1}{2}\alpha t^2[/tex]

[tex]\theta=0+\dfrac{1}{2}\times6.4\times4[/tex]

[tex]\theta=12.8\ radian[/tex]

We need to calculate no of revolutions

[tex]n = \dfrac{\theta}{2\pi}[/tex]

[tex]n=\dfrac{12.8}{2\times3.15}[/tex]

[tex]n=2.032\ revolution[/tex]

Hence, The no of revolutions is 2.032 revolution.

A 0.293-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.285 m and x2 = 0.395 m. The period of oscillation is 0.641 s. Find the frequency, f, the equilibrium position, xeq, the amplitude, A, the maximum speed, vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot.

That being said,

f=1/T=1/0.641=1.56 s

xeq=(x1+x2)/2=(-0.285+0.395)/2=0.055 m

A=(x2-x1)/2=(0.395+0.285)/2=0.340 m

x(t)=Acos(wt+ϕ)

v(t)=-wAsen(wt+ϕ)

w=2πf=2*3.14*1.56=9.8 rad/s

vmax=-wA=9.8*0.340=-3.3 m/s

a(t)=-w^2Acos(wt+ϕ)

amax=-w^2A=96.04*0.340=-32.6 m/s^2

w=(k/m)^1/2 => k=mw^2=0.293*96.04=28.1 N/m

Etot=1/2kA^2

Etot=0.5*28.1*0.116=1.63 J

The characteristics of the simple harmonic movement allow to find the results for the questions about the movement of the mass with the spring are:

     a) The frequency is: f = 1.72 Hz

     b) The equilibrium position is: x₀ = 0.088 m

     c) The amplitude is: A = 0.686 m

     d) The maximum speed is: v = 7.42 m / s

     e) the maximum acceleration is: a = 80.16 m / s²

     f) The mechanical energy is: E = 3.44 J

Given parameters

To find.

    a) Frequency.

    b) the equilibrium position.

    c) The amplitude.

    d) Maximum speed.

    e) Maximum acceleration.

    f) The total mechanical energy.

Simple harmonic motion is a periodic motion where the restoring force is proportional to the elongation. This movement is fully described by the expression

       x = A cos (wt + Ф)

       w² = [tex]\sqrt{ \frac{k}{m} }[/tex]  

Where x is the displacement, A the amplitude of the movement, w the angular velocity, t the time, Ф a phase constant that meets the initial conditions, k the spring constant and m the mass attached to the spring.

a) Frequency and period are related.

          [tex]f= \frac{1}{T}[/tex]  

Let's calculate.

        [tex]f = \frac{1}{0.581 }[/tex]  

         f = 1.72 Hz

b) The equilibrium position occurs at the midpoint of the movement

       [tex]x_o = \frac{x_f + x_o}{2}[/tex]  

       [tex]x_o = \frac{0.387 + (-0.299)}{2}[/tex]  

       x₀ = 0.088 m

c) The amplitude is the maximum elongation of the spring

        [tex]A= x_f - x_o[/tex]  

        A = 0.387 - (-0.299)

        A = 0.686 m

d) The speed is defined by the variation of the position with respect to time.

          [tex]v = \frac{dx}{dt}[/tex]

Let's make the derivatives

          v = - A w cos (wt + Ф)

The angular velocity is related to the period.

        [tex]w= \frac{\pi }{T} \\w= \frac{2 \pi }{0.581}[/tex]

        w = 10.81 rad / s

The speed is maximum when the cosine  is maximum ±1

        v = A w

        v = 0.686 10.81

        v = 7.42 m / s

e) Acceleration is defined as the change in velocity over time.

           [tex]a = \frac{dv}{dt}[/tex]  

           a = -A w² sin (wt + Ф)

The acceleration is maximum when the sine function is maximum ±1

          a = A w²

          a = 0.686 10.81²

          a = 80.16 m / s²

f) The mechanical energy is calculate for the point of maximum elongation.

          E = ½ k A²  

          k = w² m

Let's replace.

          E = ½ w² m A²

Let's calculate.

          E = ½ 10.81² 0.125 0.686²

          E = 3.44 J

In conclusion, using the characteristics of simple harmonic motion we can find the results for the questions about the motion of the mass with the spring are:

     a) The frequency is: f = 1.72 Hz

     b) The equilibrium position is: x₀ = 0.088 m

     c) The amplitude is: A = 0.686 m

     d) The maximum speed is: v = 7.42 m / s

     e) the maximum acceleration is: a = 80.16 m / s²

     f) The mechanical energy is: E = 3.44 J

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Answer:

2.13 h

Explanation:

Simple harmonic motion is:

x = A sin(2π/T t + φ) + B

where A is the amplitude, T is the period, φ is the phase shift, and B is the midline.

This can also be written in terms of cosine:

x = A cos(2π/T t + φ) + B

Here, A = d/2, T = 12.8, φ = 0, and B = d/2.  I'll use cosine so that the highest level is at t=0.

x = d/2 cos(2π/12.8 t) + d/2

When x = d - 0.250 d = 0.750 d:

0.750 d = d/2 cos(2π/12.8 t) + d/2

0.250 d = d/2 cos(2π/12.8 t)

0.500 = cos(2π/12.8 t)

π/3 = 2π/12.8 t

12.8/6 = t

t = 2.13

It takes 2.13 hours to fall 0.250 d from the highest level.

The time it takes for the water to fall a distance 0.250d from its highest level in the tide cycle can be calculated by determining a quarter of the period, which is 3.2 hours.

The time it takes for the water to fall a distance 0.250d from its highest level can be calculated using the concept of simple harmonic motion. Since the period of the tide is 12.8 hours, to find the time for the water to fall 0.250d, we need to determine the fraction of the period corresponding to this distance.

Given that the complete cycle from highest to lowest level is 12.8 hours, the time to fall 0.250d would be one-quarter of that period, which equals 3.2 hours.

Answer:

The magnitude of the electric force on an electron in a uniform electric is [tex]2.4\times10^{-16}\ N[/tex] to the west.

Explanation:

Given that,

Electric field strength = 1500 N/C

We need to calculate the electric force

Using formula of electric field

[tex]F = Eq[/tex]

E = electric field strength

q = charge of electron

Electron has negative charge.

Put the value into the formula

[tex]F=1500\times(-1.6\times10^{-19})[/tex]

[tex]F=-2.4\times10^{-16}\ N[/tex]

Negative sign shows the opposite direction of the field

Hence, The magnitude of the electric force on an electron in a uniform electric is [tex]2.4\times10^{-16}\ N[/tex] to the west.

The magnitude of the electric force on an electron in a uniform electric field of strength 1500 N/C that points due east is 2.4x10⁻¹⁶ C.

We know that electric force is given by the formula,

[tex]F = E \times q[/tex]

It is given that the electric field, E = 1500 N/C,

We also know that an electron is negatively charged and has a charge of 1.60217662 × 10⁻¹⁹ C.

[tex]F = E \times q\\\\F = 1500 \times 1.6 \times 10^{-19}\\\\F = 2.4 \times 10^{-16}\rm\ N[/tex]

Hence, the magnitude of the electric force on an electron in a uniform electric field of strength 1500 N/C that points due east is 2.4x10⁻¹⁶ C.

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Answer:

The work done on the test charge is 4.5 joules

Explanation:

It is given that,

Electric potential difference, [tex]V=1000\ V[/tex]

Charged particle, [tex]q=4.5\times 10^{-3}\ C[/tex]

Work done on the test charge is given by the product of test charge and potential difference i.e.

W = q × V

[tex]W=4.5\times 10^{-3}\ C\times 1000\ V[/tex]

W = 4.5 Joules

So, the work done on the test charge is 4.5 joules. Hence, this is the required solution.

Explanation:

It is given that,

Mass of concrete pilling, m = 50 kg

Diameter of wire, d = 1 mm

Radius of wire, r = 0.0005 m

Length of wire, L = 11.2

Young modulus of steel, [tex]Y=20\times 10^{10}\ N/m^2[/tex]

The young modulus of a wire is given by :

[tex]Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}[/tex]

[tex]Y=\dfrac{F.L}{A\Delta L}[/tex]

[tex]\Delta L=\dfrac{F.L}{A.Y}[/tex]

[tex]\Delta L=\dfrac{50\ kg\times 9.8\ m/s^2\times 11.2\ m}{\pi (0.0005\ m)^2\times 20\times 10^{10}\ N/m^2}[/tex]

[tex]\Delta L=0.034\ m[/tex]

So, the wire will stretch 0.034 meters. Hence, this is the required solution.

Answer:

Part a)

F = 0.051 N

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

Explanation:

Part a)

Electrostatic force between two charged spherical balls is given as

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we will have

[tex]q_1 = q_2 = 50 nC[/tex]

here the distance between the center of two balls is given as

[tex]r = 2.1 cm = 0.021 m[/tex]

now we will have

[tex]F = \frac{(9\times 10^9)(50 \times 10^{-9})(50 \times 10^{-9})}{0.021^2}[/tex]

[tex]F = 0.051 N[/tex]

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

Answer:

The second child must exert a force of magnitude 23.3N to keep the door from moving.

Explanation:

We have to find the moment that the first child exerts and then match it to that exercised by the second child.

F1= 17.5N

d1= 0.6m

F2= ?

d2= 0.45m

M= F * d

M1= 17.5N * 0.6m

M1= 10.5 N.m

M1=M2

M2= F2 * 0.45m

10.5 N.m= F2 * 0.45m

10.5 N.m/0.45m = F2

F2=23.3 N

The force that the second child must exert to keep the door from moving is 23.33 N.

A balanced force occurs when an object subjected to different forces are at equilibrium.

F1r1 = F2r2

(17.5 x 0.6) = F2(0.45)

F2 = 23.33 N

Thus, the force that the second child must exert to keep the door from moving is 23.33 N.

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Answer:

Work is done by the gas  = 5.92 x 10⁵ J = 592 kJ

Explanation:

Work done at fixed pressure, W = PΔV

Pressure, P = 3.7 x 10⁵ Pa

Change in volume, ΔV = 1.6 m³

Substituting the values of pressure and change in volume we will get

Work done at fixed pressure, W = PΔV =  3.7 x 10⁵ x 1.6 = 5.92 x 10⁵ J

Work is done by the gas  = 5.92 x 10⁵ J = 592 kJ

Answer:

The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.

Explanation:

Given that,

Mass = 2 kg

Radius = 0.5 m

Angular speed = 3 rad/s

Force = 10 N

(I). We need to calculate the rotational kinetic energy

Using formula of kinetic energy

[tex]K.E =\dfrac{1}{2}\timesI\omega^2[/tex]

[tex]K.E=\dfrac{1}{2}\times mr^2\times\omega^2[/tex]

[tex]K.E=\dfrac{1}{2}\times2\times(0.5)^2\times(3)^2[/tex]

[tex]K.E=2.25\ J[/tex]

(II). We need to calculate the instantaneous change rate of the kinetic energy

Using formula of kinetic energy

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

On differentiating

[tex]\dfrac{K.E}{dt}=\dfrac{1}{2}m\times2v\times\dfrac{dv}{dt}[/tex]

[tex]\dfrac{K.T}{dt}=mva[/tex]....(I)

Using newton's second law

[tex]F = ma[/tex]

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

[tex]a=\dfrac{10}{2}[/tex]

[tex]a=5 m/s^2[/tex]

Put the value of a in equation (I)

[tex]\dfrac{K.E}{dt}=mva[/tex]

[tex]\dfrac{K.E}{dt}=mr\omega a[/tex]

[tex]\dfrac{K.E}{dt}=2\times0.5\times3\times5[/tex]

[tex]\dfrac{K.E}{dt}=15\ J/s[/tex]

Hence, The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.

The rotational kinetic energy of the hoop is 2.25 J, and the instantaneous change rate of the kinetic energy is 15 W.

(a) To determine the rotational kinetic energy (RKE) of the hoop, we use the formula for RKE, which is given by:

[tex]\[ RKE = \frac{1}{2} I \omega^2 \][/tex]

where I is the moment of inertia of the hoop and [tex]\( \omega \)[/tex] is the angular speed. The moment of inertia I for a hoop (or a thin-walled cylinder) rotating about its center is:

[tex]\[ I = m r^2 \][/tex]

where m is the mass of the hoop and r is the radius. Given that  m = 2kg and ( r = 0.5 ) m, we can calculate I:

[tex]\[ I = 2 \times (0.5)^2 = 0.5 \text{ kg} \cdot \text{m}^2 \][/tex]

Now, we know [tex]\( \omega = 3 \)[/tex] rad/s. We can substitute the values into the RKE formula:

[tex]\[ RKE = \frac{1}{2} \times 0.5 \times (3)^2 = \frac{1}{2} \times 0.5 \times 9 = 2.25 \text{ J} \][/tex]

So, the rotational kinetic energy of the hoop is 2.25 J.

(b) The instantaneous change rate of the kinetic energy is equal to the power input to the system. Power P due to a torque [tex]\( \tau \)[/tex] is given by:

[tex]\[ P = \tau \omega \][/tex]

The torque [tex]\( \tau \)[/tex] can be calculated from the force F applied tangentially at the rim of the hoop:

[tex]\[ \tau = F \times r \][/tex]

Given that ( F = 10 ) N and ( r = 0.5 ) m, we find:

[tex]\[ \tau = 10 \times 0.5 = 5 \text{ N} \cdot \text{m} \][/tex]

Now, we can calculate the power:

[tex]\[ P = \tau \omega = 5 \times 3 = 15 \text{ W} \][/tex]

Therefore, the instantaneous change rate of the kinetic energy is 15 W.

Answer:

a) 17 N

b) 21 N

Explanation:

At the 3 o'clock position, the sum of the forces towards the center is:

∑F = ma

T = m v² / r

19 = m v² / r

At the 12 o'clock position, the sum of the forces towards the center is:

∑F = ma

T + mg = m v² / r

T + (0.18)(9.8) = 19

T = 17.2 N

At the 6 o'clock position, the sum of the forces towards the center is:

∑F = ma

T − mg = m v² / r

T − (0.18)(9.8) = 19

T = 20.8 N

Rounding to two significant figures, the tensions are 17 N and 21 N.

To find the tension in the stick at the twelve o’clock and six o’clock positions, consider the forces acting on the ball at each position and calculate the tension using the equation T = mg + mv^2 / r for twelve o'clock and T = mg - mv^2 / r for six o'clock.

To find the tension in the stick when the ball is at the twelve o’clock and six o’clock positions, we need to consider the forces acting on the ball at each position. At the twelve o'clock position, the tension in the stick is equal to the weight of the ball plus the centripetal force required to keep it moving in a circle. The tension can be calculated using the equation T = mg + mv^2 / r, where T is the tension, m is the mass of the ball, g is the acceleration due to gravity, v is the velocity of the ball, and r is the radius of the circle.

At the six o'clock position, the tension in the stick is equal to the weight of the ball minus the centripetal force. Using the same equation, the tension can be calculated by subtracting mv^2 / r from mg.

Substituting the given values into the equation for each position will give you the tension in the stick.

(a) [tex]2.98\cdot 10^{10} J[/tex]

The change in energy of the transferred charge is given by:

[tex]\Delta U = q \Delta V[/tex]

where

q is the charge transferred

[tex]\Delta V[/tex] is the potential difference between the ground and the clouds

Here we have

[tex]q=31 C[/tex]

[tex]\Delta V = 0.96\cdot 10^9 V[/tex]

So the change in energy is

[tex]\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J[/tex]

(b) 7921 m/s

If the energy released is used to accelerate the car from rest, than its final kinetic energy would be

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

where

m = 950 kg is the mass of the car

v is the final speed of the car

Here the energy given to the car is

[tex]K=2.98\cdot 10^{10} J[/tex]

Therefore by re-arranging the equation, we find the final speed of the car:

[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2.98\cdot 10^{10})}{950}}=7921 m/s[/tex]

Answer:

The largest equivalent resistance yu can build using these three resistors is a Serie Resistance with the value of R= 16.74 Ω

Explanation:

Adding Resistances in serie is the way to build de largest equivalent value possible.

Rt= R1+R2+R3

Rt= 6.32 + 8.13 + 2.29

Rt= 16.74Ω

Answer:[tex]4.802 m/s^2[/tex]

Explanation:

height of slope(h) =90 foot

[tex]\theta =40[/tex]

weight of snowboarder=170 pounds\approx 77.1107 kg

[tex]\mu =0.2[/tex]

as the snowboarder is sliding down the slope therefore

Now net acceleration of snowboarder is

[tex]a_{net}=gsin\theta -\mu \cdot gcos\theta[/tex]

[tex]a_{net }=9.81\times sin\left ( 40\right )-0.2\times 9.81\times cos\left ( 40\right )[/tex]

[tex]a_{net }=4.802 m/s^2[/tex]

Explanation:

It is given that,

Speed of the plane, v = 110 m/s

Radius of circle, r = 2850 m

(1) Let t is the time taken by a plane. Speed of the plane is given by :

[tex]v=\dfrac{d}{t}=\dfrac{2\pi r}{t}[/tex]

[tex]t=\dfrac{2\pi r}{v}[/tex]

[tex]t=\dfrac{2\pi \times 2850}{110}[/tex]

t = 162.79 s

(2) Mass of a driver, m = 50 kg

Height, h = 6 m

As the driver reaches ground, its potential energy point to zero. So, the change in potential energy is given by :

[tex]\Delta PE=PE_f-PE_i[/tex]

[tex]\Delta PE=0-mgh[/tex]

[tex]\Delta PE=-50\times 9.8\times 6[/tex]

[tex]\Delta PE=-2940\ J[/tex]

So, the change in potential energy is 2940 joules.

Let v is the speed. As the driver reaches water surface, its potential energy is converted to kinetic energy as :

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

[tex]v=\sqrt{\dfrac{2\times PE}{m}}[/tex]

[tex]v=\sqrt{\dfrac{2\times 2940}{50}}[/tex]

v = 10.84 m/s

So, the speed of the reactors the water is 10.84 m/s. Hence, this is the required solution.

It takes approximately 52.3 seconds for a plane, traveling at a constant speed of 110 m/s, to fly once around a circle with a radius of 2850 m. The change in potential energy for a driver of mass 50 kg jumping off a 6 m high cliff is -2940 J. The driver's velocity as they reach the water is approximately 10.8 m/s.

To find the time it takes for a plane to fly once around a circle, we need to calculate the circumference of the circle and divide it by the plane's speed. The formula for the circumference of a circle is C = 2πr, where r is the radius. In this case, the radius is given as 2850 m, so the circumference is 2π(2850) = 5700π m. Dividing the circumference by the plane's speed of 110 m/s gives us the time it takes to complete one lap around the circle: (5700π)/(110) ≈ 52.3 s.

When a driver jumps off a cliff, the force of gravity causes her potential energy to change into kinetic energy as she falls. The formula for potential energy is PE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height. In this case, the mass is given as 50 kg and the height is 6 m. Plugging these values into the formula, we get PE = (50)(9.8)(6) = 2940 J. The change in potential energy is simply the negative of the initial potential energy, since it decreases when the driver jumps off the cliff. Therefore, the change in potential energy is -2940 J.

As the driver reaches the water, all of her potential energy has been converted into kinetic energy. The formula for kinetic energy is KE = 0.5mv², where m is the mass and v is the velocity. Since the potential energy at the end of the fall is 0, the entire initial potential energy is now kinetic energy. Plugging in the mass of 50 kg, we can solve for the velocity using the formula 2940 = 0.5(50)v². Simplifying this equation, we get v² = 2940/25 = 117.6. Taking the square root of both sides, we find that v ≈ 10.8 m/s.

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