In a downtown office building, you notice each of the four sections of a rotating door has a mass of 75 kg. What is the width, in meters, of each section of the door if a force of 56 N applied to the outer edge of a section produces an angular acceleration of 0.420 rad/s2?

Answer:

Explanation:

initial velocity, u = 28 m/s

Angle of projection, θ = 30°

The acceleration in horizontal direction is zero, so the horizontal component of velocity is constant.

Horizontal component of velocity, u cos θ = 28 x Cos 30 = 24.25 m/s

At t = 2 sec, the horizontal component of velocity = 24.25 m/s

At t = 3 sec, the horizontal component of velocity = 24.25 m/s

In this exercise we have to use the knowledge about oblique launch to calculate the components of velocity for each case, so we have that:

For all times we will find a velocity equal to 24.25 m/s

organizing the information given in the statement we have that:

Knowing that the component can be written as:

[tex]u cos \theta = 28 * Cos 30 = 24.25 m/s[/tex]

So as the formula does not depend on time we have that for any value it will have a constant velocity.

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To solve this problem it is necessary to apply the concepts related to Dopler's Law. Dopler describes the change in frequency of a wave in relation to that of an observer who is in motion relative to the Source of the Wave.

It can be described as

[tex]f = \frac{c\pm v_r}{c\pm v_s}f_0[/tex]

c = Propagation speed of waves in the medium

[tex]v_r[/tex]= Speed of the receiver relative to the medium

[tex]v_s[/tex]= Speed of the source relative to the medium

[tex]f_0 =[/tex]Frequency emited by the source

The sign depends on whether the receiver or the source approach or move away from each other.

Our values are given by,

[tex]v_s = 32.2m/s \rightarrow[/tex] Velocity of car

[tex]v_r = 14.8 m/s \rightarrow[/tex] velocity of motor

[tex]c = 343m/s \rightarrow[/tex] Velocity of sound

[tex]f_0 = 523Hz \rightarrow[/tex]Frequency emited by the source

Replacing we have that

[tex]f = \frac{c + v_r}{c - v_s}f_0[/tex]

[tex]f = \frac{343 + 14.8}{343 - 32}(523)[/tex]

[tex]f = 601.7Hz[/tex]

Therefore the frequency that hear the motorcyclist is 601.7Hz

Final answer:

To determine the frequency heard by the motorcyclist, we apply the Doppler shift formula using the given speeds of the motorcycle and police car and the emitted frequency of the police car's siren. The calculated frequency is the observed frequency by the motorcyclist due to the relative motion of the two vehicles. The final answer is about 602.1 Hz

Explanation:

The subject of the question is the Doppler Effect, which is related to the change in frequency of sound waves due to the relative motion of the source and the observer. To calculate the frequency the motorcyclist hears, we use the following Doppler shift formula for sound:

f' = f(v + vo) / (v - vs)

Where:

f' is the observed frequency by the motorcyclist,

f is the emitted frequency by the police car (523 Hz),

v is the speed of sound in air, which can be assumed to be approximately 343 m/s at room temperature,

vo is the speed of the observer (motorcyclist) towards the source (14.8 m/s),

vs is the speed of the source (police car) towards the observer (32.2 m/s).

Plugging in the given values, the equation becomes:

f' = 523 Hz (343 m/s + 14.8 m/s) / (343 m/s - 32.2 m/s) = 602.089 Hz

After performing the calculations, the frequency heard by the motorcyclist can be determined. This is an application of the Doppler effect as studied in high school physics.

Answer:

The deformation in the pole due to force is 0.70 mm.

Explanation:

Given that,

Height = 20.0 m

Diameter = 4.00 cm

Force = 1100 N

We need to calculate the  area

Using formula of area

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

[tex]A=\pi\times(2.00\times10^{-2})^2[/tex]

[tex]A=0.00125\ m^2[/tex]

[tex]A=1.25\times10^{-3}\ m^2[/tex]

We need to calculate the deformation

Using formula of deformation

[tex]\Delta x=\dfrac{1}{s}(\dfrac{F}{A}\times L)[/tex]

Where, s = shear modulus

F = force

l = length

A = area

Put the value into the formula

[tex]\Delta x=\dfrac{1}{2.5\times10^{10}}\times(\dfrac{1100}{1.25\times10^{-3}}\times 20.0)[/tex]

[tex]\Delta x=0.000704\ m[/tex]

[tex]\Delta x=7.04\times10^{-4}\ m[/tex]

[tex]\DElta x=0.70\ mm[/tex]

Hence, The deformation in the pole due to force is 0.70 mm.

Answer: Question 1: Efficiency is 0.6944

Question 2: speed of similar pump is 2067rpm

Explanation:

Question 1:

Flow rate of pump 1 (Q1) = 300gpm

Flow rate of pump 2 (Q2) = 400gpm

Head of pump (H)= 55ft

Speed of pump1 (v1)= 1500rpm

Speed of pump2(v2) = ?

Diameter of impeller in pump 1= 15.5in = 0.3937m

Diameter of impeller in pump 2= 15in = 0.381

B.H.P= 6.0

Assuming cold water, S.G = 1.0

eff= (H x Q x S.G)/ 3960 x B.H.P

= (55x 300x 1)/3960x 6

= 0.6944

Question 2:

Q = A x V. (1)

A1 x v1 = A2 x V2. (2)

Since A1 = A2 = A ( since they are geometrically similar

A = Q1/V1 = Q2/V2. (3)

V1(m/s) = r x 2π x N(rpm)/60

= (0.3937x 2 x π x 1500)/2x 60

= 30.925m/s

Using equation (3)

V2 = (400 x 30.925)/300

= 41.2335m/s

To rpm:

N(rpm) = (60 x V(m/s))/2 x π x r

= (60 x 41.2335)/ 2× π × 0.1905

= 2067rpm.

Answer:

Explanation:

Tension T in the rope will create torque in solid cylinder ( axle ). If α be angular acceleration

T R = 1/2 M R²α ( M is mass and R is radius of cylinder )

= 1/2 M R² x a / R ( a is linear acceleration )

T = Ma / 2

For downward motion of the bucket

mg - T = m a ( m is mass and a is linear acceleration of bucket downwards )

mg - Ma / 2  = ma

a = mg / ( M /2 + m )

Substituting the values

a = 14.7 x 9.8 / ( 5.8+ 14.7 )

= 7 m / s²

A )

T = Ma / 2

= 5.8 x 7

= 40.6 N

B ) v² = u² + 2 a h

= 2 x 7 x 10.3

v = 12 m /s

C )

v = u + a t

12 = 0 + 7 t

t = 1.7 s

Answer:

217.43298 m/s

Explanation:

[tex]m_1[/tex] = Mass of bullet = 19 g

[tex]m_2[/tex] = Mass of bob = 1.3 kg

L = Length of pendulum = 2.3 m

[tex]\theta[/tex] = Angle of deflection = 60°

u = Velocity of bullet

Combined velocity of bullet and bob is given by

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2aL(1-cos\theta)+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (1-cos60)+0^2}\\\Rightarrow v=3.13209\ m/s[/tex]

As the momentum is conserved

[tex]m_1u=(m_1+m_2)v\\\Rightarrow u=\frac{(m_1+m_2)v}{m_1}\\\Rightarrow v=\frac{(0.019+1.3)\times 3.13209}{0.019}\\\Rightarrow v=217.43298\ m/s[/tex]

The speed of the bullet is 217.43298 m/s

The problem involves conservation of momentum and energy principles. Initially, bullet's momentum equals the final momentum of the system. The bullet's speed can be found by solving these equations, using the provided values.

This problem can be solved using principles from both conservation of momentum and conservation of energy. To find the speed of the bullet, we need to consider two scenarios: the before and after the bullet is fired into the bob. Initial momentum is the mass of the bullet multiplied by its velocity and final momentum is the combined mass of the bullet and bob at their highest point. Assuming there's no external force acting, we can have:

m_bullet * v_bullet = (m_bullet + m_bob) * v_final.

The final velocity here is the vertical component of the velocity when the pendulum reach its highest point. This can be calculated by:

v_final = sqrt(2*gravity*height).

The height can be calculated using trigonometry:

height = length - length * cos(60).

Filling all the given values into the equations will give the speed of the bullet.

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Answer:

a) the velocity is v=1.385 m/s

b) the ball has its maximum speed at 4.68 cm away from its compressed position

c)  the maximum speed is 1.78 m/s

Explanation:

if we do an energy balance over the ball, the potencial energy given by the compressed spring is converted into kinetic energy and loss of energy due to friction, therefore

we can formulate this considering that the work of the friction force is equal to to the energy loss of the ball

W fr = - ΔE = - ΔU - ΔK = Ui - Uf + Kf - Ki

therefore

Ui + Ki = Uf + Kf + W fr  

where U represents potencial energy of the compressed spring , K is the kinetic energy W fr is the work done by the friction force. i represents inicial state, and f final state.

since

U= 1/2 k x² , K= 1/2 m v²  , W fr = F*L

X= compression length , L= horizontal distance covered

therefore

Ui + Ki = Uf + Kf + W fr

1/2 k xi² + 1/2 m vi² = 1/2 k x² + 1/2 m vf² + F*L

a) choosing our inicial state as the compressed state , the initial kinetic energy is Ki=0 and in the final state the ball is no longer pushed by the spring thus Uf=0

1/2 k X² + 0 = 0 + 1/2 m v² + F*L

1/2 m v² = 1/2 k X² - F*L

v = √[(k/m)x² -(2F/m)*L] = √[(8.07N/m/5.35*10^-3 Kg)*(-0.0508m)² -(2*0.033N/5.35*10^-3 Kg)*(0.16 m)] = 1.385 m/s

b) in any point x , and since L= d-(X-x) , d = distance where is no pushed by the spring.

1/2 k X² + 0 = 1/2 k x² + 1/2 m v² + F*[d-(X+x)]

1/2 m v² =1/2 k X²-1/2 k x² - F*[d-(X-x)] = (1/2 k X²+ F*X) - 1/2k x² - F*x + F*d

taking the derivative

dKf/dx = -kx - F = 0 → x = -F/k = -0.033N/8.07 N/m = -4.089*10^-3 m = -0.4cm

at x m = -0.4 cm the velocity is maximum

therefore is 5.08 cm-0.4 cm=4.68 cm away from the compressed position

c) the maximum speed is

1/2 m v max² = (1/2 k X²+ F*X) - 1/2k x m² - F*(x m) + 0

v =√[ (k/m) (X²-xm²) + (2F/m)(X-xm) ] = √[(8.07N/m/5.35*10^-3 Kg)*[(-0.0508m)² - (-0.004m)²] + (2*0.033N/5.35*10^-3 Kg)*(-0.0508m-(-0.004m)] = 1.78 m/s

Answer:

[tex]8.2\times 10^{15}\ m[/tex]

Explanation:

[tex]\lambda[/tex] = Wavelength = 600 nm

d = Diameter of mirror = 42 m

D = Distance of object

x = Diameter of Jupiter = [tex]1.43\times 10^8\ m[/tex]

Angular resoulution is given by

[tex]\Delta\theta=1.22\frac{\lambda}{d}\\\Rightarrow \Delta\theta=1.22\frac{600\times 10^{-9}}{42}\\\Rightarrow \Delta\theta=1.74286\times 10^{-8}\ rad[/tex]

We also have the relation

[tex]\Delta\theta\approx=\frac{x}{D}\\\Rightarrow D\approx\frac{x}{\Delta\theta}\\\Rightarrow D\approx\frac{1.43\times 10^8}{1.74286\times 10^{-8}}\\\Rightarrow D\approx 8.2049\times 10^{15}\ m[/tex]

The most distant Jupiter-sized planet the telescope could resolve is [tex]8.2\times 10^{15}\ m[/tex]

Answer:

4.6939 rad/s

Explanation:

You have to use the conservation of angular momentum for both objects as an object is spinning and a collision happens. To properly solve, you have to know that the putty and the vinyl have the same angular speed after the putty lands on it and that the putty acts as a point mass so the formula of it's rotational inertia is mr^2.

Answer:

e. half of Bonnie's.

Explanation:

Jill and Bonnie move in a circular path with the same angular speed of the merry-go-round.

The tangential velocity of the body is calculated as follows:

v = ω*R

where:

v is the tangential velocity or linear velocity  (m /s)

ω is the angular speed (rad/s)

R is radius where the body is located from the center of the circular path

Data

1 rev = 2π rad

ω = 1 rev/2s = 2π rad/2s = π rad/s

R : radio of the merry-go-round

Bonnie's linear velocity (vB)

vB = ω*R  = π*R (m/s)

Jill's linear velocity (vJ)

vJ = ω*(R /2) = (1/2 )(π*R) (m/s)

Jill's linear velocity is half of Bonnie's because linear velocity depends on the distance from the center of the merry-go-round, and Jill is situated at half the radius from the center compared to Bonnie.

The question pertains to understanding the relationship between linear and angular velocities of points located at different radii on a rotating platform, such as a merry-go-round. Given that the merry-go-round makes one complete revolution every 2 seconds, the angular velocity is the same for all points on the merry-go-round. However, linear velocity depends on the radius. Bonnie, who sits on the outer rim, would have the highest linear velocity because the further you are from the center, the greater the distance you cover per rotation. Jill, sitting midway, would cover half the distance in the same time period as Bonnie, resulting in Jill having a linear velocity that is half of Bonnie's.

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Answer:

[tex]v=346.05\ m.s^{-1}[/tex]

Explanation:

Given:

initial temperature of the lead bullet, [tex]T_i=43^{\circ}C[/tex]

latent heat of fusion of lead, [tex]L_f=2.32\times 10^4\ J.kg^{-1}[/tex]

melting point of lead, [tex]T_m=327.3^{\circ}C[/tex]

We have:

specific heat capacity of lead, [tex]c=129\ J.kg^{-1}.K^{-1}[/tex]

According to question the whole kinetic energy gets converted into heat which establishes the relation:

[tex]\rm KE=(heat\ of\ rising\ the\ temperature\ from\ 43\ to\ 327.3\ degree\ C)+(heat\ of\ melting)[/tex]

[tex]\frac{1}{2} m.v^2=m.c.\Delta T+m.L_f[/tex]

[tex]\frac{1}{2} m.v^2=m(c.\Delta T+L_f)[/tex]

[tex]\frac{v^2}{2} =129\times(327.3-43)+23200[/tex]

[tex]v=346.05\ m.s^{-1}[/tex]

The minimum speed of a lead bullet that will completely melt upon stopping suddenly can be found by calculating the kinetic energy that is converted into heat via friction and comparing it with the heat required to elevate the bullet's temperature to its melting point. The formula for calculating the minimum speed of the bullet is the square root of twice the product of the latent heat of fusion and the change in temperature (√(2Lf∆T)).

The question asked pertains to the concept of kinetic energy transformations and heat generation due to friction in Physics. The kinetic energy of the bullet is transformed into thermal energy due to the sudden change in speed, causing the lead bullet to heat up and potentially melt if it is moving fast enough.

First, we calculate the change in temperature which is the difference between the melting point of lead and the initial temperature, ∆T = 327.3°C - 43.0°C = 284.3°C. Then, we utilize the formula for heat transfer Q = mLf, where m is the mass of the bullet, and Lf is the latent heat of fusion of lead. We rearrange this to find the mass of the bullet, m = Q/Lf.

Next, we use the principle of conservation of energy. All of the kinetic energy of the bullet (1/2mv²) is converted into heat (Q), leading to the equation 1/2m v² = Q. Solving for v (the bullet's speed) we get that v = √(2Q/m). Combining equations yields v = √(2Lf∆T).

This v is the minimum speed at which the bullet will completely melt upon stopping suddenly, assuming that all of its kinetic energy is converted into heat via friction and is equal to the energy required to raise the bullet's temperature to its melting point.

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Answer:

Towards left

Explanation:

Applying Fleming Right hand rule in which the middle finger denotes downward magnetic field. Thumb pointing towards you is  the direction of motion of electron beam.The direction of force on the electron indicated by the index finger is towards left. Hence the electron beam is deflected towards left.

The electron beam will be deflected to the left-hand side. This is determined by the left-hand rule, due to the negative charge on the electron.

This particular scenario of an electron beam and a magnet involves the concept of the magnetic force on a charged particle. According to the right-hand rule, when a charged particle moves in a magnetic field, the magnetic force on the particle is perpendicular to both the velocity of the particle and the magnetic field. In this case, since electrons have a negative charge, we'll be using your left hand for the rule.

Begin by extending your left hand flat, with your thumb pointing in the direction of the electron beam (towards you), and your fingers pointing in the direction of the magnetic field (downward). You'll notice that your palm would then be facing toward your left side. This indicates that the electron beam will be deflected to the left-hand side.

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Answer:

Explanation:

Let equal mass of Ne and Kr be m gm

no of moles of Ne and Kr will be m / 20  and m / 84 ( atomic weight of Ne and Kr is 20 and 84 )

Let the pressure and volume of both the gases be P and V respectively .

The temperature of Ne be T₁ and temperature of Kr be T₂.

For Ne

PV = (m / 20) x R T₁

For Kr

PV = (m / 84) x R T₂

T₁ / T₂ = 84 / 20

We know that

average KE of an atom of mono atomic gas = 3 / 2 x k T

k is boltzmann constant and T is temperature .

KEKr/KENe = T₂ / T₁

= 20 / 84

The ratio of the average kinetic energy of an atom of krypton to that of neon is 1.

The average kinetic energy of an atom is directly proportional to the temperature. In this case, the temperatures of the krypton (Kr) and neon (Ne) gases are different. Since the temperatures are different, the ratio of their average kinetic energies will be equal to the ratio of their temperatures. Therefore, the ratio KEKr/KENe is equal to Tkrypton/Tneon.

Using the given information, we can find the ratio of the temperatures: Tkrypton/Tneon = 300K/300K = 1. Therefore, the ratio of the average kinetic energy of an atom of krypton to that of neon is 1.

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Answer:

29274.93096 m/s

[tex]2.73966\times 10^{33}\ J[/tex]

[tex]-5.39323\times 10^{33}\ J[/tex]

[tex]2.56249\times 10^{33}\ J[/tex]

[tex]-5.21594\times 10^{33}[/tex]

Explanation:

[tex]r_p[/tex] = Distance at perihelion = [tex]1.471\times 10^{11}\ m[/tex]

[tex]r_a[/tex] = Distance at aphelion = [tex]1.521\times 10^{11}\ m[/tex]

[tex]v_p[/tex] = Velocity at perihelion = [tex]3.027\times 10^{4}\ m/s[/tex]

[tex]v_a[/tex] = Velocity at aphelion

m = Mass of the Earth =  5.98 × 10²⁴ kg

M = Mass of Sun = [tex]1.9889\times 10^{30}\ kg[/tex]

Here, the angular momentum is conserved

[tex]L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s[/tex]

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

[tex]K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J[/tex]

Kinetic energy at perihelion is [tex]2.73966\times 10^{33}\ J[/tex]

Potential energy is given by

[tex]P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times  10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}[/tex]

Potential energy at perihelion is [tex]-5.39323\times 10^{33}\ J[/tex]

[tex]K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J[/tex]

Kinetic energy at aphelion is [tex]2.56249\times 10^{33}\ J[/tex]

Potential energy is given by

[tex]P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}[/tex]

Potential energy at aphelion is [tex]-5.21594\times 10^{33}\ J[/tex]

Answer:

:) *_* :3 ^-^ {.}{.}

Explanation:

Answer:

[tex]a_{cm} = 9.64m/s^2[/tex]

[tex]Ff=6.42N[/tex]

Explanation:

The sum of torque on the sphere is:

[tex]m*g*sin\theta*R=I*\alpha[/tex]

[tex]m*g*sin\theta*R=2/3*m*R^2*\alpha[/tex]

[tex]m*g*sin\theta*R=2/3*m*R*a_{cm}[/tex]

Solving for a:

[tex]a_{cm}=9.64m/s^2[/tex]

Now, the sum of forces will be:

[tex]m*g*sin\theta-Ff=m*a_{cm}[/tex]

Solving for Ff:

[tex]Ff=m*g*sin\theta-m*a_{cm}[/tex]

Ff=-6.42N    The negative sing tells us that it actually points downwards.

Answer:

a) a = 3.783 m/s^2

b)  F_f = 5.045 N

Explanation:

Given:

- Mass of shell m = 2.0 kg

- Angle of slope Q = 40 degrees

- Moment of inertia of shell I = 2/3 *m*R^2

Find:

a) Find the magnitude of the acceleration a_cm of the center of mass of the spherical shell.

b) Find the magnitude of the frictional force acting on the spherical shell.

Solution:

- Draw a Free body diagram for the shell. We see that the gravitational force F_g acting parallel to the plane of the inclined surface makes the sphere to roll down. The frictional force  F_f between the inclined surface and the sphere gives the necessary torque for the sphere to roll down with out slipping. Under this conditions a sphere will roll down without slipping with some acceleration and the acceleration can be calculated from the equation of motion of the sphere:

                                      m*g*sin(Q) - F_f = m*a

- Where, The frictional force produces the torque and due to this torque the sphere gets an angular acceleration.

- Then we can write the equation for the rotational motion as:

                                      F_f*R = I*α

                                      F_f = I*α / R

- Using moment mass inertia of the shell we have:

                                      F_f = (2/3)*m*R^2*α/R

- Where the angular acceleration α is related to linear acceleration a with:

                                      α = a / R

- combing the two equations we will have friction force F_f as:

                                      F_f = (2/3)*m*R^2*a/R^2

                                      F_f = (2/3)*m*a

- Now evaluate the equation of motion:

                                      m*g*sin(Q) - (2/3)*m*a= m*a

- Simplify:

                                        (5/3)*a = g*sin(Q)

                                        a = (3/5)*g*sin(Q)

- Plug the values in:       a = (3/5)*9.81*sin(40)

                                        a = 3.783 m/s^2

- Now compute the Frictional force F_f from the expression derived above:

                                        F_f = (2/3)*m*a

- Plug values in:              F_f = (2/3)*2*3.783

                                        F_f = 5.045 N

To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

[tex]n_1sin\theta_1 = n_2sin\theta_2[/tex]

Where,

[tex]\theta =[/tex] Angle between the normal surface at the point of contact

n = Indices of refraction for corresponding media

The total internal reflection would then be given by

[tex]n_1 sin\theta_1 = n_2sin\theta_2[/tex]

[tex](1.54) sin\theta_1 = (1.33)sin(90)[/tex]

[tex]sin\theta_1 = \frac{1.33}{1.54}[/tex]

[tex]\theta = sin^{-1}(\frac{1.33}{1.54})[/tex]

[tex]\theta = 59.72\°[/tex]

Therefore the [tex]\alpha_{max}[/tex] would be equal to

[tex]\alpha = 90\°-\theta[/tex]

[tex]\alpha = 90-59.72[/tex]

[tex]\alpha = 30.27\°[/tex]

Therefore the largest value of the angle α is 30.27°

The largest value the angle α can have without any light being refracted out of the prism at face AC is 30.27°.

From the information given, the total internal reflection would be:

1.54sinb = (1.33) sin90°

sin b = (1.33 sin90° / 1.54)

b = 59.72°

Therefore, the value of the angle will be:

= 90° - 59.72°

= 30.27°

In conclusion, the correct option is 30.27°.

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The tangential acceleration of the pedal is 0.259 m/s^2. The length of chain passing over the top of the sprocket depends on the number of teeth on the sprocket, which is not provided in the question.

A) The tangential acceleration of the pedal can be determined using the equation:

at = r × α

where at is the tangential acceleration, r is the radius of the pedal (half the length of the crank arm), and α is the angular acceleration. To find the angular acceleration, we can use the equation:

α = Δω / Δt

where Δω is the change in angular velocity and Δt is the change in time. Plugging in the given values:

α = (95 rpm - 62 rpm) * 2π / 60 s = (33 * 2π) / 60 s ≈ 3.459 rad/s2

Substituting this value and the radius into the first equation, we get:

at = 0.075 m * 3.459 rad/s2 = 0.259 m/s2

Therefore, the tangential acceleration of the pedal is 0.259 m/s2.

B) The length of chain passing over the top of the sprocket during the given time interval can be calculated using the formula:

d = 2π * r * (ωf - ωi) * t / N

where d is the length of chain, r is the radius of the sprocket, ωi and ωf are the initial and final angular velocities respectively, t is the time interval, and N is the number of teeth on the sprocket. Plugging in the given values:

d = 2π * (21 cm / 2) * [(95 rpm - 62 rpm) * 2π / 60 s] * 12 s / N

The length of chain depends on the number of teeth on the sprocket. Since the number of teeth is not provided in the question, we cannot calculate the exact length of chain. However, we have all the necessary equation components to calculate it once the number of teeth is known.

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Answer:(a)12.17 rad/s

Explanation:

Given

Moment of Inertia [tex]I=0.23 kg.m^2[/tex]

Torque [tex]T=2.8 N-m[/tex]

(a)Torque is given by Product of Moment of inertia and angular acceleration

[tex]T=I\cdot \alpha [/tex]

[tex]2.8=0.23\cdot \alpha [/tex]

[tex]\alpha =\frac{2.8}{0.23}=12.17 rad/s[/tex]

(b)RPM of blades [tex]N=205 rpm [/tex]

angular velocity [tex]\omega =\frac{2\pi N}{60}[/tex]

[tex]\omega =\frac{2\pi 205}{60}=21.47 rad/s[/tex]

Rotational Kinetic Energy [tex]=\frac{I\omega ^2}{2}[/tex]

[tex]=\frac{0.23\times (21.47)^2}{2}=53.01 J[/tex]

The angular acceleration of the blades is approximately 12.1739 rad/s^2. When the blades rotate at 205 rpm, the rotational kinetic energy is approximately 0.0948 J.

(a) To find the angular acceleration of the blades, we can use the formula:

torque = moment of inertia × angular acceleration

Plugging in the given values:

torque = 2.8 N · m

moment of inertia = 0.23 kg · m2

Rearranging the formula, we get:

angular acceleration = torque / moment of inertia

Substituting the values:

angular acceleration = 2.8 N · m / 0.23 kg · m2

Solving for angular acceleration gives us:

angular acceleration ≈ 12.1739 rad/s2

(b) To find the rotational kinetic energy, we can use the formula:

rotational kinetic energy = ½ × moment of inertia × (angular velocity)2

Plugging in the given values:

moment of inertia = 0.23 kg · m2

angular velocity = 205 rpm = 205 revolutions / 60 seconds = 3.4167 rev/s

Rearranging the formula, we get:

rotational kinetic energy = ½ × moment of inertia × (angular velocity)2

Substituting the values:

rotational kinetic energy = ½ × 0.23 kg · m2 × (3.4167 rev/s)2

Solving for rotational kinetic energy gives us:

rotational kinetic energy ≈ 0.0948 J

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Answer:

0.0241875 m

Explanation:

[tex]m_1[/tex] = Mass of quarterback = 80 kg

[tex]m_2[/tex] = Mass of football = 0.43 kg

[tex]v_1[/tex] = Velocity of quarterback

[tex]v_2[/tex] = Velocity of football = 15 m/s

Time taken = 0.3 seconds

In this system as the linear momentum is conserved

[tex]m_1v_1+m_2v_2=0\\\Rightarrow v_1=-\frac{m_2v_2}{m_1}\\\Rightarrow v_1=-\frac{0.43\times 15}{80}\\\Rightarrow v_1=0.080625\ m/s[/tex]

Assuming this velocity is constant

[tex]Distance=Velocity\times Time\\\Rightarrow Distance=0.080625\times 0.3\\\Rightarrow Distance=0.0241875\ m[/tex]

The distance the quarterback will move in the horizontal direction is 0.0241875 m

Final answer:

The question is a physics problem regarding kinematics and requires calculating the horizontal distance an 80-kg quarterback moves while in the air after throwing a football. However, since the quarterback jumps vertically with zero initial horizontal speed and no horizontal force is acting after the throw, the resulting horizontal distance is zero.

Explanation:

The student is asking a question related to kinematics and the conservation of momentum in physics. Specifically, it concerns an 80-kg quarterback who jumps vertically and then throws a football horizontally. The core concept here is determining how far the quarterback will move horizontally during the time he is in the air, assuming a constant horizontal speed.

To calculate the distance (d) the quarterback moves horizontally, we can apply the equation for constant velocity motion: d = v * t. However, the information given does not directly provide the quarterback's horizontal speed after the throw; thus, we must assume that the horizontal speed of the quarterback is zero as the ball is thrown horizontally and no horizontal force on the quarterback has been mentioned.

Therefore, under the assumption that the quarterback's horizontal speed remains zero, the distance he moves horizontally d would also be zero, since he moves vertically up and down with no horizontal velocity component. If the problem had given a horizontal speed for the quarterback post-throw, we would use that given speed to calculate the distance using the equation.

Answer:[tex]10.82 kg-m^2[/tex]

Explanation:

Given

Mass of solid uniform disk [tex]M=13 kg[/tex]

radius of disk [tex]r=1.25 m[/tex]

mass of lump [tex]m=1.7 kg[/tex]

distance of lump from axis [tex]r_0=0.63[/tex]

Moment of inertia is the distribution of mass from the axis of rotation

Initial moment of inertia of disk [tex]I_1=\frac{Mr^2}{2}[/tex]

[tex]I_1=\frac{13\times 1.25^2}{2}=10.15 kg-m^2[/tex]

Final moment of inertia [tex]I_f[/tex]=Moment of inertia of disk+moment of inertia of lump about axis

[tex]I_f=\frac{Mr^2}{2}+mr_0^2[/tex]

[tex]I_f=10.15+1.7\times 0.63^2[/tex]

[tex]I_f=10.15+0.674[/tex]

[tex]I_f=10.82 kg-m^2[/tex]

The EMF is then found using Faraday's Law, resulting in ε = - μ0 n i0 A ω cos(ωt).

To find the EMF induced in a small loop of area A placed inside a long solenoid with n turns per unit length and a current i(t) = i0 sin(ωt), we use the concept of changing magnetic flux.

First, determine the magnetic field inside the solenoid. For a solenoid, the magnetic field B inside is given by:

B = μ0nI

Since the current I is changing with time, so does the magnetic field:

B(t) = μ0n i0 sin(ωt)

The flux Φ through the small loop of area A is:

Φ = B(t) * A = μ0n i0 sin(ωt) * A

The induced EMF (ε) can be found using Faraday's Law of Induction:

ε = -dΦ/dt

Taking the derivative of Φ with respect to time t:

ε = -d/dt (μ0n i0 sin(ωt) * A)

Using the chain rule, we get:

ε = - μ0n i0 A d/dt (sin(ωt))

The derivative of sin(ωt) is ω cos(ωt):

ε = - μ0n i0 A ω cos(ωt)

Therefore, the EMF induced in the loop is:

ε = - μ0n i0 A ω cos(ωt)

Answer:

René Descartes

Explanation:

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Answer:

Specific Gravity = m/[m(s)-m(os)]

Explanation:

Specific gravity, also called relative density, is the ratio of the density of a substance to the density of a reference substance. By this definition we need to find out the ratio of density of the object of mass m to the density of the surrounding liquid.

m = mass of the object

Weight in air

W (air) = mg, where g is the gravitational acceleration

Weight with submerged with only one mass

m(s)g + Fb = mg + m(b)g, consider this to be equation 1

where Fb is the buoyancy force

Weight with submerged with both masses

m(os)g + Fb’ = mg + m(b)g, consider this to be equation 2

equation 1 – equation 2 would give us

m(s)g – m(os)g = Fb’ – Fb

where Fb = D x V x g, where D is the density of the liquid the object is submerged in, g is the force of gravity and V is the submerged volume of the object

m(s)g – m(os)g = D(l) x V x g

m(s) – m(os) = D(l) x V

we know that Mass = Density x V, which in our case would be, D(b) x V, which also means

V = Mass/D(b), where D(b) is the density of the mass

Substituting V into the above equation we get

m(s) – m(os) = [D(l) x m)/ D(b)]

Rearranging to get the ratio of density of object to the density of liquid

D(b)/D(l) = m/[m(s)-m(os)], where D(b)/D(l) denotes the specific gravity

Answer:

D. 966 μm

Explanation:

b = Wien's displacement constant = [tex]2.89\times 10^{-3}\ mK[/tex]

T = Temperature

[tex]\lambda_m[/tex] = Peak wavelength

Here the Wien's displacement law is used

[tex]\lambda_m=\frac{b}{T}\\\Rightarrow \lambda_m=\frac{2.89\times 10^{-3}}{3}\\\Rightarrow \lambda_m=0.000963\\\Rightarrow \lambda_m=963\times 10^{-6}\ m=963\ \mu m\approx 966\ \mu m[/tex]

The wavelength we should tune our antenna in order to detect this radiation is 966 μm

Answer:

Option C

Explanation:

The bob of the pendulum, when the pendulum is taken into the orbiting space station will be in free fall as there is no gravity in space and also the point to which it is attached.

The movement of the bob of the pendulum about its mean position is because of its weight due to which it oscillates and in the free fall, no force acts on it due to its weight since the body is not under the attraction force due to gravity thus it experiences weightlessness and does not oscillate.

On an orbiting space station with microgravity, a pendulum will not oscillate because the pendulum and its point of attachment are in free fall, absent of the normal gravitational forces.

If a pendulum is taken into an orbiting space station, its behavior will be altered due to the absence of an external gravitational force. This situation is best explained by option c. The pendulum will no longer oscillate because the pendulum and the point to which it is attached are in free fall, also known as microgravity. In an environment of microgravity, the forces that would ordinarily cause the pendulum bob to swing are absent, thus the pendulum will essentially float instead of oscillating back and forth.

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Answer:

P = 1.89 10⁸ N / m²

Explanation:

To solve this problem we can use the definition of bulk modules

       B = - P / (ΔV/V)

The negative sign is entered for the volume module to be positive, P is the pressure and ΔV/V is the volume change fraction

In this case the volume change is 9% this is

      ΔV / V 100 = 9%

      ΔV / V = ​​0.09

      P = B ΔV / V

The bulk modulus value is that of water since it is in a liquid state and then freezes

       B = 0.21 101¹⁰ N / m²

let's calculate

       P = 0.21 10¹⁰ 0.09

      P = 1.89 10⁸ N / m²

Final answer:

When the glass flask and mercury are warmed together, the mercury expands and overflows the flask. To calculate the initial volume of the mercury, use the equation: Volume of mercury = Volume of flask + Volume overflowed.

Explanation:

When the glass flask and mercury are warmed together, both substances expand due to the increase in temperature. As a result, some of the mercury overflows the flask. To calculate the initial volume of the mercury, we can use the equation:

Volume of mercury = Volume of flask + Volume overflowed

So, the initial volume of mercury is 1000 cm³ + 8.25 cm³ = 1008.25 cm³.

Answer:yes

Explanation:

To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.

It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.

The mathematical equation that expresses this concept is

[tex]\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}[/tex]

Where

P = Pressure at each point

r = Radius

[tex]\eta =[/tex] Viscosity

l = Length

Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be

[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}[/tex]

From the problem two terms are given

[tex]R_A = \frac{R_B}{2}[/tex]

[tex]L_A = 2L_B[/tex]

Replacing we have to

[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}[/tex]

[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}[/tex]

[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}[/tex]

Therefore the ratio of the flow rate through capillary tubes A and B is 1/32

Answer:

0.04s

Explanation:

Specific heat of water c = 4.186J/g.C

The heat energy it would take to heat up 0.5 kg of water from 30C to 72 C is

[tex]E = mc\Delta T = 0.5 * 4.186 * (72 - 30) = 87.9 J[/tex]

If the power of the electric kettle is 2.2kW (or 2200W) and suppose the work efficiency is 100%, then the time it takes to transfer that power is

[tex] t = \frac{E}{P} = \frac{87.9}{2200} = 0.04 s[/tex]