A frictionless piston-cylinder device contains 4.5 kg of nitrogen at 110 kPa and 200 K. Nitrogen is now compressed slowly according to the relation PV1.5 = constant until it reaches a final temperature of 360 K. Calculate the work input during the process, in kJ.

Answer:

The height above the ground is 1279.51 km

Solution:

As per the question:

Height above the ground, [tex]h_{p} = 229.0\ km[/tex]

Speed of the satellite, [tex]v_{p} = 8.050\ km/s = 8050\ m/s[/tex]

Gravitational constant, [tex]G = 6.67\times 10^{- 11}\ m.kg[/tex]

Mass of the Earth, m = [tex]5.972\times 10^{24}\ kg[/tex]

Now,

The distance of the earth from the perigee is given by:

[tex]R_{p} = h_{p} + R_{E}[/tex]

where

[tex]R_{E} = 6371\ km[/tex]

[tex]R_{p} = 229.0 + 6371 = 6600\ km[/tex]

Now,

To calculate the distance of the earth from the earth from the apogee:

[tex]R_{A} = \frac{R_{p}}{\frac{2Gm}{v_{p}^{2}R_{p} - 1}}[/tex]

[tex]R_{A} = \frac{6600\times 1000}{\frac{2\times 6.67\times 10^{- 11}\times 5.972\times 10^{24}}{8050^{2}6600\times 1000 } - 1}}[/tex]

[tex]R_{A} = 7650.51\ km[/tex]

Height above the ground, [tex]h_{A} = R_{A} - R_{E} = 7650.51 - 6371 = 1279.51\ km[/tex]

Answer:

The initial momentum of the system is 12 kg m/s

Explanation:

Hi there!

The momentum of the system is calculated as the sum of the momenta of the objects that compose the system, in this case, Jack and the ball.

The momentum of each object is calculated as follows:

p = m · v

Where:

p = momentum

m = mass

v = velocity

The initial momentum of the system is the momentum before Jack catches the ball and is given by the sum of the momentum of the ball (pb) plus the momentum of Jack (pj) (let´s consider the right direction as positive):

Initial momentum = pb + pj

initial momentum = mb · vb + mj · vj

initial momentum = 3.0 kg · 4.0 m/s + 75 kg · 0 m/s

initial momentum = 12 kg m/s

The initial momentum of the system is 12 kg m/s

Final answer:

The initial momentum of the system, consisting of Jack, the skateboard, and the ball, is 0 kg m/s.

Explanation:

Initial momentum of the system:

To calculate the initial momentum of the system consisting of Jack, the skateboard, and the ball, we can use the principle of conservation of momentum. The initial momentum of the system is equal to the sum of the individual momenta of the objects involved.

Initial momentum = Total mass × Initial velocity = (75 kg + 3.0 kg) × 0 m/s = 0 kg m/s

Answer: 5,728.5 years

Explanation:

By measuring the current ratio of daughter to parent, that is (Dt/Pt) one can deduce the age of any sample. The assumption is that there is no daughter atom present at time,t=0 and that the daughter atoms are due to the parent decay(where none has been lost).

Lamda= ln 2÷ half life,t(1/2)

Lamda= ln2= 0.693/5730 years

Lamda=1.21 × 10^-4

Using the formula below;

Age,t= 1/lamda× (ln {1+ Dt/Pt})------------------------------------------------------------------------------------(1)

Slotting our values into equation (1) above.

Age,t= 1/lamda(ln[1+1/1])

Age,t= 1/1.2×10^-4(ln 1+1)

Age,t= 1/1.21×10^-4(ln 2)

Age,t= ln 2/ 1.21×10^-4

Age, t= 5,728.5 years.

Answer:

11,460 years

Explanation:

half-life= 5,730 years

you are looking for the 1 from the ration 1:3

multiply the half-life by two in order to get the 1

5,730 x 2 = 11,640

=11,460 years

The incandescent lightbulb emits approximately [tex]\(1.51 \times 10^{19}\)[/tex] visible-light photons per second.

To find the number of photons emitted per second, you can use the following steps:

1. Calculate the energy of one photon using the formula:

  [tex]\[ E = hf \][/tex]

  where:

  - [tex]\( E \)[/tex] is the energy of one photon,

  - [tex]\( h \)[/tex] is Planck's constant [tex](\( 6.626 \times 10^{-34} \ \text{Joule}\cdot\text{s} \))[/tex],

  - [tex]\( f \)[/tex]is the frequency of the light.

  The frequency [tex](\( f \))[/tex] can be determined using the speed of light [tex](\( c \))[/tex] and the wavelength [tex](\( \lambda \))[/tex]:

  [tex]\[ f = \frac{c}{\lambda} \][/tex]

  where:

  - [tex]\( c \)[/tex] is the speed of light [tex](\( 3.00 \times 10^8 \ \text{m/s} \))[/tex],

  - [tex]\( \lambda \)[/tex] is the wavelength of the light.

2. Find the number of photons emitted per second using the power of the visible light [tex](\( P_{\text{visible}} \))[/tex] and the energy of one photon:

 [tex]\[ \text{Number of photons per second} = \frac{P_{\text{visible}}}{E} \][/tex]

  where:

  - [tex]\( P_{\text{visible}} \)[/tex] is the power of the visible light (5 W),

  - [tex]\( E \)[/tex] is the energy of one photon.

Let's perform the calculations:

[tex]\[ f = \frac{c}{\lambda} \][/tex]

[tex]\[ f = \frac{3.00 \times 10^8 \ \text{m/s}}{600 \times 10^{-9} \ \text{m}} \][/tex]

[tex]\[ f = 5.00 \times 10^{14} \ \text{Hz} \][/tex]

Now, calculate the energy of one photon:

[tex]\[ E = hf \][/tex]

[tex]\[ E = (6.626 \times 10^{-34} \ \text{Joule}\cdot\text{s}) \times (5.00 \times 10^{14} \ \text{Hz}) \][/tex]

[tex]\[ E = 3.313 \times 10^{-19} \ \text{Joule} \][/tex]

Finally, calculate the number of photons emitted per second:

[tex]\[ \text{Number of photons per second} = \frac{5 \ \text{W}}{3.313 \times 10^{-19} \ \text{Joule/photon}} \][/tex]

[tex]\[ \text{Number of photons per second} \approx 1.51 \times 10^{19} \ \text{photons/s} \][/tex]

So, the incandescent lightbulb emits approximately [tex]\(1.51 \times 10^{19}\)[/tex] visible-light photons per second.

Answer:

1. [tex]X_{cm}=-8.57m[/tex]

2. [tex]Y_{cm}=-14.29m[/tex]

3. [tex]V_{cm} = 16.66m/s[/tex]

4. α = 59.05°

5. [tex]V_{cm} = 16.66m/s[/tex]

6. α = 59.05°

Explanation:

The position of the center of mass 1s before the collision is:

[tex]X_{cm}=\frac{X_c*mc+X_t*m_t}{m_c+m_t}[/tex]

where

[tex]X_c=-20m[/tex];  [tex]m_c=1500kg[/tex];  

[tex]X_t=0m[/tex];   [tex]m_t=2000kg[/tex];

Replacing these values:

[tex]X_{cm}=-8.57m[/tex]

[tex]Y_{cm}=\frac{Y_c*mc+Y_t*m_t}{m_c+m_t}[/tex]

where

[tex]Y_c=0m[/tex];  [tex]m_c=1500kg[/tex];  

[tex]Y_t=-25m[/tex];   [tex]m_t=2000kg[/tex];

Replacing these values:

[tex]Y_{cm}=-14.29m[/tex]

The velocity of their center of mass is:

[tex]V_{cm-x}=\frac{V_{c-x}*mc+V_{t-x}*m_t}{m_c+m_t}[/tex]

where

[tex]V_{c-x}=20m/s[/tex];  [tex]m_c=1500kg[/tex];  

[tex]V_{t-x}=0m/s[/tex];   [tex]m_t=2000kg[/tex];

Replacing these values:

[tex]V_{cm-x}=8.57m/s[/tex]

[tex]V_{cm-y}=\frac{V_{c-y}*mc+V_{t-y}*m_t}{m_c+m_t}[/tex]

where

[tex]V_{c-y}=0m[/tex];  [tex]m_c=1500kg[/tex];  

[tex]V_{t-y}=-25m[/tex];   [tex]m_t=2000kg[/tex];

Replacing these values:

[tex]V_{cm-y}=-14.29m[/tex]

So, the magnitude of the velocity is:

[tex]V_{cm}=\sqrt{V_{cm-x}^2+V_{cm-y}^2}[/tex]

[tex]V_{cm}=16.66m/s[/tex]

The angle of the velocity is:

[tex]\alpha =atan(V_{cm-y}/V_{cm-x})[/tex]

[tex]\alpha=59.05\°[/tex]

Since on any collision, the velocity of the center of mass is preserved, then the velocity after the collision is the same as the previously calculated value of 16.66m/s at 59.0° due north of east

Calculations using the physics concept of center of mass determine the positions and velocities of the system of two vehicles before and after collision. The x and y coordinates of the center of mass are found to be -12m and -15m respectively, with a combined vCM direction of 180 degrees relative to the positive x-axis.

To answer the questions presented, it's important to recall the concept of center of mass in physics. The center of mass of a system of particles (like two cars in this case) is the point at which its mass may be considered to be concentrated, when calculating moments and linear momentum. The positions of center of mass are given by:

xCm = (m1*x1 + m2*x2) / (m1 + m2)

yCm = (m1*y1 + m2*y2) / (m1 + m2)

And the center of mass velocity can be found using:

vCm = (m1*v1 + m2*v2) / (m1 + m2)

Assuming east as positive x-direction and north as positive y-direction, substituting the given values into the equations above, we find that:

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Answer:

The gauge pressure is 6670.8 Pascal

Explanation:

Given the height is 0.68 m

The Gauge pressure is determine by the formula,

Gauge pressure , P = hdg

Where h = 0.68 m

d = 1000

g = 9.81

Substituting the values in the formula of Gauge pressure, we get

Gauge Pressure = hdg =[tex]0.68 \times 1000 \times 9.81[/tex]

Gauge Pressure = 6670.8

Therefore, the gauge pressure is 6670.8 Pascal

Answer:

The final velocity of the cars is 6.894 m/s

The kinetic energy lost is 117,441 J

Explanation:

Using the conservation of the linear momentum :

[tex]P_i = P_f[/tex]

Where [tex]P_i[/tex] is the inicial linear momentum and [tex]P_f[/tex] is the final linear momemtum.

The linear momentum is calculated by:

[tex]P = MV[/tex]

where M is the mass and V is the velocity.

First we identify the directions of the velocity of both cars:

The conservation of the linear momentum is made on direction x, so:

[tex]P_i = P_f\\M_{1}V_{1ix} +M_{2}V_{2ix} = V_{sx}(M_1+M_2)[/tex]

where [tex]M_1[/tex] is the mass of the car 1, [tex]V_{1ix}[/tex] is the car 1's  inicial velocity in the axis x,  [tex]M_2[/tex] is the mass of the car 2, [tex]V_{2ix}[/tex] is the car 2's  inicial velocity in the axis x and [tex]V_{sx}[/tex] is the velocity of the car 1 and car 2 after the collition.

The car 1 just move in the axis y so, it dont have horizontal velocity ([tex]V_{1ix}[/tex] = 0)

now the equation is:

[tex]M_2V_{2ix} = V_{sx}(M_1 +M_2)[/tex]

Replacing the values, we get:

(700 kg)(21 m/s) = [tex]V_{sx}[/tex]( 1800 kg + 700 kg)

solving for [tex]V_{sx}[/tex]:

[tex]V_{sx} = \frac{700(21)}{1800+700}[/tex]

[tex]V_{sx}[/tex] = 5.88 m/s

Now we do the conservation of the linear momentum on direction y:

[tex]P_i = P_f\\M_{1}V_{1iy} +M_{2}V_{2iy} = V_{sy}(M_1+M_2)[/tex]

where [tex]M_1[/tex] is the mass of the car 1, [tex]V_{1iy}[/tex] is the car 1's  inicial velocity in the axis y,  [tex]M_2[/tex] is the mass of the car 2, [tex]V_{2iy}[/tex] is the car 2's  inicial velocity in the axis y and [tex]V_{sy}[/tex] is the velocity of the car 1 and car 2 together after the collition.

The car 2 just move in the axis x so, it don't have horizontal velocity ([tex]V_{2iy}[/tex] = 0)

now the equation is:

[tex]M_1V_{1iy} = V_{sy}(M_1 +M_2)[/tex]

Replacing the values, we get:

[tex]1800(5 m/s) = V_{sy}(1800 +700)[/tex]

solving for [tex]V_{sx}[/tex]:

[tex]V_{sy} = \frac{1800(5)}{1800+700}[/tex]

[tex]V_{sy}[/tex] = 3.6 m/s

Now, we have the two components of the velocity and using pythagorean theorem we find the answer as:

[tex]V_f = \sqrt{5.88^2+3.6^2}[/tex]

[tex]V_f =[/tex] 6.894 m/s

Finally we have to find the kinetic energy lost, so the kinetic energy is calculated by:

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

so, ΔK = [tex]K_f -K_i[/tex]

where [tex]K_f[/tex] is the final kinetic energy and [tex]K_i[/tex] is the inicial kinetic energy.

then:

[tex]K_i = \frac{1}{2}(1800)(5)^2+\frac{1}{2}(700)(21)^2= 176,850 J[/tex]

[tex]K_f = \frac{1}{2}(1800+700)(6.894)^2 = 59,409 J[/tex]

so:

ΔK =  59409 - 176850 = -117441 J

Final answer:

The final velocity of the cars is -9.48 m/s due south. The amount of kinetic energy lost in the collision is 49275 J.

Explanation:

To find the final velocity of the cars, we can use the conservation of momentum principle. Since the cars stick together after the collision, their combined mass is 1800 kg + 700 kg = 2500 kg. We can calculate the momentum of each car before the collision and add them together. Car A's momentum is 1800 kg * (-5.00 m/s) = -9000 kg·m/s, and Car B's momentum is 700 kg * (-21.0 m/s) = -14700 kg·m/s. The combined momentum of the cars is -9000 kg·m/s + (-14700 kg·m/s) = -23700 kg·m/s. Since momentum is conserved, the combined momentum of the cars after the collision is also -23700 kg·m/s. Dividing this momentum by the combined mass of the cars gives us the final velocity: -23700 kg·m/s / 2500 kg = -9.48 m/s. The magnitude of the final velocity is 9.48 m/s, and the direction is south since Car A originally had a southward velocity.

To calculate the amount of kinetic energy lost in the collision, we need to find the initial and final kinetic energies. The initial kinetic energy is given by the formula KE = 0.5 * mass * velocity^2. For Car A, the initial kinetic energy is 0.5 * 1800 kg * (5.00 m/s)^2 = 11250 J. For Car B, the initial kinetic energy is 0.5 * 700 kg * (21.0 m/s)^2 = 51450 J. The total initial kinetic energy is 11250 J + 51450 J = 62700 J.

The final kinetic energy is given by the formula KE = 0.5 * mass * velocity^2. For the cars combined, the final kinetic energy is 0.5 * 2500 kg * (9.48 m/s)^2 = 111975 J.

The kinetic energy lost in the collision is the difference between the initial kinetic energy and the final kinetic energy, which is 62700 J - 111975 J = 49275 J.

Final answer:

Io remains volcanically active due to tidal heating caused by Jupiter's gravitational forces.

Explanation:

Io, the innermost of Jupiter's Galilean moons, is known for its high level of volcanism, making it the most volcanically active body in the solar system. Despite its small size, Io remains volcanically active due to the effect of gravity, specifically tidal heating caused by the massive gravity of Jupiter. Jupiter's strong gravitational forces pull Io into an elongated shape, resulting in tidal heating that generates enough internal heat to drive volcanic activity on the moon.

Answer:

height solid / height hollow = 3/4

Explanation:

the height reached is directly proportional to the original KE of the object.

Total KE solid

=[tex]1/2mv^2+1/2\times1/2m r^2(v^2 / r^2 )=3/4mv^2[/tex]

Total KE hollow = [tex]1/2mv^2+1/2mr^2(v^2 / r^2 )=mv^2[/tex]

height solid / height hollow = 3/4

Answer:

It's held together by the nuclear force.

Explanation:

There are more elemental forces than just the electromagnetic one. In this case, it is the nuclear force (called also strong force) the one that holds the nucleus together because it is stronger than the electromagnetic force over such short distances as the one inside the atomic nucleus.

Answer:

The strong nuclear force, which is attractive over short distances like the nucleus, and stronger than electricity, holds the nucleus together.

Explanation:

Answer:[tex]5.67\times 10^{3} m/s[/tex]

Explanation:

Given

mass of satellite [tex]m=155 kg[/tex]

Satellite is orbiting 5995 km above Earth surface

mass of Earth [tex]M=5.97\times 10^{24} kg[/tex]

Radius of Earth [tex]R=6370 km[/tex]

here [tex]r=R+5995=6370+5995=12,365 km[/tex]

Gravitational Force will Provide the centripetal Force

[tex]\frac{GMm}{r^2}=\frac{mv^2}{r}[/tex]

[tex]v=\sqrt{\frac{GM}{r}}[/tex]

[tex]v=\sqrt{\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{12365\times 10^{3}}}[/tex]

[tex]v=\sqrt{32.203\times 10^6}[/tex]

[tex]v=5.67\times 10^3m/s=5674.7 km/s[/tex]            

The work done to move a positive test charge from a point at a radial distance of 5R from the center of a charged sphere to a point at a radial distance of 3R is given by the difference in the electric potentials at these points times the charge of the test charge.

The work done, W, in moving a small positive test charge, q0, in an electric field produced by another charged object is given by the expression W = q0 x (Vf - Vi), where Vf and Vi are> the final and initial electric potentials, respectively. The electric potential, V, at a point located a distance r from the center of a conducting sphere carrying a charge q is V = 1/4πε0 x q/r. So, the work done to move the test charge from r = 5R to r = 3R is W = q0 x {[1/4πε0 x q/(3R)] - [1/4πε0 x q/(5R)]}, which simplifies to W = q0q/4πε0R x (5/15 - 3/25).

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The work required to move a small positive test charge from a radius of 5R to 3R on a conducting sphere of radius R carrying positive charge q is calculated to be W = q*q0/(30πϵ0R) using Coulomb's law and the work-energy theorem.

The work needed to move a charge in an electric field is given by the integral of force times distance. The force on a charge in an electric field is given by Coulomb's law: F = k*q*q0/r^2, where k is Coulomb's constant = 1/4π*ϵ0. The work done in moving a charge from r1 to r2 is given by the integral from r1 to r2 of dr, which results in k*q*q0*(1/r1 - 1/r2).

In this case, with r1=5R and r2=3R, the work is W = k*q*q0*(1/5R - 1/3R), which simplifies to W = 2k*q*q0/(15R), or W = q*q0/(30πϵ0R) when substituting k = 1/4π*ϵ0.

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Answer:

The man will be able to move the crate.

Explanation:

[tex]\mu_s[/tex] = Coefficient of static friction between the crate and the floor = 0.2

[tex]\mu'_s[/tex] = Coefficient of static friction between the man's shoes and the floor = 0.4

[tex]m_c[/tex] = Mass of crate = 150 kg

[tex]m_p[/tex] = Mass of man = 85 kg

g = Acceleration due to gravity = 9.81 m/s²

Horizontal force in order to move the crate is given by

[tex]F_h=\mu_sm_cg\\\Rightarrow F_h=0.2\times 150\times 9.81\\\Rightarrow F_h=294.3\ N[/tex]

Maximum force that the man can apply

[tex]F_m=\mu'_sm_pg\\\Rightarrow F_m=0.4\times 85\times 9.81\\\Rightarrow F_m=333.54\ N[/tex]

Here it can be seen that [tex]F_m>F_h[/tex].

So, the man will be able to move the crate.

Final answer:

To determine whether the 85-kg man can move the 150-kg crate, we need to compare the force of static friction between the crate and the floor with the force exerted by the man. If the force of static friction is greater than the force exerted by the man, the crate will not move. If it is less than or equal to the force exerted by the man, the crate will move.

Explanation:

The force of static friction can be calculated using the equation:

fs = μsN

where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force.

In this case, the normal force can be calculated as:

N = mass × gravity

where the mass is the total mass of the crate and the man, and gravity is the acceleration due to gravity.

If the force of static friction is greater than the force exerted by the man, the crate will not move. If the force of static friction is less than or equal to the force exerted by the man, the crate will move.

In this case, we have:

Mass of crate = 150 kg

Mass of man = 85 kg

Coefficient of static friction between crate and floor (μs) = 0.2

Coefficient of static friction between man's shoes and floor (μ's) = 0.4

Force exerted by man = mass of man × acceleration due to gravity

The normal force can be calculated as the sum of the weight of the crate and the weight of the man:

Normal force (N) = (mass of crate + mass of man) × acceleration due to gravity

The force of static friction between the crate and floor can be calculated as:

fs = μs × N

If this force is greater than the force exerted by the man, the crate will not move. If it is less than or equal to the force exerted by the man, the crate will move.

Answer:

a)[tex]t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec[/tex]

b)[tex]\theta_1=\frac{w_1^2}{2\alpha}rad[/tex]

c)[tex]t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec[/tex]

Explanation:

1) Basic concepts

Angular displacement is defined as the angle changed by an object. The units are rad/s.

Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:

[tex]w=\frac{\Delat \theta}{\Delta t}[/tex]

Angular acceleration is the rate of change of the angular velocity respect to the time

[tex]\alpha=\frac{dw}{dt}[/tex]

2) Part a

We can define some notation

[tex]w_o=0\frac{rad}{s}[/tex],represent the initial angular velocity of the wheel

[tex]w_1=?\frac{rad}{s}[/tex], represent the final angular velocity of the wheel

[tex]\alpha[/tex], represent the angular acceleration of the flywheel

[tex]t_1[/tex] time taken in order to reach the final angular velocity

So we can apply this formula from kinematics:

[tex]w_1=w_o +\alpha t_1[/tex]

And solving for t1 we got:

[tex]t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec[/tex]

3) Part b

We can use other formula from kinematics in order to find the angular displacement, on this case the following:

[tex]\Delta \theta=wt+\frac{1}{2}\alpha t^2[/tex]

Replacing the values for our case we got:

[tex]\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2[/tex]

And we can replace [tex]t_1[/tex]from the result for part a, like this:

[tex]\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2[/tex]

Since [tex]\theta_o=0[/tex] and [tex]w_o=0[/tex] then we have:

[tex]\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}[/tex]

And simplifying:

[tex]\theta_1=\frac{w_1^2}{2\alpha}rad[/tex]

4) Part c

For this case we can assume that the angular acceleration in order to stop applied on the wheel is [tex]\alpha_1 =-5\alpha \frac{rad}{s}[/tex]

We have an initial angular velocity [tex]w_1[/tex], and since at the end stops we have that [tex]w_2 =0[/tex]

Assuming that [tex]t_2[/tex] represent the time in order to stop the wheel, we cna use the following formula

[tex]w_2 =w_1 +\alpha_1 t_2[/tex]

Since [tex]w_2=0[/tex] if we solve for [tex]t_2[/tex] we got

[tex]t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}[/tex]

And from part a) we can see that [tex]w_1=\alpha t_1[/tex], and replacing into the last equation we got:

[tex]t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec[/tex]

Answer:

Explanation:

ω = [tex]\sqrt{\frac{k}{m} }[/tex]

k = 2.5 N/m

m = 10 kg

[tex]\omega = \sqrt{\frac{2.5}{10} }[/tex]

ω = .5 rad /s

x(t) = A cos(ωt + φ₀)

When t = 0 , x(t) = 0

0 = A cos(ωx 0 + φ₀)

cos φ₀ = 0

φ₀ = π /2

x(t) = A cos(ωt +π /2 )

Putting the value of ω

x(t) = A cos(.5 t +π /2 )

Differentiating on both sides

dx(t)/dt = - .5 A sin(.5 t +π /2 )

v(t) = - .5 A sin(.5 t +π /2 )

Given t =0 , v(t) = -5 m/s

-5 = - .5 A sin(.5 x0 +π /2 )

-5 = - .5 A sinπ /2

A = 10 m

x(t) = 10 cos( .5 t +π /2 )

b )

when t = π ( 3.14 s )

x(t) = -  10 m

when t = 2π ( 6.28s )

x(t) = 0

when t = 3π ( 9.42 s )

x(t) =  10 m

and so on

Final answer:

The position as a function of time for the block in the given scenario is x(t) = 0 for any value of t.

Explanation:

To find the form x(t) for the position as a function of time for the block in the given scenario, we first need to determine the constants A, ω, and φ0. We are given that at time t = 0, the amplitude of the motion is 0 and the velocity is -5 m/s.

Using the equation x(t) = A cos(ωt + φ0), we can determine the values of A, ω, and φ0. Since the amplitude is 0, we have A = 0. The velocity is given by v(t) = - Umax sin(ωt + φ0), where Umax = √(k/m)A. Plugging in the given velocity of -5 m/s, we can solve for Umax and find its value to be 5 m/s.

Using the equation ω = √(k/m), we can substitute in the given values to find ω. Therefore, ω = √(2.5 N/m /10 kg) = 0.5 rad/s. Finally, we can substitute the values of A = 0, ω = 0.5 rad/s, and φ0 = 0 into the equation x(t) = A cos(ωt + φ0) to find the form of x(t) as x(t) = 0 cos(0.5t + 0) = 0 for any value of t.

To develop this problem, we simply have to make a relationship between speeds. This is because the echo of the sound is traveling a distance equal to that which the light travels.

We must calculate how much is that relationship of round trips between the two, as well

[tex]N = \frac{c}{s}[/tex]

Where

c = Speed of light

s = Speed of sound

By definition we know that

[tex]c = 3*10^8m/s[/tex]

[tex]s = 342,2m/s[/tex](Normal conditions)

Then,

[tex]N = \frac{3*10^8m/s}{342,2m/s}[/tex]

[tex]N = 876680.3[/tex]

Therefore the flash of the gunshot travel the round-trip distance between the mirror around to 876680.3 times before the echo of the gunshot is heard.

To determine how many times the flash of the gunshot travels the round-trip distance between the mirrors before the echo is heard, we need to consider the time it takes for sound to travel from the gun to the first mirror, and then from the first mirror to the second mirror and back.

To determine how many times the flash of the gunshot travels the round-trip distance between the mirrors before the echo is heard, we need to consider the time it takes for sound to travel from the gun to the first mirror, and then from the first mirror to the second mirror and back.

The total round-trip distance between the mirrors is twice the distance between the mirrors. Let's say this distance is 'd'.

The time it takes for sound to travel from the gun to the first mirror and back is given by 2d/v, where 'v' is the speed of sound.

So, the number of times the flash of the gunshot travels the round-trip distance before the echo is heard is given by the time taken for the sound divided by the time taken by the flash to travel the same distance.

Therefore, the answer to the question is the speed of sound divided by the speed of light (ignoring any delays due to the reflection in the mirrors).

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The object at 12,000 K emits more light in the ultraviolet region of the spectrum compared to the object at 3000 K.

When comparing the two objects, we can see that the object at 12,000 K emits more light in the ultraviolet region of the spectrum compared to the object at 3000 K. This is because as the temperature of an object increases, the frequency of maximum emission also increases. The object at 12,000 K has a higher frequency of maximum emission, which falls in the ultraviolet region of the spectrum.

Answer:

Frequency will be 743.09 Hz

Explanation:

We have given maximum speed [tex]v_m=2.66\times 10^{-3}m/sec[/tex]

Amplitude [tex]A=5.7\times 10^{-7}m[/tex]

We have to find the frequency of the vibration

We know that angular frequency is given by

[tex]\omega =\frac{v_m}{A}=\frac{2.66\times 10^{-3}}{5.7\times 10^{-7}}=4666.666rad/sec[/tex]

[tex]2\pi f=4666.66[/tex]

[tex]f=743.09Hz[/tex]

Answer:

Their kinetic energies have the same magnitude and sign.

Explanation:

Hi there!

Kinetic energy is not a vector, then it has no direction and therefore it does not matter the sense of movement of the car relative to a system of reference. Mathematically it would be also impossible to obtain a negative kinetic energy. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

where m is the mass of the car (always positive) and v is its speed (not velocity, remember that the speed is the magnitude of the velocity vector, that´s why the kinetic energy is not a vector. I agree that the "v" in the formula is confusing).

So, even if we use a negative speed (that would be wrong), the kinetic energy will be positive because the speed is squared.

Then, if the cars have the same mass and speed, they will have the same kinetic energy, magnitude and sign (positive).

Answer:395.6 m/s

Explanation:

Given

mass of bullet [tex]m=5 gm[/tex]

mass of wood block [tex]M=1 kg[/tex]

Length of string [tex]L=2 m[/tex]

Center of mass rises to an height of [tex]0.38 cm[/tex]

initial velocity of bullet [tex]u=450 m/s[/tex]

let [tex]v_1[/tex] and [tex]v_2[/tex] be the velocity of bullet and block after collision

Conserving momentum

[tex]mu=mv_1+Mv_2[/tex] -------------1

Now after the collision block rises to an height of 0.38 cm

Conserving Energy for block

kinetic energy of block at bottom=Gain in Potential Energy

[tex]\frac{Mv_2^2}{2}=Mgh_{cm}[/tex]

[tex]v_2=\sqrt{2gh_{cm}} [/tex]

[tex]v_2=\sqrt{2\times 9.8\times 0.38}[/tex]

[tex]v_2=0.272 m/s[/tex]

substitute the value of [tex]v_2[/tex] in equation 1

[tex]5\times 450=5\times v_1+1000\times 0.272[/tex]

[tex]v_1=395.6 m/s[/tex]

Answer:

C) the magnitude of the acceleration is a minimum.

Explanation:

As we know that ,the general equation of the simple harmonic motion given as

The displacement x given as

x=X sinω t

Then the velocity v will become

v= X ω cosωt

The acceleration a

a= - X ω² sinω t

The speed of the particle will be maximum when cosωt will become 1 unit.

It means that sinωt will become zero.So acceleration and displacement will be minimum.

Therefore when speed is maximum then acceleration will be minimum.

At the mean position the speed of the particle is maximum that is why kinetic energy also will be maximum and the potential energy will be minimum.

Therefore option C is correct.

In simple harmonic motion, the speed is greatest when the displacement is a maximum.

In simple harmonic motion, the speed is greatest at the point in the cycle when the displacement is a maximum. When an object is at its maximum displacement, it has the maximum potential energy, and as it moves towards the equilibrium position, the potential energy is converted into kinetic energy, leading to an increase in speed.

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Answer:

[tex]h=6.377\times10^{-34}kgm^2/s[/tex]

Explanation:

The maximum kinetic energy of the photoelectrons is given by the formula [tex]K_M=hf-\phi[/tex].

We have two situations where for [tex]f_1=547.5\times10^{12}Hz[/tex] we get [tex]K_{M1}=1.26\times10^{-19}J[/tex] and for [tex]f_2=738.8\times10^{12}Hz[/tex] we get [tex]K_{M2}=2.48\times10^{-19}J[/tex], so we have:

[tex]K_{M1}=hf_1-\phi[/tex]

[tex]K_{M2}=hf_2-\phi[/tex]

We can eliminate [tex]\phi[/tex] by substracting the first equation to the second:

[tex]K_{M2}-K_{M1}=hf_2-\phi-(hf_1-\phi)=h(f_2-f_1)[/tex]

Which means:

[tex]h=\frac{K_{M2}-K_{M1}}{f_2-f_1}=\frac{2.48\times10^{-19}J-1.26\times10^{-19}J}{738.8\times10^{12}Hz-547.5\times10^{12}Hz}=6.377\times10^{-34}kgm^2/s[/tex]

Answer:

The phase angle is 0.0180 rad.

(c) is correct option.

Explanation:

Given that,

Voltage = 12 V

Angular velocity = 50 Hz

Capacitance [tex]C= 20\times10^{-2}\ F[/tex]

Inductance [tex]L=20\times10^{-3}\ H[/tex]

Resistance [tex]R=  50\ Omega[/tex]

We need to calculate the impedance

Using formula of impedance

[tex]z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}[/tex]

[tex]z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}[/tex]

[tex]z=50.00[/tex]

We need to calculate the phase angle

Using formula of phase angle

[tex]\theta=\cos^{-1}(\dfrac{R}{z})[/tex]

[tex]\theta=\cos^{-1}(\dfrac{50}{50.00})[/tex]

[tex]\theta=0.0180\ rad[/tex]

Hence, The phase angle is 0.0180 rad.

Answer:

a. the absolute value of the change in the momentum of the small car is 0.078  

b. the velocity of the larger car after the collision is 0.1513 m/s

Explanation:

The linear momentum P is calculated as:

P = MV

Where M is the mass and V the velocity

Therefore, for calculated the change of the linear momentum of the small cart, we get:

[tex]P_{fc}-P_{ic} =[/tex]ΔP

where [tex]P_{ic}[/tex] in the inicial momentum and [tex]P_{fc}[/tex] is the final momentum of the small cart. Replacing the values, we get:

0.10 kg (0.76) -0.10(1.54) = -0.078 kg m/s

The absolute value: 0.078 kg m/s

On the other hand, using the law of the conservation of linear momentum, we get:

[tex]P_i = P_f[/tex]

Where [tex]P_i[/tex] is the linear momentum of the sistem before the collision and [tex]P_f[/tex] is the linear momentum after the collision.

[tex]P_i=P_{ic}\\P_f=P_{fc} + P_{ft}[/tex]

Where [tex]P_{fc}[/tex] is the linear momentum of the small cart after the collision and [tex]P_{ft}[/tex] is the linear momentum of the larger cart after the collision

so:

(0.10 kg)(1.54 m/s) = (0.10 kg)(-0.76 m/s) + (1.52 kg)(V)

Note: we choose the first direction of the small car as positive.

Solving for V:

[tex]\frac{0.10(1.54)+0.10(0.76)}{1.52} = V[/tex]

V = 0.1513 m/s

Final answer:

The magnitude of the change in momentum for the small cart is 0.078 kg·m/s. The speed of the larger cart after the collision is 0 m/s.

Explanation:

(a) What is the magnitude (absolute value) of the change in momentum for the small cart?

To find the change in momentum for the small cart, we can use the formula:

Change in momentum = mass × change in velocity

The mass of the small cart is 0.10 kg and its initial velocity is 1.54 m/s. After the collision, its velocity changes to -0.76 m/s (since it recoils in the opposite direction). Therefore, the change in velocity is: (0.76 m/s - 1.54 m/s) = -0.78 m/s.

Substituting the values into the formula, we get:

Change in momentum = 0.10 kg × (-0.78 m/s) = -0.078 kg·m/s

Since the question asks for the magnitude (absolute value) of the change in momentum, the answer is 0.078 kg·m/s.

(b) What is the speed (absolute value of the velocity) of the larger cart after the collision?

In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since the larger cart is initially at rest, its momentum before the collision is zero. Therefore, its momentum after the collision must also be zero.

Since momentum is equal to mass × velocity, the mass of the larger cart is 1.52 kg and its final velocity after the collision is denoted as Vf. We can set up the equation:

1.52 kg × Vf = 0 kg·m/s

Solving for Vf, we find that the velocity of the larger cart after the collision is also zero. Therefore, the answer is 0 m/s.

The spring's force constant is [tex]\( 15.30 \, \text{N/m} \).[/tex]

To find the spring's force constant, we can use the formula for the period of oscillation of a mass-spring system:

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

where [tex]\( T \)[/tex] is the period of oscillation, m is the mass attached to the spring, and [tex]\( k \)[/tex] is the spring constant.

Given:

[tex]\[ m = 0.200 \, \text{kg} \]\[ \text{Time for one complete oscillation } = 2.60 \, \text{s} \][/tex]

First, let's find the period [tex]\( T \)[/tex] using the given time for one complete oscillation:

[tex]\[ T = 2.60 \, \text{s} \][/tex]

Now, we can rearrange the formula for the period to solve for the spring constant [tex]\( k \):[/tex]

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

Substitute the known values:

[tex]\[ k = \frac{4\pi^2 \cdot 0.200 \, \text{kg}}{(2.60 \, \text{s})^2} \][/tex]

Calculate:

[tex]\[ k = \frac{4\pi^2 \cdot 0.200 \, \text{kg}}{6.76 \, \text{s}^2} \]\[ k = \frac{7.854 \, \text{kg} \cdot \text{m}^{-1}}{6.76 \, \text{s}^2} \]\[ k \approx 15.30 \, \text{N/m} \][/tex]

So, the spring's force constant is [tex]\( 15.30 \, \text{N/m} \).[/tex]

Complete Question:
In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring’s force constant

Answer:

1000 Nm

2000 Nm

1.00007 seconds

Explanation:

I = Moment of inertia = 5 kgm²

[tex]\alpha[/tex] = Angular acceleration

[tex]\omega_f[/tex] = Final angular velocity

[tex]\omega_i[/tex] = Initial angular velocity

t = Time taken

Torque is given by

[tex]\tau=I\alpha\\\Rightarrow \tau=5\times 200\\\Rightarrow \tau=1000\ Nm[/tex]

The torque of the disc would be 1000 Nm

If [tex]\alpha=400\ rad/s^2[/tex]

[tex]\tau=I\alpha\\\Rightarrow \tau=5\times 400\\\Rightarrow \tau=2000\ Nm[/tex]

The torque of the disc would be 2000 Nm

From equation of rotational motion

[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{3820\times \frac{2\pi}{60}-0}{400}\\\Rightarrow t=1.00007\ s[/tex]

It would take 1.00007 seconds to reach 3820 rpm

To solve this problem it is necessary to apply the continuity equations for which it is defined that the proportion of Area in the initial section is equal to the final section. In other words,

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

Where,

[tex]A_i =[/tex] Cross sectional area at each section

[tex]v_i =[/tex]Velocities of fluid at each section

The total area of the branch is eighteen times of area of smaller artery. The average cross-sectional area of each artery is [tex]0.7cm^2.[/tex]

Therefore the Cross-sectional area at the end is

[tex]A_2= 18*0.7cm^2[/tex]

[tex]A_2 = 12.6cm^2[/tex]

Applying the previous equation we have then

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

[tex](1.7cm^2) v_1 = (12.6cm^2)v_2[/tex]

The ratio of the velocities then is

[tex]\frac{v_1}{v_2} = \frac{1.7}{12.6}[/tex]

[tex]\frac{v_1}{v_2} = 0.135[/tex]

Therefore the factor by which the velocity of blood will reduce when it enters the smaller arteries is 0.1349

To solve this problem it is necessary to apply the concepts related to Current and Load.

The current in terms of the charge of an electron can be expressed as

[tex]i = \frac{q}{t}[/tex]

Where,

q = Charge

t = time

At the same time the Charge is the amount of electrons multiplied by the amount of these, that is

q = ne

Replacing in the first equation we have to

[tex]i = \frac{q}{t}[/tex]

[tex]i = \frac{ne}{t}[/tex]

Clearing n,

[tex]n = \frac{it}{e}[/tex]

Here the time is one second then

[tex]n = \frac{i}{e}[/tex]

[tex]n = \frac{4.1}{1.6*10^{-9}}[/tex]

[tex]n = 2.56*10^{19}electrons[/tex]

Therefore the number of electrons per second are passing any cross sectional area of the wire are [tex]2.56*10^{19}electrons[/tex]

Answer:

Length of the pendulum will be 3.987 m

Explanation:

We have given time period of the pendulum T = 8 sec

Acceleration due to gravity [tex]g=9.81m/sec^2[/tex]

We have to find the length of the simple pendulum

We know that time period of the simple pendulum is given by

[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]

[tex]8=2\times 3.14 \sqrt{\frac{l}{9.81}}[/tex]

[tex]l=3.987m[/tex]

So length of the pendulum will be 3.987 m

Answer:B

Explanation:

For the ball dropped From a tall ball building the direction of acceleration is downward and its magnitude is [tex]9.8 m/s^2[/tex]

For the ball thrown upward towards the dropped ball the direction and magnitude of acceleration is same i.e. downwards as gravity always acting downward with constant magnitude.

Thus option B is correct

In this case, the acceleration of both balls is the same (Option B).

Acceleration is a physical property to measure the movement of an object in a given period of time.

In conclusion, in this case, the acceleration of both balls is the same (Option B).

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