When ________ is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system? pressure and volume temperature temperature and volume pressure volume

Answer:

510.4 N

Explanation:

u = 13.2 m /s, v = 25.7 m/s, t = 8.6 s, m = 352 kg

Use first equation of motion

v = u + a t

a = (25.7 - 13.2) / 8.6 = 1.45 m/s^2

Use Newton's second law

F = m a = 352 x 1.45 = 510.4 N

The range of the marble when fired horizontally from 1.8m above the ground can be calculated using the equations of motion in physics. First, the time of flight is found using the vertical motion and then the range is calculated using the time of flight and the initial velocity determined from the vertical launch. The marble's range is approximately 8.4m.

To solve this problem, we need to make use of the concept of projectile motion in physics. The most crucial part in solving this type of problem is to break the motion into its horizontal and vertical components.

First, we find the time the projectile is in the air using the vertical motion. Ignoring air resistance, the time a projectile is in the air is determined by the initial vertical velocity and the height from which it drops. Here, the height is given as 1.8m and we can use the equation h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. After calculating, we find that the time the marble is in the air is about 0.6 seconds.

Now, we can use the time to find the horizontal distance traveled by the marble, a.k.a the range. The range is given by R = vt, where v is the horizontal velocity, which is the same as the initial vertical velocity. From the problem, we know the marble reached a height of 9.0m when shot vertically, which we can use to find the initial velocity using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height. We find that the initial velocity is about 14 m/s.

So, the range R = vt = 14m/s * 0.6s = 8.4m. Therefore, the marble's range when fired horizontally from 1.8m above the ground is approximately 8.4m.

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The marble's range if it is fired horizontally from 1.8 m above the ground is 7.968 m.

At the highest point, the velocity of the gun will be zero. The initial velocity of the shot can be determined using the kinematic equation,

[tex]v^2 = u^2 + 2gs[/tex]

here, v = 0 m/s

g = 9.8 m/s² (downward, hence taken negative)

s = 9.0 m (upwards, hence taken positive.)

Using proper sign convention, we get:

[tex](0 \hspace{0.8 mm} m/s)^2 = u^2 - 2 \times (9.8 \hspace{0.8 mm} m/s^2) \times (9.0 \hspace{0.8 mm} m )[/tex]

u = 13.28 m/s

Now, when the marble is fired at a height of 1.8 m above the ground, the time taken by the marble to reach the ground can be determined by the kinematic equation:

[tex]h = ut + \frac{1}{2}gt^2[/tex],

using u = 0 m/s (as we will consider the time of fall of the marble from highest point), g = 9.8 m/s² (downward, hence taken to be negative), h = 1.8 m (downwards, hence taken positive), we get:

[tex]- 1.8 \hspace{0.8 mm} m = (0 \hspace{0.8 mm} m/s)t - \frac{1}{2}(9.8 \hspace{0.8 mm} m/s^2)t^2[/tex]

or, [tex]1.8 \hspace{0.8 mm} m = (4.9 \hspace{0.8 mm} m/s^2)t^2[/tex]

or, t = 0.60 s

Range of the marble will be determined by the horizontal velocity of the marble. It will be the maximum horizontal distnace covered by the marble as it falls down in time t = 0.60 s.

R = [tex]v_x[/tex]t

Since, the horizontal velocity is not influenced by any acceleration and will remain constant.

∴ R = [tex]v_x[/tex]t = 13.28 m/s × 0.60 s = 7.968 m

Answer:

20.26 Nm

Explanation:

r = 1.23 m , m = 93.6 kg, w = 0, f0 = 369 rpm = 369 / 60 = 6.15 rps

w0 = 2 x 3.14 x 6.15 = 38.622 rad/s

t = 2.25 min = 2.25 x 60 = 135 second

Moment of inertia = 1/2 m r^2 = 0.5 x 93.6 x 1.23 x 1.23 = 70.8 kg m^2

use first equation of motion for rotational motion

w = w0 + α t

0 = 38.622 - α x 135

α = 0.286 rad/s^2

torque = moment of inertia x angular acceleration

Torque = 70.8 x 0.286 = 20.26 Nm

Answer:

You will have 4.5 million dollar

Explanation:

The thickness of a $1 bill is 0.11 mm

So we have

              1 $ = 0.1 mm

              0.1 mm = 1 $

              0.0001 m =  1 $

              1 m = 10000 $

           450 m = 450 x 10000 = 4500000 $

So you will have 4.5 million dollar

Answer:

4090909

Explanation:

Thickness of one bill = 0.11 mm

Total thickness = 450 m

No of $1 bills = total thickness / thickness of one bill

No of $1 bills = 450 / 0.11 × 10^-3

= 4090909

Answer:

[tex]T_2 = 309 N[/tex]

[tex]T_1 = 489 N[/tex]

Explanation:

As we know that total tension in both the ropes is counter balancing the weight of scaffold and worker both

so here we will have

[tex]T_1 + T_2 = (m + M)g[/tex]

now we have

[tex]T_1 + T_2 = 198 + 600 = 798 N[/tex]

now we also know that net torque due to both tension force in the string with respect to the position of worker must be zero so that platform will remain in equilibrium and horizontal in position

so here we will have

[tex]T_1(1.12) + (198)(1.6 - 1.12) = T_2(3.2 - 1.12)[/tex]

[tex]T_1+ 84.86 = 1.86 T_2[/tex]

now from above two equations we will have

[tex](1.86 T_2 - 84.86) + T_2 = 798 [/tex]

[tex]T_2 = 309 N[/tex]

also we have

[tex]T_1 = 489 N[/tex]

Final answer:

To calculate the tension in each rope supporting a scaffold with a window washer, the principles of torques and equilibrium are applied. By considering the system's equilibrium, equations based on the torques created by the worker's weight, the scaffold's weight, and the tensions in the ropes can be set up and solved.

Explanation:

To solve for the tension in each rope when a window washer stands on a scaffold, we use the concept of torques and equilibrium. The entire system is in equilibrium, meaning the sum of torques around any pivot point is zero. Considering the window washer weighing 600 N and standing 1.12 m from one end of a 3.2 m scaffold that weighs 198 N, we choose the pivot point at one end of the scaffold for easier calculations.

Let T1 be the tension in the rope closest to the worker and T2 be the tension in the other rope. The force due to the worker creates a clockwise torque, and the force due to the scaffold creates a clockwise torque as well, whereas the tensions in the ropes create counterclockwise torques. The torques are calculated by multiplying the force by the distance from the pivot point.

To find T1 and T2, we set up the equilibrium condition for torques and solve the equations considering the weight of the worker and the scaffold acting at their respective centers of mass. The specific calculations require numerical values that depend on the distances given and the gravitational force.

Ultimately, by solving these equations, we find the tensions T1 and T2 that support the system in equilibrium.

Answer:

Gravitational force, [tex]F=1.003\times 10^{-11}\ N[/tex]

Explanation:

It is given that,

Mass of roast beef sandwich, m₁ = 0.05 kg

Mass of hungry tarantula, m₂ = 87 gm = 0.087 kg

Distance between two objects, d = 17 cm = 0.17 m

We need to find the gravitational force between them. It is given by :

[tex]F=G\dfrac{m_1m_2}{d^2}[/tex]

[tex]F=6.67\times 10^{-11}\times \dfrac{0.05\ kg\times 0.087\ kg}{(0.17\ m)^2}[/tex]

[tex]F=1.003\times 10^{-11}\ N[/tex]

So, the gravitational force between two objects is [tex]F=1.003\times 10^{-11}\ N[/tex]. Hence, this is the required solution.

Answer:

75 rotations

Explanation:

f0 = 0, f = 3000 rpm = 50 rps, t = 3 s

(a) use first equation of motion for rotational motion

w = w0 + α t

2 x 3.14 x 50 = 0 + α x 3

α = 104.67 rad/s^2

(b) Let θ be the angular displacement

use second equation of motion for rotational motion

θ = w0 t + 1/2 α t^2

θ = 0 + 0.5 x 104.67 x 3 x 3

θ = 471.015 rad

The angle turn in one rotation is 2 π radian.

Number of rotation = 471.015 / (2 x 3.14) = 75 rotations

Answer:

90 degree

Explanation:

According to the formula of torque

torque = force x displacement x Sine of angle between force and displacement

So, for the maximum torque, the value of Sin theta should be maximum.

the maximum value of Sin theta is 1.

that means the value of theta is 90 degree.

Answer:

v = 7121.3 m/s

Explanation:

As we know that the centripetal force for the space shuttle is due to gravitational force of earth due to which it will rotate in circular path with constant speed

so here we will have

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

here we know that

r = orbital radius = 6370 km + 1482 km

[tex]r = 7.852 \times 10^6 m[/tex]

also we know that

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

now we will have

[tex]v^2 = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{7.852 \times 10^6}[/tex]

[tex]v^2 = 5.07 \times 10^7[/tex]

[tex]v = 7121.3 m/s[/tex]

The speed of the space shuttle that orbited the earth at an altitude of 1482km  will be    [tex]V=7121.3\dfrac{m}{s}[/tex]

As we know that the centripetal force for the space shuttle is due to the gravitational force of the earth due to which it will rotate in a circular path with constant speed

so here we will have

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

[tex]V^2=\dfrac{GM}{r}[/tex]

here we know that

[tex]\rm r= orbital \ radius =6370+1482=7852 \ km[/tex]

[tex]r=7.852\times10^6\ m[/tex]

mass of earth [tex]M=5.97\times10^{24}\ kg[/tex]

Gravitational constant  [tex]G=6.67\times10^{-11}[/tex]

By putting all the values we get

[tex]V^2=\dfrac{(6.67\times10^{-11} )(5.97\times10^{24})}{7.852\times10^{6}}[/tex]

[tex]V^2=5.07\times10^7[/tex]

[tex]V=7121.3 \dfrac{m}{s}[/tex]

Thus the speed of the space shuttle that orbited the earth at an altitude of 1482km  will be    [tex]V=7121.3\dfrac{m}{s}[/tex]

Answer:

A probe charge

Explanation:

As we know that electric field intensity is the force experienced by the probe charge which is placed in the electric field region

Here we can say it as

[tex]E = \frac{F}{q}[/tex]

so here that probe charge should be very small so that it will not disturb the electric field in the space.

If the probe charge is of large magnitude then the field will get disturbed and the intensity which is to be measured is different from its actual value.

Also the sign of the probe charge is taken to be positive.

so correct answer here will be

A probe charge

Final answer:

The electric field of a charge is defined by the force experienced by a probe charge placed within the electric field, so the correct option will be D probe charge

Explanation:

The electric field of a charge is defined by the force on a probe charge. An electric field is a vector field that affects the space surrounding a source charge and exerts a force on any test charge (probe charge) within that region. The direction of the electric field represents the direction a positive probe charge would move if placed within the field.

It's calculated as the force per unit charge exerted on a test charge and can be represented mathematically by the equation F = qE, where F is the force experienced by the test charge, q represents the charge of the test charge, and E is the electric field strength.

Answer:

4.3 N/m

Explanation:

m = 230 g = 0.230 kg, x = 8.2 cm

in 13 oscillations, time taken = 19 s

In 1 oscillation, time taken = 19 / 13 = 1.46 s

By the use of formula of time period , Let k be the spring constant.

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

[tex]1.46 = 2\pi \sqrt{\frac{0.230}{k}}[/tex]

0.054 = 0.230 / k

k = 4.26 N/m

k = 4.3 N/m

The accelerating potential needed for electrons of a particular wavelength can be found by rearranging the de Broglie wavelength equation. The energy of photons of the same wavelength can be found using Planck's equation, by substituting the given wavelength.

The accelerating potential needed to produce electrons of a wavelength (de Broglie wavelength) can be calculated using the equation: λ = h / √(2mVq), where h is Planck's constant, m is the mass of the electron, V is the accelerating voltage, and q is the charge of the electron. For the wavelength of 5.20 nm, one can rearrange and solve for V to get V = h² / (2mqλ²).

Now, for the energy of photons having the same wavelength as the electrons, we can use Planck's equation, E = hv = hc/λ. Here, 'h' is Planck's constant, 'v' is the frequency of light, 'c' is the speed of light and 'λ' is the wavelength. Substituting λ = 5.20 nm, we get the energy of photons in terms of electron-volts (eV).

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Answer:

a) [tex]v_{max} = 2\ \textup{m/s}[/tex]

b) [tex]v_{avg} = 1\ \textup{m/s}[/tex]

c) Q = 1.256 × 10⁻³ m³/s

Explanation:

Given:

The velocity profile as:

[tex]u(r) = 2(1-\frac{r^2}{R^2} )[/tex]

Now, the maximum velocity of the flow is obtained at the center of the pipe

i.e r = 0

thus,

[tex]v_{max}=u(0) = 2(1-\frac{0^2}{R^2} )[/tex]

or

[tex]v_{max} = 2\ \textup{m/s}[/tex]

Now,

[tex]v_{avg} = \frac{v_{max}}{2}\ \textup{m/s}[/tex]

or

[tex]v_{avg} = \frac{2}}{2}\ \textup{m/s}[/tex]

or

[tex]v_{avg} = 1\ \textup{m/s}[/tex]

Now, the flow rate is given as:

Q = Area of cross-section of pipe × [tex]v_{avg}[/tex]

or

Q = [tex]\frac{\pi D^2}{4}\times v_{avg}[/tex]

or

Q = [tex]\frac{\pi 0.04^2}{4}\times 1[/tex]

or

Q = 1.256 × 10⁻³ m³/s

For fully developed laminar pipe flow in a circular pipe, the maximum velocity is 2 m/s, the average velocity is (4/3) m/s, and the volume flow rate is (16/3)π cm^3/s. The velocity profile equation and the formulas for maximum velocity, average velocity, and volume flow rate are explained in detail.

For fully developed laminar pipe flow in a circular pipe, the velocity profile is given by the equation u(r) = 2(1-r^2/R^2) in m/s, where R is the inner radius of the pipe. To find the maximum velocity, we need to substitute r = R into the equation. The maximum velocity u_max is then equal to 2(1-1^2/R^2), which simplifies to 2 m/s.

The average velocity can be found by integrating the velocity profile equation over the entire cross-sectional area of the pipe. The cross-sectional area is given by A = πR^2, so the average velocity v_avg is equal to (1/A) times the integral of 2(1-r^2/R^2) over the range r = 0 to r = R. Simplifying the integral and dividing by A, we get v_avg = (4/3) m/s.

The volume flow rate Q can be calculated by multiplying the cross-sectional area A by the average velocity v_avg. Using the given diameter of the pipe, we can find the radius R = 2 cm. Substituting the values into the equation Q = (πR^2)(4/3), we get Q = (16/3)π cm^3/s.

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Answer:

Velocity is 1.73 m/s along 54.65° south of east.

Explanation:

Let unknown velocity be v, mass of billiard ball be m and east direction be positive x axis.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = m x 2i + m x (-1)i = m i

Final momentum = m x v + m x 1.41 j = mv + 1.41 m j

Comparing

mi = mv + 1.41 m j

v = i - 1.41 j

Magnitude of velocity

      [tex]v=\sqrt{1^2+(-1.41)^2}=1.73m/s[/tex]        

Direction,  

       [tex]\theta =tan^{-1}\left ( \frac{-1.41}{1}\right )=-54.65^0[/tex]             

Velocity is 1.73 m/s along 54.65° south of east.

Using the law of conservation of momentum, one can deduce the speed and direction of the first ball after the collision. It's found to be traveling east at 1 m/s.

The scenario described is an example of a two-dimensional collision. In such collision, the law of conservation of momentum applies both in the east-west direction (x-axis) and the north-south direction (y-axis).

With the given velocities before the collision, the total momentum in the x-axis before the collision is: momentum_east = mass*velocity = m*2.00 m/s, and momentum_west = m*(-1.00) m/s. Therefore, total momentum in the x-axis before the collision = m*2.00 m/s + m*(-1.00) m/s = m m/s.

After the collision, the first ball keeps moving in the east-west direction (since the second ball is deflected north), but we don't know its velocity, let's call it v1. Applying the conservation of momentum after the collision in the x-direction, we get: total momentum = m*v1 + 0 (since the second ball no longer moves in the east-west direction) = m m/s. From this, we can solve for v1 and find that v1 = 1 m/s east. Thus, the first billiard ball is traveling east at 1 m/s after the collision.

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Answer:

[tex]d = L_1 + L_2 + \frac{(m_1 + m_2)g}{K_1} + \frac{m_2g}{K_2}[/tex]

Explanation:

For first spring the total extension is given as

[tex]F_{net} = K_1 x_1[/tex]

here net force is due to weight of both masses

[tex](m_1 + m_2)g = K_1 x_1[/tex]

so extension of first spring is

[tex]x_1 = \frac{(m_1 + m_2)g}{K_1}[/tex]

now similarly the extension in second spring is only due to the weight of mass m2

so here we will have

[tex]m_2g = K_2 x_2[/tex]

[tex]x_2 =\frac{m_2g}{K_2}[/tex]

so the total distance from the ceiling for mass m2 is given as

[tex]d = L_1 + L_2 + \frac{(m_1 + m_2)g}{K_1} + \frac{m_2g}{K_2}[/tex]

The total distance of m₂ from the ceiling is [tex]d = L_1 + L_2 + \frac{g(m_1 + m_2)}{k_1} + \frac{m_2g}{k_2}[/tex]

The given parameters;

The extension of the first spring due to mass m₁ and m₂ is calculated as;

[tex]F = kx\\\\ g(m_1+m_2) = k_1x_1\\\\x_1 = \frac{g(m_1+m_2)}{k_1}[/tex]

The extension of the second spring due to mass (m₂) is calculated as;

[tex]m_2g = k_2x_2\\\\x_2 = \frac{m_2g}{k_2}[/tex]

The total distance of m₂ from the ceiling is calculated as follows;

[tex]d = L_1 + L_2 + x_1 + x_2\\\\d = L_1 + L_2 + \frac{g(m_1 + m_2)}{k_1} + \frac{m_2g}{k_2}[/tex]

Thus, the total distance of m₂ from the ceiling is [tex]d = L_1 + L_2 + \frac{g(m_1 + m_2)}{k_1} + \frac{m_2g}{k_2}[/tex]

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Answer:

Displacement is 565.69 m at 45° west of north

Explanation:

Let north represent positive y axis and east represent positive x axis.

We have journey started from 600 N. Michigan Avenue, and walked 3 blocks toward north, 4 blocks toward west, and 1 block toward north to a train station.

3 blocks toward north = 300 j m

4 blocks toward west = -400 i m

1 blocks toward north = 100 j m

Total displacement = -400 i + 400 j m

Magnitude

     [tex]s=\sqrt{(-400)^2+400^2}=565.69m[/tex]

Direction,

     [tex]\theta =tan^{-1}\left ( \frac{400}{-400}\right )=135^0[/tex]

     Direction is 45° west of north.

Displacement is 565.69 m at 45° west of north

(a) 7.9 s

The period of a wave is time that passes between two consecutive crests (or two consecutive troughs).

In this case, we are told that five crests pass in a time of 39.5 s. Therefore we can find the period by using the proportion:

[tex]\frac{5}{39.5 s}=\frac{1}{T}[/tex]

Where T is the period. Re-arranging the equation, we find

[tex]T=\frac{(39.5)(1)}{5}=7.9 s[/tex]

(b) 0.127 Hz

The frequency of a wave is equal to the reciprocal of the period:

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

where

f is the frequency

T is the period

For this wave, we have T = 7.9 s, so its frequency is

[tex]f=\frac{1}{7.9 s}=0.127 Hz[/tex]

(c) 37.9 m

The wavelength of a wave is the distance between two consecutive crests (or two consecutive troughs). For this wave, the distance between two successive crests is 37.9 m, so the wavelength of the wave is

[tex]\lambda=37.9 m[/tex]

(d) 4.81 m/s

The speed of a wave is given by

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

where

[tex]\lambda[/tex] is the wavelength

f is the frequency

For the wave in the problem, we have

[tex]\lambda=37.9 m\\f=0.127 Hz[/tex]

Therefore, the speed of the wave is

[tex]v=(37.9)(0.127)=4.81 m/s[/tex]

Answer:

Option B

Explanation:

The net emf in the circuit

E = 12 - 10 = 2 V

Total effective resistance,

r = 1 + 1 = 2 ohm

By using Ohm's law

E = I × R

I = 2 / 2 = 1 A

To find the current in the circuit, use Ohm's Law and solve two equations.

The current in the circuit can be found using Ohm's Law: V = IR, where V is the voltage, I is the current, and R is the resistance.

In this case, the emf of the first battery is 12 V with an internal resistance of 1 Ω. Let's assume the current in the circuit is I1. So, the terminal voltage across the first battery can be calculated using Ohm's Law: 12 = I1 * (1 + 1).

Similarly, for the second battery with emf of 10 V and internal resistance of 1 Ω, the terminal voltage can be calculated as 10 = (I1 - I) * (1 + 1), where I is the current flowing in the second battery.

Now, we need to solve these two equations to find the value of I, which is the current in the circuit.

Answer:

0.115 m

Explanation:

Consider the motion of tongue during acceleration :

v₀ = initial velocity of the tongue = 0 m/s

v = final velocity of the tongue = 5 m/s

a = acceleration = 2500 m/s²

d = distance traveled during acceleration phase

Using the equation

v² = v₀² + 2 a d

5² = 0² + 2 (2500) d

d = 0.005 m

Consider the motion of tongue after it attains constant speed

d' = distance traveled during constant velocity

v = constant velocity = 5 m/s

t = time of travel = 22 ms = 0.022 s

using the equation

d' = v t

d' = 5 x 0.022

d' = 0.11 m

D = Total distance traveled by tongue

Total distance traveled by tongue is given as

D = d + d'

D = 0.005 + 0.11

D = 0.115 m

Answer:

a) The velocity is zero when the object is at the highest vertical point (peak). b) Yes, at the peak, the velocity for an instant gets to zero and then the direction of motion and the velocity will be downwards.

C) No, the direction of ‘g’ will be opposite in direction when the object is on the way down when compared to what it was during the upward motion.

Explanation:

a) Since gravity is acting on the object and trying to pull it down to earth, the ball will slow down to a velocity of zero when the ball starts its free fall downward.

b) When the direction of the object changes, its velocity will also change direction.

c) When the ball has a positive velocity, gravity is pulling it in the opposite direction, so it has a negative acceleration trying to slow it down. When the ball's velocity is negative (downward free fall), the acceleration will be positive at the instant when the ball hits the ground. The only way to stop the ball from falling is a positive force upward, which results in a positive acceleration when the ball hits the ground.

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(a) The velocity of the object will be zero at the maximum height

(b) The velocity of the object changes direction.

(c) The acceleration due to gravity will not have the same sign on the way up as on the way down.

(A) The velocity of the object decreases as the object ascends upwards and eventually becomes zero at the maximum height. As the object descends downwards, the velocity increases and becomes maximum before the object hits the ground.

(B) The velocity of the object is always directed in the direction of the motion of the object. As the object moves upward, the direction of the object is upward and as the object moves downwards the direction of the object is downwards. Thus, the velocity of the object changes direction.

(c) The upward motion of the object is opposite direction to acceleration due to gravity and the sign of acceleration due to gravity becomes negative.

As the object moves downwards, it will be moving in the same direction as the acceleration due to gravity. The sign of acceleration due to gravity is positive.

Thus, the acceleration due to gravity will not have the same sign on the way up as on the way down.

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Answer:

The soda is being sucket out at a rate of 3.14 cubic inches/second.

Explanation:

R= 2in

S= π*R²= 12.56 inch²

rate= 0.25 in/sec

rate of soda sucked out= rate* S

rate of soda sucked out=  3.14 inch³/sec

Final answer:

Soda is being pulled out of a cylindrical glass at the rate of -3.14 cubic inches per second.

Explanation:

The question asks at what rate soda is being absorbed from a cylindrical glass. Given the cylindrical glass's dimensions and the rate at which the depth of soda decreases, we need to find the volume rate of change.

He formula for the volume of a cylinder is V = πr2h, where r is the radius and h is the height of the cylinder. To find the rate at which volume changes, we differentiate with respect to time to obtain dV/dt = πr2(dh/dt). Plugging in the values, we have r = 2 inches and dh/dt = -0.25 inches/second (negative because the depth is decreasing).

Using the given values, the rate at which soda is being pulled out of the glass is

dV/dt = π(2 in)2(-0.25 in/sec) = π(4 in2)(-0.25 in/sec) = -3.14 in3/sec.

Therefore, the rate at which the soda is being pulled out of the glass is -3.14 in3/sec.

Answer:

The final temperature is 270.5°C

Explanation:

Given that,

Change length = 0.18 m

Length = 80 m

Coefficient of thermal expansion [tex]\alpha=12\times10^{-6}\ C^{\circ}[/tex]

Temperature = 83°C

We need to calculate the final temperature

Using formula thermal expansion

[tex]\delta l=l\alpha\times\delta t[/tex]

Where, [tex]\Delta l[/tex] =change in length

[tex]\Delta t[/tex] = change in temperature

[tex]\alpha[/tex] = coefficient of thermal expansion

Put the value into the formula

[tex]0.18=80\times12\tmes10^{6}\times(T_{f}-T_{i})[/tex]

[tex]T_{f}-83=\dfrac{0.18}{80\times12\times10^{-6}}[/tex]

[tex]T_{f}=187.5+83=270.5^{\circ} C[/tex]

Hence, The final temperature is 270.5°C

Answer:

The number of turns is 64.

Explanation:

Given that,

Magnetic field = 0.050 T

Area of coil = 100 cm²

Frequency = 60 Hz

Output voltage emf= 12 V

We need to calculate the number of turns

Using formula of induced emf

[tex]emf =NAB\omega[/tex]

[tex]N=\dfrac{emf}{A\times B\times2\pi f}[/tex]

[tex]N=\dfrac{12}{0.01\times0.050\times2\times3.14\times60}[/tex]

[tex]N =63.6 = 64\ turns[/tex]

Hence, The number of turns is 64.

Answer:

You need 63.66 turns.

Explanation:

The number of turns of a magnetic field is given by the following formula:

[tex]N = \frac{V}{S*T*2\pi f}[/tex]

In which N is the number of turns, V is the maximum output voltage, S is the area of the rotating coil, in square meters and T is the measure of the magnetic field and f is the frequency.

In this problem, we have that:

Suppose that you wish to construct a simple ac generator having an output of 12 V maximum when rotated at 60 Hz. This means that [tex]V = 12[/tex] and [tex]f = 60[/tex].

A uniform magnetic field of 0.050 T is available. This means that [tex]T = 0.050[/tex].

If the area of the rotating coil is 100 cm2, how many turns do you need?

This means that [tex]S = 0.01[/tex]m². So:

[tex]N = \frac{V}{S*T*2\pi f}[/tex]

[tex]N = \frac{12}{0.01*0.05*120\pi}[/tex]

[tex]N = 63.66[/tex]

You need 63.66 turns.

The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.

C = Q/ΔV

C is the capacitance

Q is the stored charge

ΔV is the potential difference

Rearrange the equation:

ΔV = Q/C

We also know the capacitance of a parallel-plate capacitor is given by:

C = κε₀A/d

C is the capacitance

κ is the capacitor's dielectric constant

ε₀ is the electric constant

A is the area of the plates

d is the plate separation

If we substitute C:

ΔV = Qd/(κε₀A)

We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.

Answer:

[tex]F = 92.45 N[/tex]

Explanation:

As we know that the force between two charge particles is given by

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

here we know that

[tex]q_1 = 3.55 \mu C[/tex]

[tex]q_2 = 5.45 \mu C[/tex]

now the distance between the two charges is

r = 4.34 cm

now from the formula of electrostatic force we will have

[tex]F = \frac{(9\times 10^9)(3.55 \mu C)(5.45 \mu C)}{0.0434^2}[/tex]

[tex]F = 92.45 N[/tex]

Answer:

velocity maximum = 21.6 ft /s

frequency = 16.88 Hz

Explanation:

Amplitude, A = 0.24 in = 0.02 ft

maximum acceleration, a = 225 ft/s\

The formula for maximum acceleration is

a = ω² A

225 = ω² x 0.02

ω² = 11250

ω = 106.06 rad/s

Maximum velocity, v = ω A

v = 106.06 x 0.02 = 2.1 ft/s

Let f be the frequency

ω = 2 x 3.14 x f

f = 106.06 / (2 x 3.14) = 16.88 Hz

velocity maximum = 21.6 ft /s

frequency = 16.88 Hz

Amplitude, A = 0.24 in = 0.02 ft

maximum acceleration, a = 225 ft/s

The formula for maximum acceleration is

a = ω² A

225 = ω² x 0.02

ω² = 11250

ω = 106.06 rad/s

Maximum velocity, v = ω A

v = 106.06 x 0.02 = 2.1 ft/s

Let f be the frequency

ω = 2 x 3.14 x f

f = 106.06 / (2 x 3.14) = 16.88 Hz

Answer:

The sound intensity of train is 1000 times greater than that of the library.

Explanation:

We have expression for sound intensity level,

            [tex]L=10log_{10}\left ( \frac{I}{I_0}\right )[/tex]

A train whistle has a sound intensity level of 70 dB

We have

           [tex]70=10log_{10}\left ( \frac{I_1}{I_0}\right )[/tex]

A library has a sound intensity level of about 40 dB

We also have

           [tex]40=10log_{10}\left ( \frac{I_2}{I_0}\right )[/tex]

Dividing both equations

           [tex]\frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000[/tex]

The sound intensity of train is 1000 times greater than that of the library.

According to Newton's Law of Gravitation, the Gravity Force is:

[tex]F=\frac{GMm}{{r}^{2}}[/tex]     (1)

This expression can also be written as:

[tex]F=GMm{r}^{-2}[/tex]     (2)

If we derive this force [tex]F[/tex] respect to the distance [tex]r[/tex] between the two masses:

[tex]\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr[/tex]     (3)

Taking into account [tex]GMm[/tex] are constants:

[tex]\frac{dF}{dr}dFdr=-2GMm{r}^{-3}[/tex]     (4)

Or

[tex]\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}[/tex]     (5)

In other words, this means how much does the Gravity Force changes with the distance between the two bodies.

More precisely this change is inversely proportional to the distance elevated to the cubic exponent.

As the distance increases, the Force decreases.

This is because Gravity is an attractive force, as well as, a central conservative force.

This means it does not depend on time, and both bodies are mutually attracted to each other.

In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit [tex]N/km[/tex]:

[tex]\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}[/tex]     (5)

We have a force that decreases with a rate 1 [tex]\frac{dF_{1}}{dr}dFdr=4N/km[/tex] when [tex]r=20000km[/tex]:

[tex]4N/km=-2\frac{GMm}{{(20000km)}^{3}}[/tex]     (6)

Isolating [tex]-2GMm[/tex]:

[tex]-2GMm=(4N/km)({(20000km)}^{3})[/tex]     (7)

In addition, we have another force that decreases with a rate 2 [tex]\frac{dF_{2}}{dr}dFdr=X[/tex] when [tex]r=10000km[/tex]:

[tex]XN/km=-2\frac{GMm}{{(10000km)}^{3}}[/tex]     (8)

Isolating [tex]-2GMm[/tex]:

[tex]-2GMm=X({(10000km)}^{3})[/tex]     (9)

Making (7)=(9):

[tex](4N/km)({(20000km)}^{3})=X({(10000km)}^{3}[/tex]       (10)

Then isolating [tex]X[/tex]:

[tex]X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}[/tex]  

Solving and taking into account the units, we finally have:

[tex]X=-32N/km[/tex]>>>>This is how fast this force changes when [tex]r=10000 km[/tex]

Answer:

6.29 m/s option (A)

Explanation:

theta = 45 degree, H = 1.01 m

let v be the launch speed

Use the formula for the maximum height for the projectile

H = v^2 Sin^θ / 2g

1.01 = v^2 x Sin^2(45) / (2 x 9.8)

1.01 = 0.0255 v^2

v^2 = 39.59

v = 6.29 m/s

The initial velocity of the grasshopper is 6.29 m/s.

The Initial velocity of the grasshopper is calculated from the following kinematic equation.

[tex]H = \frac{v_0^2 sin^2 \theta}{2g}[/tex]

where;

[tex]v_0^2 = \frac{2gH}{sin^2\theta} \\\\v_0^2 = \frac{2 \times 9.8 \times 1.01 }{(sin45)^2} \\\\v_0^2 = 39.6\\\\v_0 = 6.29 \ m/s[/tex]

Thus, the initial velocity of the grasshopper is 6.29 m/s.

Learn more about initial velocity of a projectile here: https://brainly.com/question/12870645