A rock is thrown from the top of a 20-m building at an angle of 53° above the horizontal. If the horizontal range of the throw is equal to the height of the building, with what speed was the rock thrown? What is the velocity of the rock just before it strikes the ground?

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]

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:

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:

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:

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:

The change in the area of the circle is [tex]8.88\times10^{-7}\ m^2[/tex]

Explanation:

Given that,

Radius = 0.250 cm

Temperature = 20.0°C

Final temperature =800.0°C

Coefficient of linear thermal expansion for lead[tex]\alpha =29.0\times10^{-6}\ /\°C [/tex]

We calculate the change in temperature,

[tex]\Delta T=800.0-20.0=780^{\circ}[/tex]

Now, We calculate the area of the disc

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

Put the value into the formula

[tex]A=3.14\times(2.5\times10^{-3})^2[/tex]

[tex]A =1.9625\times10^{-5}\ m^2[/tex]

We need to calculate the areal expansion

[tex]\Delta A=2\alpha\times A\times\Delta T[/tex]

[tex]\Delta A=2\times29.0\times10^{-6}\times1.9625\times10^{-5}\times780[/tex]

[tex]\Delta A=8.88\times10^{-7}\ m^2[/tex]

Hence, The change in the area of the circle is [tex]8.88\times10^{-7}\ m^2[/tex]

Final answer:

The change in area of the lead circular disc can be calculated using the formulas for linear thermal expansion. The change in radius is calculated using the coefficient of linear expansion and the change in temperature. The change in area is then calculated using the change in radius and the original radius.

Explanation:

To determine the change in area of the circular disc made of lead, we need to calculate the change in its radius using the coefficient of linear thermal expansion. The formula for linear thermal expansion is given by AL = a * L * AT, where AL is the change in length, a is the coefficient of linear expansion, L is the original length, and AT is the change in temperature.

In this case, we are interested in the change in radius, so we can use the formula AR = a * R * AT, where AR is the change in radius and R is the original radius.

Substituting the given values, we have:

AR = (29.0 × 10^-6/°C) * (0.250 cm) * (800.0 °C - 20.0 °C)

AR = 0.00145 cm

The change in area can be calculated using the formula AE = 2π * R * AR. Substituting the values, we have:

AE = 2π * (0.250 cm) * (0.00145 cm)

AE = 2.31 x 10^-3 cm²

Answer:

63.6 %

Explanation:

TH = 1100 K , Tc = 400 k

Efficiency is given by

η = 1 - Tc / TH

η = 1 - 400 / 1100

η = 1 - 0.36

η = 0.636

η = 63.6 %

Answer:

11.515 Joule

Explanation:

Volume of aluminium = V = 4.89×10⁻³ m³

Coefficient of volume expansion for aluminum = α = 69×10⁻⁶ /°C

Initial temperature = 19.1°C

Final temperature = 357°C

Pressure of air = 1.01×10⁵ Pa

Change in temperature = ΔT= 357-19.1 = 337.9 °C

Change in volume

ΔV = αVΔT

⇒ΔV = 69×10⁻⁶×4.89×10⁻³×337.9

⇒ΔV = 114010.839×10⁻⁹ m³

Work done

W = PΔV

⇒W = 1.01×10⁵×114010.839×10⁻⁹

⇒W = 11.515 J

∴ Work is done by the expanding aluminum is 11.515 Joule

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

The heat needed to evaporate 31.5 g of boiling water is approximately 7,150 J.

To calculate the heat needed to completely evaporate water, we use the equation Q = m × ΔHvap, where Q is the heat, m is the mass, and ΔHvap is the heat of vaporization.

The heat of vaporization for water is approximately 40.7 kJ/mol. To find the heat for 31.5 g of water, we need to convert the mass to moles by dividing by the molar mass of water (18.015 g/mol).

Using the equation, Q = (31.5 g ÷ 18.015 g/mol) × (40.7 kJ/mol × 1000 J/kJ), the answer is approximately 7,150 J.

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Final answer:

To evaporate 31.5 g of boiling water at 100°C, 70,875 joules of heat are needed. This is determined by multiplying the mass by the heat of vaporization for water, which is 2,250 J/g.

Explanation:

To calculate the amount of heat required to completely evaporate 31.5 g of boiling water at 100°C, the heat of vaporization of water is needed. The heat of vaporization of water is 2,250 J/g. To find the total heat we will multiply the mass of the water by the heat of vaporization:

Heat required = mass × heat of vaporization

Heat required = 31.5 g × 2,250 J/g

Heat required = 70,875 J

Therefore, 70,875 joules of heat would be needed to completely evaporate 31.5 g of boiling water at 100°C.

Answer:

32 J

Explanation:

Power is V²/R = 20²/1500 = 4/15 . . . watts

In 120 seconds, the energy is ...

(4/15 J/s)×(120 s) = 32 J

In a 2-minute period 32 joules are dissipated as heat.

We calculate the power dissipation rate of the 1.5-kΩ resistor under a constant potential difference of 20- V, which is 0.27 Watts or 0.27 Joules per second. Multiplying this rate by the time interval of two minutes (or 120 seconds), we find the total energy dissipated as heat, which is 32.4 Joules.

To understand the amount of energy dissipated during the given time interval, we first need to determine the power of the 1.5-kΩ resistor. This can be done using the equation P = V²/R, where V is the voltage, and R is the resistance. For this case, we have P = (20 V) ² / (1.5 kΩ) = 0.27 Watts.

Since the power dissipation rate is 0.27 Watts, which is equivalent to 0.27 Joules per second, we can now find out the energy dissipated during the two-minute interval using the equation E = Pt. Here, t represents the time in seconds, so t = 2 minutes * 60 seconds/minute = 120 seconds. Therefore, substituting the known values, we get E = (0.27 J/s)(120 s) = 32.4 Joules.

So, this is the total energy dissipated as heat by the 1.5- kΩ resistor under a constant potential difference of 20- V during the two-minute interval.

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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:

13 sec

Explanation:

Hello

by definition the acceleration is a vector derived magnitude that indicates the speed variation per unit of time

[tex]a=\frac{V_{f}-V_{i}}{t_{f}-t_{i}}\\a(t_{f}-t_{i})={V_{f}-V_{i}  }}\\t_{f}-t_{i}=\frac{V_{f}-V_{i}}{a} \\\\t_{f}=\frac{V_{f}-V_{i}}{a}+t_{i} \\\\[/tex]

Let

Vi=4.0 m/s

a=1.0 m/s2

Vf=17.0 m/s

ti= (0)sec

t2= unknown

[tex]t_{f}=\frac{V_{f}-V_{i} }{a}+t_{i}\\t_{f}=\frac{17-4 }{1}+0\\t_{f}=\ 13\ sec\\[/tex]

Answer: 13 sec

I hope it helps

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.

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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:

[tex]F_{net} = 220.8 N[/tex]

Explanation:

It is pulled by three forces as given below

1. Jack pulls directly ahead of the donkey with a force of 61.3 N,

2. Jill pulls with 83.9 N in a direction 45° to the left, and

3. Jane pulls in a direction 45° to the right with 137 N.

Now net force directly in forward direction given as

[tex]F_x = 61.3 N + 83.9 cos45 + 137cos45[/tex]

[tex]F_x = 217.5 N[/tex]

Now similarly in perpendicular to this we have

[tex]F_y = 137 sin45 - 83 sin45 [/tex]

[tex]F_y = 38.2 N[/tex]

Now net force is given by them

[tex]F_{net} = \sqrt{F_x^2 + F_y^2}[/tex]

[tex]F_{net} = \sqrt{217.5^2 + 38.2^2}[/tex]

[tex]F_{net} = 220.8 N[/tex]

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:

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:

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.

Answer:

W = 2.158 eV

fo = 5.23 x 10^14 Hz

Explanation:

f = 7.9 x 10^14 Hz, Vo = 1.1 V

Let W be the work function.

Use the Einstein equation

Energy = W + eVo

hf = W + eVo

where, h is the Plank's constant and e be the electronic charge.

W = hf - eVo

W = (6.6 x 10^-34 x 7.9 x 10^14) - (1.6 x 10^-19 x 1.1)

W = 5.214 x 10^-19 - 1.76 x 10^-19 = 3.454 x 10^-19 J

W = (3.454 x 10^-19) / (1.6 x 10^-19) = 2.158 eV

Let fo be the cut off frequency

W = h fo

fo = W / h = (3.454 x 10^-19) / (6.6 x 0^-34) = 5.23 x 10^14 Hz

Answer:

temperature

Explanation:

When temperature is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system.

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

Enthalpy, the sum of the internal energy and the product of the pressure and volume of a thermodynamic system. Enthalpy is an energy-like property or state function—it has the dimensions of energy (and is thus measured in units of joules or ergs),

and its value is determined entirely by the temperature, pressure, and composition of the system and not by its history. In symbols, the enthalpy, H, equals the sum of the internal energy, E, and the product of the pressure, P, and volume, V, of the system: H = E + PV.

According to the law of energy conservation, the change in internal energy is equal to the heat transferred to, less the work done by, the system.

If the only work done is a change of volume at constant pressure, the enthalpy change is exactly equal to the heat transferred to the system

Hence when pressure is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system

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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:

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:

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:

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

Answer:

[tex]f_{min} = 574.3 Hz[/tex]

[tex]f_{max} = 604.5 Hz[/tex]

Explanation:

As per Doppler's effect of sound we know that when source and observer moves relative to each other then the frequency of sound observed by the observer is different from real frequency of sound.

As the source is moving here in this case so the frequency is given as

[tex]f = f_o\frac{v}{v\pm v_s}[/tex]

part a)

for lowest frequency we will have

[tex]f_{min} = 589(\frac{343}{343 + R\omega})[/tex]

here we know that

R = 54.6 cm

[tex]\omega = 16.1 rad/s[/tex]

now we have

[tex]f_{min} = 589(\frac{343}{343 + 0.546(16.1)})[/tex]

[tex]f_{min} = 574.3 Hz[/tex]

part b)

for maximum frequency we will have

[tex]f_{max} = 589(\frac{343}{343 - R\omega})[/tex]

here we know that

R = 54.6 cm

[tex]\omega = 16.1 rad/s[/tex]

now we have

[tex]f_{max} = 589(\frac{343}{343 - 0.546(16.1)})[/tex]

[tex]f_{max} = 604.5 Hz[/tex]

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:

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:

1.77 x 10^-8 C

Explanation:

Let the surface charge density of each of the plate is σ.

A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2

d = 2 mm

E = 2.5 x 10^6 N/C

ε0 = 8.85 × 10-12 C2/N ∙ m2

Electric filed between the plates (two oppositively charged)

E = σ / ε0

σ = ε0 x E

σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2

The surface charge density of each plate is ± σ / 2

So, the surface charge density on each = ± 22.125 x 10^-6 / 2

                                                                 = ± 11.0625 x 10^-6 C/m^2  

Charge on each plate = Surface charge density on each plate x area of each plate

Charge on each plate = ± 11.0625 x 10^-6  x 16 x 10^-4 = ± 1.77 x 10^-8 C