A golfer hits a shot to a green. The ball leaves the club at a speed of 20 m/s at an angle 32° above the horizontal. It rises to its maximum height and then falls down to the green. What is the speed of the ball at maximum height? Ignore air resistance.

(a) 107.2 Ω

The voltage in the circuit is written in the form

[tex]V=V_0 sin(\omega t)[/tex]

where in this case

V0 = 35.0 V is the maximum voltage

[tex]\omega=100 rad/s[/tex]

is the angular frequency

Given the inductance: L = 165 mH = 16.5 H, the reactance of the inductance is:

[tex]X_L = \omega L=(100)(0.165)=16.5 \Omega[/tex]

Given the capacitance: [tex]C=99\mu F=99\cdot 10^{-6} F[/tex], the reactance of the capacitor is:

[tex]X_C = \frac{1}{\omega C}=\frac{1}{(100)(99\cdot 10^{-6})}=101.0 \Omega[/tex]

And given the resistance in the circuit, R = 66.0 Ω, we can now find the impedance of the circuit:

[tex]Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{66^2+(16.5-101.0)^2}=107.2\Omega[/tex]

(b) 0.33 A

The maximum current can be calculated by using Ohm's Law for RLC circuit. In fact, we know the maximum voltage:

V0 = 35.0 V

The equivalent of Ohm's law for an RLC circuit is

[tex]I=\frac{V}{Z}[/tex]

where

Z=107.2 Ω

is the impedance. Substituting these values into the formula, we find:

[tex]I=\frac{35.0}{107.2}=0.33 A[/tex]

(c) 100 rad/s

The equation for the current is

[tex]I=I_0 sin(\omega t-\Phi)[/tex]

where

I0 = 0.33 A is the maximum current, which we have calculated previously

[tex]\omega[/tex] is the angular frequency

[tex]\Phi[/tex] is the phase shift

In an RLC circuit, the voltage and the current have the same frequency. Therefore, we can say that

[tex]\omega=100 rad/s[/tex]

also for the current.

(d) -0.907 rad

Here we want to calculate the phase shift [tex]\Phi[/tex], which represents the phase shift of the current with respect to the voltage. This can be calculated by using the equation:

[tex]\Phi = tan^{-1}(\frac{X_L-X_C}{R})[/tex]

Substituting the values that we found in part a), we get

[tex]\Phi = tan^{-1}(\frac{16.5-101.0}{66})=-52^{\circ}[/tex]

And the sign is negative, since the capacitive reactance is larger than the inductive reactance in this case. Converting in radians,

[tex]\theta=-52 \cdot \frac{2\pi}{360}=-0.907 rad[/tex]

So the complete equation of the current would be

[tex]I=0.33sin(100t+0.907)[/tex]

The impedance of the circuit is 107.2 Ω.

The maximum current is 0.33 A.

The numerical value for ω in the equation i = Imax sin (ωt − ϕ). rad/s is 100 rad/s.

The numerical value for ϕ in the equation i = Imax sin (ωt − ϕ) is -0.907 rad.

1. To find the voltage, we already know that V0=35V

And the voltage in the circuit is V= V0sin(wt)

Hence, the angular frequency, w= 100 rad/s

Given the inductance is 16.5Ω and the capacitance is 101Ω.

And the resistance is 66Ω

Hence, the impedance of the circuit is [tex]\sqrt{66^{2} + (16.5-101)^2} =107.2[/tex]Ω

2. To find the maximum current, since Z= 107.2Ω,

Recall, I = V/Z

Put the values into the formula

I = 0.33A.

3. To find the numerical value for ω in the equation i = Imax sin (ωt − ϕ)
We recall that I0 = 0.33A

In an RLC circuit, the voltage and the current have the same frequency. Therefore, we can say that

w= 100 rad/s.

4. To find the numerical value for ϕ in the equation i = Imax sin (ωt − ϕ)

Because the sign is in the negative and since the capacitive reactance is larger than the inductive reactance in this case.

Converting in radians,

θ = -52. 2π/360

= -0.907rad.

Hence, the complete equation would be:

I = 0.33sin(100t + 0.907)

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Answer:

28.87 volt

Explanation:

Let the capacitance of the capacitor is C.

Energy, U = 3 J, V = 50 volt

U = 1/2 C V^2

3 = 0.5 x C x 50 x 50

C = 2.4  x 10^-3 F

Now U' = 1 J

U' = 1/2 C x V'^2

1 = 0.5 x 2.4 x 10^-3 x V'^2

V' = 28.87 volt

Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

Given that,

Mass of large ball = 3.0 kg

Mass of steel ball = 1.0 kg

Velocity = 3.0 kg

After collision,

Velocity = 2.0 m/s

Using conservation of momentum

[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}[/tex]

[tex]3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}[/tex]

[tex]-3i+2j=3.0\times v_{2}[/tex]

[tex]v_{2}=-i+0.66j[/tex]

The direction of the momentum

[tex]tan\theta=\dfrac{0.66}{-1}[/tex]

[tex]\theta=tan^{-1}\dfrac{0.66}{-1}[/tex]

[tex]\theta=-33.42^{\circ}[/tex]

The direction of the momentum with respect to east

[tex]\theta=180-33.42=146.58^{\circ}[/tex]

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Answer:

The combined speed of the woman and the skateboard is 6.67 m/s.

Explanation:

It is given that,

Mass of woman, m₁ = 60 kg

Speed of woman, v = 10 m/s

The woman jumps onto a giant skateboard of mass, m₂ = 30 kg

Let V is the combined speed of the woman and the skateboard. It can be calculated using the conservation of momentum as :

[tex]60\ kg\times 10\ m/s+30(0)=(60\ kg+30\ kg)V[/tex]

On solving the above equation we get V = 6.67 m/s

So, the combined speed of the woman and the skateboard is 6.67 m/s. Hence, this is the required solution.

After a minute of flight, the plane's altitude changes due to its climb. The speed at which the distance from the plane to the radar station is increasing equals the resultant of its horizontal and vertical speed components. This can be computed as approximately 1.26 km or 1260 meters.

The student's question pertains to both kinematics and trigonometry in Physics. In this scenario, the plane is climbing at an angle, while its horizontal speed is constant. The speed at which the distance from the plane to the radar station increases involves understanding the principle of vector addition and the concept of resultant velocity.

We can construct a right-angled triangle where one side is the horizontal speed component (= 150 km/h), the other side is the vertical speed component (altitude change over time, given by climbing speed = altitude/duration), and the hypotenuse is the resultant velocity, i.e., the speed at which the distance from the plane to the radar station is increasing.

After a minute, the altitude gains due to the climb is 3 km + (150 km/h * sin(30°) * 1/60 hr) = 3.0375 km, where sin(30°) represents the vertical ratio of the velocity. Radar station distance change can be calculated using Pythagoras theorem. In one minute, the plane travels horizontally by 150 km/h * 1/60 hr = 2.5 km. Thus, the change in distance is sqrt{(3.0375 km)^2 + (2.5 km)^2 } - 3 km (original altitude), which approximately equals 1.26 km or 1260 meters when rounded to the nearest whole number.

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The distance from the plane to the radar station is increasing at a rate of approximately 5 km/h one minute later.

Let's define the variables:

v = 150 km/h (speed of the plane)

h = 3 km (altitude of the radar station)

θ = 30° (angle of ascent)

We need to find the rate at which the distance from the plane to the radar station is increasing 1 minute (or 1/60 hours) after the plane passes over the station.

We'll use the following steps:

Determine the horizontal and vertical components of the plane's velocity:

v_x = v × cos(θ) = 150 km/h × cos(30°)

v_x = 150 × (√3 / 2) ≈ 129.9 km/h

v_y = v × sin(θ) = 150 km/h × sin(30°)

v_y = 150 ×0.5 = 75 km/h

d_x = v_x × (1/60) hours

d_x = 129.9 km/h × (1/60) = 2.165 km

New altitude, h_new = h + (v_y × (1/60))

h_new = 3 km + (75 × (1/60)) = 3 + 1.25 km = 4.25 km

Using the Pythagorean Theorem: d = sqrt(d_x² + h_new²)

d = square root of (2.165² + 4.25²)

d = square root of (4.687 + 18.06) = √22.75 = 4.77 km

The rate of distance increase is approximately 4.77 km/h rounded to the nearest whole number, which is 5 km/h

Answer:

A) Calculate the distance

To find the image distance for a diverging lens with a focal length of -30.0 cm and an object placed 18.0 cm in front, use the lens formula to calculate di = -77.14 cm, indicating a virtual image. The magnification equation yields a magnification of 4.285, indicating an upright image.

To calculate the image distance and magnification for a diverging lens, we can use the lens formula and the magnification equation. Given that a diverging lens has a focal length of -30.0 cm, and an object is placed 18.0 cm in front of this lens, we first need to use the lens formula:

1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance.

Substituting the values, we get:

1/(-30) = 1/18 + 1/di

Following the steps:

Solving the equation for 1/di gives us 1/di = 1/(-30) - 1/18.

Finding a common denominator and subtracting the fractions we get 1/di = (-2 - 5)/(-540), which simplifies to 1/di = -7/540.

Therefore, di = -540/7 = -77.14 cm.

The negative sign indicates that the image is virtual and located on the same side of the lens as the object.

For magnification (m), use the equation m = - di/do:

m = - (-77.14)/18

m = 77.14/18

m = 4.285

The positive magnification value indicates that the image is upright compared to the object.

Answer:

26.67 Volt

Explanation:

number of electrons, n = 5 x 10^21

R = 20 ohm

t = 10 min = 10 x 60 = 600 sec

V = ?

q = n x charge of one electron

q = n x e

q = 5 x 10^21 x 1.6 x 10^-19 = 800 C

i = q / t = 800 / 600 = 4 / 3 A

According to Ohm's law

V = i x R

V = 4 x 20 / 3 = 26.67 Volt

The voltage or potential difference across the resistor is 26.67 Volt.

Given the data in the question;

Ohm’s law states that the potential difference between two points is directly proportional to the current flowing through the resistance.

Hence

[tex]V = IR[/tex]

Where V is the voltage or potential difference, potential difference, I is the current and R is the resistance.

First we determine the total charge.

[tex]Q = n * e[/tex]

Where Q is the charge, n is the number of electrons and e is the charge on one electron. ( [tex]e = 1.6 * 10^{-19}C[/tex] )

Hence

[tex]Q = (5*10^{21}) * (1.6 * 10^{-19C)}\\\\Q = 800C[/tex]

Since current is the electric charge transferred per unit time.

[tex]I = \frac{Q}{t}[/tex]

Hence

[tex]I = \frac{800C}{600s}\\ \\I = 1.33A[/tex]

Now, we input our values into the above expression

[tex]V = I * R\\\\V = 1.33A * 20Ohms\\\\V = 26.67Volt[/tex]

Therefore, the voltage or potential difference across the resistor is 26.67 Volt.

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Answer:

i = 0.2 A

Explanation:

R = resistance of the resistor = 1000 Ω

V = Potential difference across the capacitor = Potential difference across the resistor = 200 V

i = initial current just after the capacitor is connected to resistor

Using ohm's law

[tex]i = \frac{V}{R}[/tex]

Inserting the values

[tex]i = \frac{200}{1000}[/tex]

i = 0.2 A

The initial current flowing through a 1000-Ω resistor when a 20.0-μF capacitor charged to 200 V is connected to it is 200 mA.

To find the initial current, we use Ohm's law, which states I = V / R, where I is the current, V is the voltage, and R is the resistance. At the moment the capacitor is connected to the resistor, the voltage across the resistor is the same as the voltage on the capacitor, so we have I = 200 V / 1000 Ω. The initial current is therefore 0.2 A, or 200 mA.

Answer:

The drift velocity of electrons in a copper wire is [tex]1.756\times10^{-4}\ m/s[/tex]

Explanation:

Given that,

Density [tex]\rho=8.93\ g/cm^3[/tex]

Mass [tex]M=63.5\ g/mol[/tex]

Radius = 0.625 mm

Current = 3 A

We need to calculate the drift velocity

Using formula of drift velocity

[tex]v_{d}=\dfrac{J}{ne}[/tex]

Where, n = number of electron

j = current density

We need to calculate the current density

Using formula of current density

[tex] J=\dfrac{I}{\pi r^2}[/tex]

[tex] J=\dfrac{3}{3.14\times(0.625\times10^{-3})^2}[/tex]

[tex] J=2.45\times10^{6}\ A/m^2[/tex]

Now, we calculate the number of electron

Using formula of number of electron

[tex]n=\dfrac{\rho}{M}N_{A}[/tex]

[tex]n=\dfrac{8.93\times10^{6}}{63.5}\times6.2\times10^{23}[/tex]

[tex]n=8.719\times10^{28}\ electron/m^3[/tex]

Now put the value of n and current density into the formula of drift velocity

[tex]v_{d}=\dfrac{2.45\times10^{6}}{8.719\times10^{28}\times1.6\times10^{-19}}[/tex]

[tex]v_{d}=1.756\times10^{-4}\ m/s[/tex]

Hence, The drift velocity of electrons in a copper wire is [tex]1.756\times10^{-4}\ m/s[/tex]

Answer:

The tension of the rope is T= 172.52 N

Explanation:

m= 21 kg

α= 21º

μ= 0.285

a= 2.09 m/s²

g= 9.81 m/s²

W= m*g

W=206.01 N

Fr= μ*W*cos(α)

Fr= 54.81 N

Fx= W * sin(α)

Fx= 73.82 N

T-Fr-Fx = m*a

T= m*a + Fr + Fx

T= 172.52 N

The magnitude of the force of motion of an object pulled by a rope up an inclined plane is the tension in the rope less the frictional forces as well as the component of the weight of the object acting along the plane

The tension in the rope is approximately 172.53 Newtons

Known:

The mass of the box, m = 21.0 kg

The angle of inclination of the ramp, θ = 21.0°

The coefficient of kinetic friction, μk= 0.285

The acceleration of the box up the ramp, a = 2.09 m/s²

The acceleration due to gravity, g = 9.81 m/s²

Required:

The tension in the rope, T

Solution;

The normal reaction of the box, N = m × g × cos(θ)

∴ N = 21.0 × 9.81 × cos(21.0°) ≈ 192.33

The normal reaction, N ≈ 192.33 N

The frictional force, [tex]F_f[/tex] = μ × N

∴ [tex]F_f[/tex] = 0.285 × 192.33 ≈ 54.81405

The frictional force, [tex]F_f[/tex] ≈ 54.81405 N

The force pulling the box, F = m×a = T - [tex]F_f[/tex] - The component of the weight acting along the plane

The component of the weight acting along the plane = m·g·sin(θ)

∴ m×a = T - [tex]F_f[/tex] - m·g·sin(θ)

T = m×a + [tex]F_f[/tex] + m·g·sin(θ)

Which gives;

T = 21.0 × 2.09 + 54.81405 + 21.0 × 9.81 × sin(21.0°) ≈ 172.53

The tension in the rope, T ≈ 172.53 N

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Answer:

1.2 x 10¹¹ kgm²/s

Explanation:

m = mass of the airplane = 39043.01

r = altitude of the airplane = 9.2 km = 9.2 x 1000 m = 9200 m

v = speed of airplane = 335 m/s

L = Angular momentum of airplane

Angular momentum of airplane is given as

L = m v r

Inserting the values

L = (39043.01 ) (335) (9200)

L =  (39043.01 ) (3082000)

L = 1.2 x 10¹¹ kgm²/s

The final image position is 4.34 cm to the right of the second lens, and the final image height is 2.4 cm.

First, we will consider the object in front of the first lens (with a focal length of 40 cm). The object is placed 14 cm in front of this lens.

For the first lens, the lens equation is given by:

[tex]\[\frac{1}{f_1} = \frac{1}{d_o} + \frac{1}{d_i}\][/tex]

where [tex]\(f_1\)[/tex] is the focal length of the first lens, [tex]\(d_o\)[/tex] is the object distance from the first lens, and [tex]\(d_i\)[/tex] is the image distance from the first lens.

Rearranging the equation to solve for [tex]\(d_i\)[/tex], we get:

[tex]\[d_i = \frac{f_1 \cdot d_o}{d_o - f_1}\][/tex]

Plugging in the values:

[tex]\[d_i = \frac{40 \cdot 14}{14 - 40}\] \[d_i = \frac{560}{-26}\] \[d_i = -21.54 \text{ cm}\][/tex]

This means the image formed by the first lens is 21.54 cm to the left of the first lens (virtual image).

Next, we calculate the height of this image using the magnification equation:

[tex]\[m_1 = -\frac{d_i}{d_o}\] \[m_1 = -\frac{-21.54}{14}\] \[m_1 = 1.54\][/tex]

The height of the image formed by the first lens is:

[tex]\[h_i = m_1 \cdot h_o\][/tex]

where [tex]\(h_o\)[/tex] is the object height (2.0 cm).

So, [tex]\(h_i = 1.54 \cdot 2.0\)[/tex]

[tex]\[h_i = 3.08 \text{ cm}\][/tex]

Now, this image becomes the object for the second lens (with a focal length of 20 cm). The distance between the two lenses is 16 cm, so the object distance for the second lens is the image distance from the first lens plus 16 cm.

For the second lens, the object distance [tex]\(d_o'\)[/tex] is:

[tex]\[d_o' = -21.54 + 16\] \[d_o' = -5.54 \text{ cm}\][/tex]

Using the lens equation for the second lens:

[tex]\[\frac{1}{f_2} = \frac{1}{d_o'} + \frac{1}{d_i'}\][/tex]

where [tex]\(f_2\)[/tex] is the focal length of the second lens, [tex]\(d_o'\)[/tex] is the object distance from the second lens, and [tex]\(d_i'\)[/tex] is the image distance from the second lens.

Rearranging the equation to solve for [tex]\(d_i'\)[/tex], we get:

[tex]\[d_i' = \frac{f_2 \cdot d_o'}{d_o' - f_2}\][/tex]

Plugging in the values:

[tex]\[d_i' = \frac{20 \cdot (-5.54)}{-5.54 - 20}\] \[d_i' = \frac{-110.8}{-25.54}\] \[d_i' = 4.34 \text{ cm}\][/tex]

This means the final image is 4.34 cm to the right of the second lens (real image).

To find the height of the final image, we use the magnification equation for the second lens:

[tex]\[m_2 = -\frac{d_i'}{d_o'}\] \[m_2 = -\frac{4.34}{-5.54}\] \[m_2 = 0.78\][/tex]

The height of the final image is:

[tex]\[h_i' = m_2 \cdot h_i\] \[h_i' = 0.78 \cdot 3.08\] \[h_i' = 2.4 \text{ cm}\][/tex]

Answer:

1. Yes, it can occur adiabatically.

2. The work required is: 86.4kJ

Explanation:

1. The internal energy of a gas is just function of its temperature, and the temperature changes between the states, so, the internal energy must change, but how could it be possible without heat transfer? This process may occur adiabatically due to the energy balance:

[tex]U_{2}-U_{1}=W[/tex]

This balance tell us that the internal energy changes may occur due to work that, in this case, si done over the system.

2. An internal energy change of a gas may be calculated as:

[tex]du=C_{v}dT[/tex]

Assuming [tex]C_{v}[/tex] constant,

[tex]U_{2}-U_{1}=W=m*C_{v}(T_{2}-T_{1})[/tex]

[tex]W=0.72*1*(420-300)=86.4kJ[/tex]

Answer:

1.16

Explanation:

Let the angle is theta.

tan θ = perpendicular / base

tan θ = 5.9 / 5.1  = 1.16

Answer:

a) Force must be applied to the input piston = 10N

b) The input piston moved by 7.5 m

Explanation:

a) For a hydraulic lift we have equation

                  [tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]

   F₁ = ?

   d₁ = 10 cm = 0.1 m

   F₂ = 250 N

   d₂ = 50 cm = 0.5 m

Substituting

   [tex]\frac{F_1}{\frac{\pi \times 0.1^2}{4}}=\frac{250}{\frac{\pi \times 0.5^2}{4}}\\\\F_1=10N[/tex]

Force must be applied to the input piston = 10N

b) We have volume of air compressed is same in both input and output.    

   That is        A₁x₁ = A₂x₂

   A is area and x is the distance moved

   x₂ = 0.3 m

   Substituting

             [tex]\frac{\pi \times 0.1^2}{4}\times x_1=\frac{\pi \times 0.5^2}{4}\times 0.3\\\\x_1=7.5m[/tex]

The input piston moved by 7.5 m

Answer:

376.9911ft²/minute

Explanation:

In the given question the rate of chage of radius in given as

[tex]\frac{\mathrm{d}r }{\mathrm{d} t}[/tex]=5ft per minute

we know ares of circle A=pi r^{2}

differentiating w.r.t. t we get

[tex]\frac{\mathrm{d} A}{\mathrm{d} t}=2\pi r\frac{\mathrm{d}r }{\mathrm{d} t}[/tex]

Now, we have find [tex]\frac{\mathrm{d}A }{\mathrm{d} t} at r=12 feet[/tex]

[tex]\frac{\mathrm{d} A}{\mathrm{d} t}=2\times\pi\times12\times5=120\pi=376.9911ft^{2}/minute[/tex]

The rate at which the area of the ripple is changing when the radius is 12 feet is 120π square feet per minute.

The concept in question is related to mathematical calculus and the principle of rates. The area of a circle is represented by the formula A = πr², with 'A' representing the area and 'r' the radius. To determine how the area changes with changes in the radius, we differentiate this function with respect to time, resulting in dA/dt = 2πr (dr/dt). We know the radius is increasing at a rate of 5 feet per minute (dr/dt = 5 ft/min). At the instant when the radius is 12 feet, we simply substitute into our differentiated equation to find that dA/dt = 2π(12ft)(5ft/min) = 120π ft²/min.

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Answer:

Current, I = 2.45 T

Explanation:

It is given that,

Magnetic field, B = 0.92 T

Length of wire, l = 2.6 m

Mass, m = 0.6 kg

We need to find the minimum current needed to levitate the wire. It is given by balancing its weight to the magnetic force i.e.

[tex]Ilb=mg[/tex]

[tex]I=\dfrac{mg}{lB}[/tex]

[tex]I=\dfrac{0.6\ kg\times 9.8\ m/s^2}{2.6\ m\times 0.92\ T}[/tex]

I = 2.45 A

So, the minimum current to levitate the wire is 2.45 T. Hence, this is the required solution.

The minimum current needed to levitate the wire in the given magnetic field is approximately 2.58 Amps, determined by setting the magnetic force acting on the wire equal to the gravitational force and solving for current.

The minimum current necessary to levitate the wire in a uniform magnetic field can be determined by equating the magnetic force acting on the wire to the gravitational force acting on it. The magnetic force exerted on a current-carrying wire in a magnetic field is given by F = IℓBsinθ, where F is the force, I is the current, ℓ is the length of the wire, B is the magnetic field strength, and θ is the angle between the current and the magnetic field. Given the wire is levitating, the angle θ is 90°, meaning sinθ is 1. Additionally, the gravitational force is F = mg, where m is the mass of the wire and g is the acceleration due to gravity. Setting the magnetic force equal to the gravitational force gives IℓB = mg, which we can solve for I to get I = mg/(ℓB). Using the given values, I = (0.60 kg * 9.8 m/s²) / (2.6 m * 0.92 T) = 2.58 A. So, the minimum current needed to levitate the wire is approximately 2.58 Amps.

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Answer:

Work done, W = −287500 Joules

Explanation:

It is given that,

Mass of the car, m = 920 kg

The car is travelling at a speed of, u = 90 km/h = 25 m/s

We need to find the amount of work must be done to stop this car. The final velocity of the car, v = 0

Work done is also defined as the change in kinetic energy of an object i.e.

[tex]W=\Delta K[/tex]

[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]

[tex]W=\dfrac{1}{2}\times 920\ kg(0-(25\ m/s)^2)[/tex]

W = −287500 Joules

Negative sign shows the work done is done is opposite direction.

Answer:

0.31 m

Explanation:

m = mass of the block = 1.5 kg

H = height from which the block is released on ramp = 0.81 m

k = spring constant of the spring = 250 N/m

x = maximum compression of the spring

using conservation of energy

Spring potential energy gained by spring = Potential energy lost by block

(0.5) k x² = mgH

(0.5) (250) x² = (1.5) (9.8) (0.81)

x = 0.31 m

Final answer:

The maximum compression of the spring, when a block of mass 1.5 kg is released from rest at a height of 0.81 m on a frictionless ramp onto a spring with spring constant 250.0 N/m, is approximately 30.7 cm.

Explanation:

To determine the maximum compression of the spring, x, we use the conservation of energy principle. The gravitational potential energy (PE) at the height H will convert to the elastic potential energy (spring PE) in the spring when the block of mass m compresses the spring at the foot of the ramp.

Initially, the block has gravitational potential energy given by PE = mgh, where g = 9.81 m/s2 is the acceleration due to gravity. At the point of maximum compression, all this energy will be stored as elastic potential energy in the spring given by spring PE = ½kx2.

Setting the initial PE equal to the spring PE and solving for x, we get:

mgh = ½kx2

1.5 kg × 9.81 m/s2 × 0.81 m = ½ × 250.0 N/m × x2

x2 = (1.5 kg × 9.81 m/s2 × 0.81 m) / (0.5 × 250.0 N/m)

x2 = (11.7915 kg·m2/s2) / (125.0 N/m)

x2 = 0.094332 m2

x = √0.094332 m2

x ≈ 0.307 m or 30.7 cm

Therefore, the maximum compression of the spring is approximately 30.7 cm.

Answer:

New speed of water is 2.8 m/s.

Explanation:

It is given that,

Speed of water, v₁ = 2.3 m/s

Diameter of pipe, d₁ = 3.2 cm

Radius of pipe, r₁ = 1.6 cm = 0.016 m

Area of pipe, [tex]A_1=\pi(0.016)^2=0.000804\ m^2[/tex]

If the pipe narrows its diameter, d₂ = 2.9 cm

Radius, r₂ = 0.0145 m

Area of pipe, [tex]A_2=\pi(0.0145)^2=0.00066\ m^2[/tex]

We need to find the new speed of the water. It can be calculated using equation of continuity as :

[tex]v_1A_1=v_2A_2[/tex]

[tex]v_2=\dfrac{v_1A_1}{A_2}[/tex]

[tex]v_2=\dfrac{2.3\ m\times 0.000804\ m^2}{0.00066\ m^2}[/tex]

[tex]v_2=2.8\ m/s[/tex]

So, the new speed of the water is 2.8 m/s. Hence, this is the required solution.

Answer:

318.5 x 10^4 Pa

Explanation:

weight of woman = m g = 65 x 9.8 = 637 N

Area of both the heels = 1 x 2 = 2 cm^2 = 2 x 10^-4 m^2

Pressure is defined as the thrust acting per unit area.

P = F / A

Where, F is the weight of the woman and A be the area of heels

P = 637 / (2 x 10^-4) = 318.5 x 10^4 Pa

Answer:

So recoil speed of the Earth will be

[tex]v = 2.25 \times 10^{-24} m/s[/tex]

Explanation:

Here if we assume that during collision if ball will lose very small amount of energy and rebound with same speed

then the impulse given by the ball is

[tex]Impulse = m(v_f - v_i)[/tex]

[tex]Impulse = (0.155)(43.6 - (-43.6))[/tex]

[tex]Impulse = 13.52 Ns[/tex]

so impulse received by the Earth is same as the impulse given by the ball

so here we will have

[tex]Impulse = mv[/tex]

[tex]13.52 = (6 \times 10^{24})v[/tex]

[tex]v = 2.25 \times 10^{-24} m/s[/tex]

The recoil speed of the Earth is [tex]2.25*10^-24 m/s.[/tex]

Collision can be regarded as the forceful coming together of bodies.

The impulse of the ball can be calculated as;

[tex]Impulse= MV[/tex]

where V=( V1 -Vo)

V1= final velocity=(43m/s)

V0= initial velocity= (-43.6m/s)

m= mass of baseball

Hence,[tex]V= [43.6-(-43.6)]= 87.2m/s[/tex]

Then Impulse given by the ball=[tex]( 0.155*87.2)= 13.53Ns.[/tex]

We can now calculate the recoil speed of the  earth as;

[tex]Impluse= MVV= Impulse/ mass = 13.52/(6*10^24)[/tex]

=2.25*10^-24 m/s

Therefore, the recoil speed of the Earth is 2.25*10^-24 m/s

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Final answer:

Hyperopia, also known as farsightedness, is the condition where close objects are blurry while distant objects are clearly visible. It is corrected with converging lenses. This is the opposite of myopia, which affects distance vision.

Explanation:

The ability to clearly see objects at a distance but not close up is properly called hyperopia, which is a vision problem where close objects are out of focus but distant vision is unaffected; also known as farsightedness. Farsighted individuals have difficulty seeing close objects clearly because their eyes do not converge light rays from a close object enough to make them meet on the retina. To correct this vision problem, converging lenses are used, which help to increase the power of the eyes, allowing for clear vision of nearby objects. This contrasts with myopia (nearsightedness), where distant objects are out of focus while close vision is typically unaffected.

Answer:

efficiency = 5.4%

Explanation:

Efficiency of heat engine is given as

[tex]\eta = \frac{W}{Q_{in}}[/tex]

now we will have

[tex]W = Q_1 - Q_2[/tex]

so we will have

[tex]\eta = 1 - \frac{Q_2}{Q_1}[/tex]

now we know that

[tex]\frac{Q_2}{Q_1} = \frac{T_2}{T_1}[/tex]

so we have

[tex]\eta = 1 - \frac{T_2}{T_1}[/tex]

[tex]\eta = 1 - \frac{273+6}{273+22}[/tex]

[tex]\eta = 0.054[/tex]

so efficiency is 5.4%

The vertical height between the climbers can be calculated from the projectile motion's kinematic formula: h = (v₀² sin²θ) / 2g. Here, the first aid kit has reached its trajectory's peak when caught, and its vertical velocity component is zero, making it ideal to implement the equation.

In your question, you want to know the vertical height between the two climbers. This is a question in projectile motion in physics. When an object is thrown with an initial velocity at an angle, it creates a trajectory as it travels. When the climber above catches the first aid kit, it's mentioned that it is moving horizontally, which means its vertical speed is zero at that very moment. Hence the first aid kit has reached the maximum height of its trajectory or the peak where the vertical component of the velocity is zero.

We can calculate the vertical distance or height (h) using a formula from kinematics. The formula is: h = (v₀² sin²θ) / 2g. Where v₀ is the initial velocity, θ is the angle of elevation, and g is the acceleration due to gravity (which is ~9.8 m/s² on Earth). Plugging in the values we get: h = (5.70 m/s)² × sin²(45.7 °) / (2 * 9.8 m/s²). Solve this to get the height.

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The vertical height between the two climbers is approximately 0.8616 meters.

The vertical height between the two climbers is given by the equation for the vertical displacement in projectile motion:

[tex]\[ h = \frac{{v_{0y}^2}}{{2g}} \][/tex]

First, we need to find the initial vertical component of the velocity[tex](\( v_{0y} \))[/tex]. Since the initial velocity[tex](\( v_0 \))[/tex] is given as 5.70 m/s at an angle of 45.7 we can use the sine function to find [tex]\( v_{0y} \):[/tex]

[tex]\[ v_{0y} = v_0 \sin(\theta) \][/tex]

[tex]\[ v_{0y} = 5.70 \, m/s \times \sin(45.7) \][/tex]

Now, we calculate the sine of 45.7 and multiply it by the initial velocity:

[tex]\[ v_{0y} = 5.70 \, m/s \times 0.721 \][/tex]

[tex]\[ v_{0y} = 4.1117 \, m/s \][/tex]

Next, we use the equation for vertical displacement to find the height h :

[tex]\[ h = \frac{{(4.1117 \, m/s)^2}}{{2 \times 9.81 \, m/s^2}} \][/tex]

[tex]\[ h = \frac{{16.9046 \, m^2/s^2}}{{19.62 \, m/s^2}} \][/tex]

[tex]\[ h = 0.8616 \, m \][/tex]

Answer:

Charge, [tex]q=2.98\times 10^{-8}\ C[/tex]

Explanation:

It is given that,

Value of electric field, [tex]E=7.5\times 10^5\ V/m[/tex]

Area of parallel plates, [tex]A=45\ cm^2=0.0045\ m^2[/tex]

Distance between two parallel plates, d = 2.45 mm = 0.00245 m

For a parallel plate capacitor, the capacitance is given by :

[tex]C=\dfrac{\epsilon_oA}{d}[/tex].......(1)

Since, [tex]E=\dfrac{V}{d}[/tex]

V = E . d ............(2)

And [tex]C=\dfrac{q}{V}[/tex].....(3)

From equation (1), (2) and (3) we get :

[tex]\dfrac{q}{V}=\dfrac{\epsilon_oA}{d}[/tex]

[tex]q=\epsilon_o EA[/tex]

[tex]q=8.85\times 10^{-12}\ F/m\times 7.5\times 10^5\ V/m\times 0.0045\ m^2[/tex]

[tex]q=2.98\times 10^{-8}\ C[/tex]

So, the charge on the each plate is [tex]2.98\times 10^{-8}\ C[/tex]. Hence, this is the required solution.

To determine the charge on each plate for a desired electric field in a capacitor with parallel plates, use the formula Q = E x A x ε_0. Convert the area to m^2, insert all values including the permittivity of free space, and solve for Q.

To find the charge required on each of the parallel plates to create a specific electric field, you can use the relationship between the electric field (E), the charge (Q) on the plates, and the permittivity of free space (ε0). The electric field between two parallel plates is given by the equation:

E = Q / (ε0 × A)

where E is the electric field, Q is the charge on each plate, A is the area of the plates, and ε0 is the permittivity of free space (ε0 = 8.854 x 10−12 C2/N·m2). Rearranging this for Q gives:

Q = E × A × ε0

First, convert the area from cm2 to m2 by multiplying by (10−4)2. Then, plug in the values:

Q = (7.50 × 105 V/m) × (0.45 × 10−4 m2) × (8.854 × 10−12 C2/N·m2)

Calculate Q to find the charge required on each plate.

Answer:

Heat generated in the rod is 8.33 watts.

Explanation:

It is given that,

Length of rod, l = 2 m

Area of cross section, [tex]A=2\ mm\times 2\ mm=4\ mm^2=4\times 10^{-6}\ m^2[/tex]

Resistivity of rod, [tex]\rho=6\times 10^{-8}\ \Omega-m[/tex]

Potential difference, V = 0.5 V

The value of resistance is given by :

[tex]R=\rho\dfrac{l}{A}[/tex]

[tex]R=6\times 10^{-8}\ \Omega-m\times \dfrac{2\ m}{4\times 10^{-6}\ m^2}[/tex]

R = 0.03 ohms

Let H is the rate at which heat is generated in the rod . It is given by :

[tex]\dfrac{H}{t}=I^2R[/tex]

Since, [tex]I=\dfrac{V}{R}[/tex]

[tex]\dfrac{H}{t}=\dfrac{V^2}{R}[/tex]

[tex]\dfrac{H}{t}=\dfrac{(0.5)^2}{0.03}[/tex]

[tex]\dfrac{H}{t}=8.33\ watts[/tex]

So, the at which heat is generated in the rod is 8.33 watts. Hence, this is the required solution.

Answer:

(a) 4323.67 N

(b) 421.13 degree

Explanation:

A = 2580 N 36 degree South of east

B = 2270 N due South

A = 2580 ( Cos 36 i - Sin 36 j) = 2087.26 i - 1516.49 j

B = 2270 (-j) = - 2270 j

A + B = 2087.26 i - 1516.49 j - 2270 j

A + B = 2087.26 i - 3786.49 j

(a) Magnitude of A + B = [tex]\sqrt{2087.26^{2}+(-3786.49)^{2}}[/tex]

Magnitude of A + B = 4323.67 N

(b) Direction of A + B

Tan theta = - 3786.49 / 2087.26

theta = - 61.13 degree

Angle from ast = 360 - 61.13 = 421.13 degree.

To determine the magnitude and direction of the resultant force acting on a tree stump when two forces are applied, one must resolve the forces into components, sum the components, then apply the Pythagorean theorem for magnitude and inverse tangent for direction.

We need to resolve each force into its horizontal (east-west) and vertical (north-south) components. For Force A, we can use trigonometry to find the components:

Force B is already pointing due south, so its components are:

The net force components are:

With these components, the magnitude of the resultant force can be calculated using the Pythagorean theorem:

Fnet = √(Fnet,x² + Fnet,y²)

To find the direction of the resultant force relative to due east, we can use the inverse tangent function (tan-1):

θ = tan-1(Fnet,y / Fnet,x)

This angle should be positive, as we are measuring it with respect to due east.

Answer:

1.33 x 10^26 photons

Explanation:

λ = 12 cm = 0.12 m

Δ T = 5 degree C

m = 11 g = 0.011 kg

c = 4 J/g K = 4000 J / kg K

Let the number of photons are n.

Energy of one photon = h c / λ

Where, h is the Plank's constant and c be the velocity of light.

Energy of n photons, E = n h c / λ

This energy is used to raise the temperature of eye, so

n h c / λ = m c Δ T

n = m c Δ T λ / h c

n = (0.011 x 4000 x 5 x 0.12) / (6.63 x 10^-34 x 3 x 10^8)

n = 1.33 x 10^26

Thus, there are 1.33 x 10^26 photons.

To estimate the number of photons that must be absorbed to raise the temperature of an eye by 5.00 °C, calculate the energy required and convert it into the number of photons using Planck's constant and the frequency of the microwaves.

To estimate the number of photons that must be absorbed to raise the temperature of an eye by 5.00 °C, we need to calculate the amount of energy required and then convert it into the number of photons. Here are the steps:

Therefore, approximately 3.30 × 10^11 photons with a wavelength of 12 cm must be absorbed to raise the temperature of an eye by 5.00 °C.

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Answer:

2.12 rpm

Explanation:

[tex]a_c= Centripetal\ acceleration=3.00\ m/s^2\\r=radius=\frac {120}{2}=60\ m\\a_c=\frac {v^2}{r}\\\Rightarrow v^2=a_c\times r\\\Rightarrow v^2=3\times 60=180\\\Rightarrow v=\sqrt{180}=13.41\ m/s\\[/tex]

[tex]v=\omega r\\\Rightarrow \omega=\frac {v}{r}\\\Rightarrow \omega=\frac {13.41}{60}\\\Rightarrow \omega=0.2236\ rad/s\\1\ rad/s=9.55\ rpm\\\Rightarrow 0.2236\ rad/s=2.12\ rpm\\\therefore Wheel\ rotations=2.12\ rpm[/tex]

Final answer:

To generate an artificial gravity of 3.00 m/s² on a space station with a 120 m diameter, it must rotate at approximately 2.14 revolutions per minute.

Explanation:

To find the rate of the wheel's rotation in revolutions per minute (rpm) that will provide an artificial gravity of 3.00 m/s² for a space station with a diameter of 120 m, we use the formula for centripetal acceleration:

a = ω² × r

where a is the centripetal acceleration, ω is the angular velocity in radians per second, and r is the radius of the circle.

Since we are looking for ω in revolutions per minute and not radians per second, we will need to convert our solution. First, we find ω in radians per second by rearranging the formula:

ω = √(a/r)

Given a = 3.00 m/s² and the radius r = 60 m (half the diameter),

ω = √(3.00 m/s² / 60 m) = √0.05 s²/m = 0.224 rad/s

One revolution is 2π radians and there are 60 seconds in a minute, so we can find the number of revolutions per minute by

rev/min = ω × (60 s/min) / (2π rad/rev)

Plugging in our value of ω:

rev/min = 0.224 rad/s × (60 s/min) / (2π rad/rev) ≈ 2.14 rev/min

Therefore, the space station must rotate at approximately 2.14 revolutions per minute to produce the desired artificial gravity.