You are riding a ferris wheel while sitting on a scale. A ferris wheel with radius 9.7 m and a period 32s. Find the scale reading for a 60kg person at the bottom of the ferris wheel and the top of the ferris wheel, assuming it moves at a constant rate.

Answer and Explanation:

The answer is attached below

The velocity of both the objects after the collision is [tex]\frac{25}{11} m/s[/tex].

(a) the relative velocity of the masses is zero.

(b) the compression of the spring is 0.185m.

The given case is an example of an inelastic collision in which two objects after the collision move together. The momentum of the system is conserved, therefore,

[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]

here, m₁ = 5kg , m₂ = 6kg

u₁ = 5 m/s , initial velocity of mass m₁, and

u₂= 0, initial speed of the mass m₂

[tex]5\times5+0=(5+6)v\\\\v=\frac{25}{11}m/s[/tex]

(a) When the spring is fully compressed both the masses move with the same velocity, therefore the relative speed of the masses is zero.

(b) from the law of conservation of energy:

the initial kinetic energy of the masses is converted into final kinetic energies and the potential energy of the spring:

[tex]\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u^2_2=\frac{1}{2} (m_1+m_2)v^2+\frac{1}{2}kx^2[/tex]

where x is the compression of the spring

[tex]\frac{1}{2}\times5\times5^2+0=\frac{1}{2}(5+6)\times(\frac{25}{11})^2=\frac{1}{2}\times2000x^2\\\\x=0.185m[/tex]

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Answer:

Explanation:

Total charge passed

= 1.25 x 60 x 60 x 5.1 C

= 22950 C

Equivalent mass of copper

= 63.5 / 2

= 31.75 g

96500 coulomb is required to obtain 31.75 g of copper

22950 C will release (31.75 / 96500) x 22950

= 7.55 g of copper .

Answer:

Ф = 4.98 °

Explanation:

Given:

R = 1.75 km = 1.75 * 10^3

g = 9.8 m/s^2

v = 120 km/h ==> 33.33 m/s

required:

Ф

solution:

The angle(Ф) is given by

Ф = arctan(v^2/R*g)

   = arctan(33.33^2/9.8*1.75 * 10^3)

   = 4.98 °

note:

there maybe error in calculation but method is correct.

Answer:

The number of polarizer needed so transmitted light has at least 12% intensity = 17

Explanation:

Given :

Angle between incident light and optic axis of polarizer = 20°

Given that, the transmission axis of each additional analyzer is rotated 20° relative to the transmission axis of the previous one

According to the malus law,

The intensity of the transmitted light passes through the polarizer is proportional to the square of the cosine of angle between the transmission axis to the optic axis.

⇒  [tex]I = I_{o} cos^{2} \alpha[/tex]

Where, [tex]I =[/tex] transmitted intensity through polarizer, [tex]I_{o} =[/tex] incident intensity of the light.

Given in question, all the time [tex]\alpha =[/tex] 20°

By calculation ∴ [tex]cos^{2} 20 = 0.883[/tex]

After 1st polarizer,

∴ [tex]I_{1} = 0.883I_{o}[/tex]

Now we need to multiply all the time 0.883 until we get 0.12 (relative 20° angle given in question)

After 17th polarizer we get 0.1205 ≅ 0.12

[tex]I_{17} = 0.883^{17} = 0.1205 \times 100 = 12[/tex]% [tex]I_{o}[/tex]

Means we get 12% intensity after 17th polarizing disk.

Final answer:

The correct statement about a wheel with four forces of the same magnitude but different directions acting tangentially at its rim is that both the net force on the wheel and the net torque about the CM are zero.

Explanation:

When four forces of the same magnitude but differing directions act tangentially at the rim of a uniform wheel free to spin about its center, the net force acting on the wheel may be zero if the forces cancel each other out. However, the direction and point of application of these forces are critical in determining the net torque. If the forces are applied in such a way that they create rotational effects that cancel each other, the net torque about the center of mass (CM) will also be zero. Therefore, the correct statement about the wheel is:

D. The net force on the wheel and the net torque on the wheel about the CM are zero.

To put it simply, the forces are arranged in a way that causes them to balance out, leading to no linear acceleration (net force is zero), and they are positioned symmetrically around the center of mass, producing no rotational acceleration (net torque is zero).

Answer:

[tex]L=4.8*10^{17}m[/tex]

Explanation:

Given data

Power P=4.00×10²W

Radius r=6.5×10⁴m

Voltage V=120V

To find

Length of wire L

Solution

We know that resistance of wire can be obtained from

[tex]P=\frac{V_{2}}{R}\\ R=\frac{V_{2}}{P}[/tex]

We also know that R=pL/A solving the length noting that A=πr²

and using p=100×10⁻⁸Ω.m we find that

So

[tex]L=\frac{RA}{p}\\ L=\frac{\frac{(V^{2})}{P}(\pi r^{2}) }{p} \\L=\frac{V^{2}(\pi r^{2})}{pP}\\ L=\frac{(120V)^{2}\pi (6.5*10^{4} m)^{2} }{100*10^{-8}(4.00*10^{2} W) }\\ L=4.8*10^{17}m[/tex]

Final answer:

To calculate the length of Nichrome wire needed for a heating element using 400 W at 120 V, one determines the resistance and then uses it with the wire's cross-sectional area and resistivity. Approximately 41.45 meters of wire is required.

Explanation:

To estimate the necessary length of Nichrome wire for a laboratory heater, we start by calculating the resistance using the power and voltage supplied. The formula for power (P) in terms of voltage (V) and resistance (R) is P = V2 / R. Given P = 400 W and V = 120 V, the resistance can be calculated as follows:

R = V2 / P = (1202) / 400 = 36 Ω.

Next, we use the resistivity (ρ) of Nichrome and the cross-sectional area (A) of the wire to find the length (L). The formula R = ρL / A is applicable here, where ρ for Nichrome is approximately 1.10×10-6 Ω·m and A = πr2. The radius (r) given is 6.5×10-4 m, so:

A = π(6.5×10-4)2 = 1.33×10-6 m2.

Substituting the values into the formula, we get:

L = (R · A) / ρ = (36 · 1.33×10-6) / (1.10×10-6) ≠ 41.45 m.

Thus, approximately 41.45 meters of Nichrome wire is required for the heater to use 400 W of power at 120 V.

Answer:

58.8 N

Explanation:

Let 'F₁' be the force by front arm and 'F₂' be the force by back arm.

Given:

Mass of the rod (m) = 3.00 kg

Length of the pole (L) = 4.50 m

Acceleration due to gravity (g) = 9.8 m/s²

Distance of 'F₁' from one end of pole (d₁) = 0.750 m

'F₂' acts on the end. So, distance between 'F₁' and 'F₂' = 0.750 m

Now, weight of the pole acts at the center of pole.

Now, distance of center of pole from 'F₁' is given as:

d₂ = (L ÷ 2) - d₁

[tex]d_2=\frac{4.50}{2}-0.75=1.5\ m[/tex]

Now, as the pole is held horizontally straight, the moment about the point of application of force 'F₁' is zero for equilibrium of the pole.

So, Anticlockwise moment  = clockwise moment

[tex]F_2\times d_1=mg\times d_2\\\\F_2=\frac{mg\times d_2}{d_1}[/tex]

Plug in the given values and solve for 'F₂'. This gives,

[tex]F_2=\frac{3.00\ kg\times 9.8\ m/s^2\times 1.5\ m}{0.75\ m}\\\\F_2=\frac{44.1}{0.75}=58.8\ N[/tex]

Therefore, the force exerted by the back arm on the pole is 58.8 N vertically down.

Answer:

Explanation:

a ) The speed of light will be 3 x 10⁸ m /s

b ) Speed of light will again be  3 x 10⁸ m /s . for an observer on the ground. It is so because speed of light is independent of moving frame of reference . It is absolute . It can not be changed by changing frame of reference .

The behavior of light differs because the speed of bullet will be less in case of stationary train . For moving train , speed of bullet will be increased by amount equal to speed of train for an observer on the ground.

The velocity at which the long jumper landed can be computed by separating the jump into horizontal and vertical components, using the initial take-off angle and the horizontal distance. The horizontal and vertical velocities are combined to find the magnitude of the total landing velocity, and the arctangent of their ratio gives the direction.

To determine the velocity at which the long jumper landed, we need to consider the projectile motion of the jump. There are two components of velocity to consider: the horizontal (vx) and the vertical (vy) at the point of landing.

First, the horizontal velocity (vx) can be found by dividing the total horizontal distance by the time of flight (t). The equation of horizontal motion is:

vx = d / t, where d is the horizontal distance.

The vertical velocity (vy) of the jumper when he lands will be the same magnitude but opposite in direction to the vertical velocity at take-off due to symmetry in projectile motion, assuming no air resistance. The vertical velocity at take-off can be calculated using the initial jump angle (Ө) and the initial velocity (v0).

Using the initial angle of 30°, the vertical component of the initial velocity at take-off (v0y) is:

v0y = v0 * sin(Ө).

Since the vertical motion is subject only to acceleration due to gravity (g), the final vertical velocity at landing (vy) will be v0y (but in the opposite direction).

The total landing velocity (v) is then found by combining these two components using the Pythagorean theorem:

v = √(vx^2 + vy^2).

The direction of the landing velocity is given by the angle made with the horizontal, found using the arctangent of the ratio of vy over vx:

θ = arctan(vy / vx).

These calculations would provide the magnitude and direction of the landing velocity.

Answer:

The Reasons for which petroleum is the chosen fuel for transportation in the U.S. include number 1) and 2).

Explanation:

The main reasons why oil is the most used fuel in the USA is because of its energy potential with a high combustion power and a better octane rating compared to other fuels. Thus, fossil fuel derived from oil has a high energy value per unit volume and an ability to quickly start or stop providing energy.

Answer:

Explanation:

This does not violate Newton's 1st law because the net force would still be 0 in order to produce uniform motion (aka constant velocity). The other forces acting on the vehicles is air resistance which is non-zero. So we need car internal force to counter balance this force, which require extra gas for the car.

Applying extra gas to a car moving at a constant velocity does not violate Newton's 1st Law as the additional force is required to counteract other forces, such as air resistance and road friction, allowing the car to maintain its velocity. This is, in fact, a confirmation of Newton's 1st Law.

Even while driving at a constant velocity on a straight highway, one must occasionally apply a bit of extra gas, which means applying an extra external force.

However, this does not violate Newton's 1st Law, or the law of inertia, which states that an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

The reason for this is the existence of other forces acting on the car that work against the forward motion. These can include air resistance (or drag), friction from the road, and uphill forces if the road is inclined.

The extra force (gas) provided helps overcome these forces, allowing the car to maintain a constant velocity. So, this activity is actually a confirmation of Newton's first Law, not a violation.

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Explanation:

Formula for path difference is as follows.

             x = 2tn

and,     refractive index (n) = [tex]\frac{\lambda}{2t}[/tex]

Thickness is calculated as follows.

           Thickness (t) = [tex]\frac{volume}{area}[/tex]

       Area = [tex]\pi r^{2}[/tex]

                = [tex]\pi \times (\frac{d}{2})^{2}[/tex]

                = [tex]\frac{0.49 \times 10^{-6}}{3.14 \times 0.01 m}[/tex]

                = [tex]1.56 \times 10^{-8}[/tex] m

Now, the refractive index will be calculated as follows.

For drop,      n = [tex]\frac{\lambda}{2t}[/tex]

For B drop,    n = [tex]\frac{\lambda}{26t}[/tex]

So,     n = [tex]\frac{640 \times 10^{-9}}{26 \times 1.56 \times 10^{-8}}[/tex]

            = [tex]\frac{640 \times 10^{-9}}{40.56 \times 10^{-8}}[/tex]

            = 1.5

Thus, we can conclude that index of refraction of the oil is 1.5.

Answer:

intensity at a point on the circle at an angle of 4.45° from the center line is 1.77 W/m².

Explanation:

See attached picture.

The intensity at a point on the circle at an angle of 4.45° from the centerline is approximately 4.00 W/m².

To find the intensity at a point on the circle at an angle of 4.45° from the centerline, we need to consider the interference pattern created by the two transmitters.

1. Wavelength Calculation:

The wavelength [tex](\(\lambda\))[/tex] of the transmitted signal can be calculated using the frequency [tex](\(f\))[/tex] and the speed of light c.

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

Given:

- [tex]\( c = 3 \times 10^8 \)[/tex] m/s (speed of light)

- [tex]\( f = 1575.42 \times 10^6 \)[/tex] Hz (frequency)

[tex]\[ \lambda = \frac{3 \times 10^8}{1575.42 \times 10^6} \approx 0.1902 \text{ meters} \][/tex]

2. Phase Difference Calculation:

The phase difference [tex](\(\Delta \phi\))[/tex] between the two waves at a point on the circle is determined by the path difference [tex](\(\Delta L\))[/tex] travelled by the waves.

The path difference is given by:

[tex]\[ \Delta L = d \sin(\theta) \][/tex]

where:

- d is the distance between the two transmitters (5.18 mm = 0.00518 m)

- [tex]\(\theta\)[/tex] is the angle from the centerline (4.45°)

[tex]\[ \Delta L = 0.00518 \sin(4.45^\circ) \approx 0.00518 \sin(0.0776) \approx 0.00518 \times 0.0775 \approx 0.0004019 \text{ meters} \][/tex]

The phase difference is:

[tex]\[ \Delta \phi = \frac{2 \pi \Delta L}{\lambda} = \frac{2 \pi \times 0.0004019}{0.1902} \approx 0.0133 \text{ radians} \][/tex]

3. Intensity Calculation:

The intensity at a given point due to two sources interfering constructively or destructively can be found using:

[tex]\[ I = I_0 \left(1 + \cos(\Delta \phi)\right) \][/tex]

Here, [tex]\(I_0\)[/tex] is the intensity when both waves interfere constructively at the centerline [tex](\(\theta = 0^\circ\))[/tex], which is given as 2.00 W/m^2.

[tex]\[ I = 2.00 \left(1 + \cos(0.0133)\right) \][/tex]

Since [tex]\(\cos(0.0133) \approx 0.9999\)[/tex]:

[tex]\(\cos(0.0133) \approx 0.9999\)[/tex]

Thus, the intensity at a point on the circle at an angle of 4.45° from the centerline is approximately 4.00 W/m².

Answer:

[tex]\eta_{th} = 8.247\%[/tex]

Explanation:

The maximum possible efficiency for the floating power plant is given by the Carnot's Efficiency:

[tex]\eta_{th} = \left(1-\frac{278.15\,K}{303.15\,K} \right)\times 100\%[/tex]

[tex]\eta_{th} = 8.247\%[/tex]

Answer:

The answer to this question is attached.

The pressure gradient required to accelerate the air inside a pipe can be calculated using Euler's simplified equation for fluid motion. At standard temperature and pressure, you would need a pressure gradient of approximately -102.3 Pa/m to accelerate air at 274 ft/s².

To calculate the pressure gradient, or dp/ds, required to accelerate the air inside a pipe, we can use Euler's equation for fluid motion, which in its simplified form is dp/ds = -ρa, where ρ represents the density of the fluid (in this case, air) and a is the acceleration. For standard temperature and pressure, the density of the air is approximately 1.225 kg/m³. It is important to note that the acceleration provided is in feet per second squared, so we would need to convert that to meters per second squared to match units with the density. The conversion factor is 1 ft/s² = 0.3048 m/s², resulting in an acceleration of approximately 83.515 m/s². By substituting these values into the equation, we get dp/ds = -(1.225 kg/m³)(83.515 m/s²) = -102.3 kg/(m·s²), or -102.3 Pa/m, since a Pascal (Pa) is equivalent to a kg/(m·s²).

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Answer:

[tex]\Delta U = 2.2126039 x 10^{12} J[/tex]

Explanation:

While the satellite is in the space vehicle, it has the next potential energy

[tex]U = -\frac{GmMe}{r}[/tex]

             where G is the gravitational constant

                         m is the satellite's mass in kilograms

                        Me is the earth's mass

                        r is the orbit's radius from to the earth's center in meters

[tex]U = - \frac{6.67x10^{-11}*2721.554*5.972x10^{24} }{482803}[/tex]

[tex]U = -2.2423x10^{12} J[/tex]

The additional energy required is the difference between this energy and the energy that the satellite would have in an orbit with an altitude of 22000 mi

[tex]U = -\frac{6.67x10^{-11}*2721.554*5.792x10^{24} }{35405568}[/tex]

[tex]U = -29696124610.3 J[/tex]

Then

[tex]\Delta U = 2.2126039 x 10^{12} J[/tex]

Answer:

L = 8.66 m

Explanation:

The length measured by the moving observer is related to the true length is given by

L = L₀ √1 - (v²/c²)

Where L₀ is the length of the spaceship as measured by person A, v is the speed of spaceship of person A and c is the speed of light c = 3.8x10ᵃ m/s

L = 10√1 - (0.5c)²/c²

L = 10√1 - (0.5*3.8x10ᵃ)²/3.8x10ᵃ ²

L = 8.66 m

Therefore, the Person B would measure the spaceship length to be 8.66 m

Answer: 0.03 N/C

Explanation:

We use the current density formula to solve this question.

I/A = σ * E

Where,

I = current flowing in the circuit = 0.3 A

A = cross sectional area of the wire = 1 mm²

σ = resistivity of the wire = 1*10^7 Ω^-1·m^-1

E = strength of the electric field required

I/A = σ *E

E = I/(A * σ)

First we convert area from mm to m, so that, 1*10^-3 mm = 1*10^-6 m

E = 0.3 A / (1*10^-6 m * 1*10^7 Ω^-1·m^-1)

E = 0.3 A / 10 Ω^-1

E = 0.03 N/C

The required electric field to drive given current through the wire is 0.03 V/m.

To calculate the strength of the electric field required to drive a current of 0.3 amperes through a platinum wire, we can use the relationship between current density, electrical conductivity, and electric field.

The current density (J) is given by:

J = I / A

where I = 0.3 A and A = 1 mm² = 1 × 10⁻⁶ m².

Substituting the values, we get:

J = 0.3 A / (1 × 10⁻⁶ m²) = 3 × 10⁵A/m²

Next, we use Ohm's law in the form that relates current density to the electric field:

J = σE

where σ is the electrical conductivity of platinum, σ = 1.0 × 10⁷ S/m (S = 1/Ω), and E is the electric field.

Rearranging for the electric field, we get:

E = J / σ

Substituting the values, we get:

E = (3 × 10⁵ A/m²) / (1.0 × 10⁷ S/m) = 3 × 10⁻² V/m

Therefore, the strength of the very small electric field required to drive the current through the wire is 0.03 V/m.

Answer:

Valid

Explanation:

to determine if the claim is valid, we compare the efficiency of the device to that of a Carnot engine.

The following data were given

High Temperature = 920R,

Low temperature =490R

work=4.5hp =4.5*2544.5=11450.25Btu/h

low heat Ql heat= 15000Btu/h

High heat Qh=work +Ql=11450.25Btu/h+15000Btu/h=26450.25Btu/h

Next we cal calculate the efficiency of the Carnot engine

[tex]E_Carnot=1-\frac{T_L}{T_H}\\ E_Carnot=1-\frac{490}{920}\\ E_Carnot=0.467[/tex]

Hence the maximum efficiency at the given temperature is 47%

Next we calculate the efficiency of the device

[tex]E_device=\frac{work}{Q_H} \\E_device=\frac{11450.25}{26450.25} \\E_device=0.433\\[/tex]

which is 43%

since the maximum efficiency of 47% is not exceeded, we can conclude that the claim is valid

Answer:

Magnetic energy stored in the inductor when all of the energy in the circuit is in the inductor = 0.049 mJ

If all the energy is then transferred into the capacitor, the voltage drop across the capacitor = 0.00572 V = 0.01 V (expressed to the hundredths value)

Explanation:

In an RLC circuit with maximum current of 7mA = 0.007 A

When all of the energy is stored in the inductor, maximum current will flow through it,

Hence E = (1/2) LI²

L = inductance of the inductor = 2 H

E = (1/2) (2)(0.007²) = 0.000049 J = 0.049 mJ

When all the energy in the circuit is in the capacitor, this energy will be equal to the energy calculated above.

And for a capacitor, energy is given as

E = (1/2) CV²

E = 0.000049 J, C = 3 F, V = ?

0.000049 = (1/2)(3)(V²)

V = 0.00572 V = 0.01 V

Answer:

(a) The initial speed required is 13116 m/s

(b) The escape speed is 10394 m/s

This problem involves the application of newtons laws of gravitation. The forces in action here are conservative and as a result mechanical energy is conserved.

The full calculation can be found in the attachment below.

Explanation:

In both parts (a) and (b) the energy conservation equation were used. Assumption was made that when the object is very far from the planet the distance from the planet's center approaches infinity and the gravitational potential energy approaches zero.

The calculation can be found below.

Answer:

Resistance, [tex]R=0.529\ \Omega[/tex]

Explanation:

Given that,

Voltage of the battery, V = 9 volts

Current produced in the circuit, I = 17 A

We need to find the resistance when shorted by a wire of negligible resistance. It is a case of Ohm's law. The voltage is given by :

[tex]V=IR[/tex]

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

[tex]R=\dfrac{9\ V}{17\ A}[/tex]

[tex]R=0.529\ \Omega[/tex]

So, the resistance in the circuit is 0.529 ohms. Hence, this is the required solution.

Answer:

(a) v = 15m/a

(b) No they won't feast because the rock can only rise to a height of 11.5m which is less than 15m.

Explanation:

Please see the attachment below for film solution.

The rock does not reach the fruit because the maximum height it achieves is 11.48 meters, which is below the required 15 meters. Therefore, the contestants will go hungry.

Let's solve the problem step by step. First, we'll determine the speed with which the rock leaves the spring, and then we'll check if this speed is enough to reach the fruit hanging in the tree.

Part (a): Speed of the rock leaving the spring

Using the principle of conservation of energy, the potential energy stored in the spring when compressed will be converted into the kinetic energy of the rock when the spring is released.

The potential energy stored in the spring [tex]\( E_{\text{spring}} \)[/tex] is given by:

[tex]\[E_{\text{spring}} = \frac{1}{2} k x^2\][/tex]

Substituting the given values:

[tex]\[E_{\text{spring}} = \frac{1}{2} \times 1000 \, \text{N/m} \times (0.30 \, \text{m})^2\][/tex]

[tex]\[E_{\text{spring}} = \frac{1}{2} \times 1000 \times 0.09\][/tex]

[tex]\[E_{\text{spring}} = 45 \, \text{J}\][/tex]

This energy will be converted into kinetic energy [tex]\( E_{\text{kinetic}} \)[/tex] of the rock:

[tex]\[E_{\text{kinetic}} = \frac{1}{2} m v^2\][/tex]

Equating the spring potential energy to the kinetic energy:

[tex]\[45 \, \text{J} = \frac{1}{2} \times 0.4 \, \text{kg} \times v^2\][/tex]

Solving for ( v ):

[tex]\[45 = 0.2 \times v^2\][/tex]

[tex]\[v^2 = \frac{45}{0.2}\][/tex]

[tex]\[v^2 = 225\][/tex]

[tex]\[v = \sqrt{225}\][/tex]

[tex]\[v = 15 \, \text{m/s}\][/tex]

So, the speed of the rock as it leaves the spring is [tex]\( 15 \, \text{m/s} \).[/tex]

Part (b): Checking if the rock can reach the fruit

To determine if the rock can reach the height of 15 m, we use the following kinematic equation for vertical motion:

[tex]\[h = v_0 t - \frac{1}{2} g t^2\][/tex]

First, determine the time ( t ) it takes to reach the maximum height where the vertical velocity becomes zero:

[tex]\[v = v_0 - g t\][/tex]

At the maximum height, [tex]\( v = 0 \):[/tex]

[tex]\[0 = 15 - 9.81 t\][/tex]

[tex]\[t = \frac{15}{9.81}\][/tex]

[tex]\[t \approx 1.53 \, \text{s}\][/tex]

Now, calculate the maximum height [tex]\( h_{\text{max}} \):[/tex]

[tex]\[h_{\text{max}} = v_0 t - \frac{1}{2} g t^2\][/tex]

[tex]\[h_{\text{max}} = 15 \times 1.53 - \frac{1}{2} \times 9.81 \times (1.53)^2\][/tex]

[tex]\[h_{\text{max}} = 22.95 - 11.47\][/tex]

[tex]\[h_{\text{max}} = 11.48 \, \text{m}\][/tex]

The maximum height reached by the rock is approximately 11.48 m, which is less than the 15 m needed to reach the fruit.

Answer:

Explanation:

Let the position of balance point be x distance from 1.5 µC on the right side . Balance point can not be on the left side or between the charges because in that case electric field by both the charges will be in the same direction .

For equilibrium of field

k x 4.75 µC / ( .35 +x )² -  k x  1.5  µC / x ² = 0

4.75 / ( .35 +x )² = 1.5 / x²

( .35 +x )² / x² = 4.75 / 1.5

( .35 +x )² / x² = 3.167

( .35 +x ) / x =  1.78

.35 + x = 1.78 x

.78 x = .35

x = .45 m ( + ve ) to the right of 1.50  µC

b ) When both the charges are positive , balance point will lie in between them because , electric field will be in opposite direction .

For equilibrium of field

k x 4.75 µC / ( .35 - x )² =   k x  1.5  µC / x ²

4.75 / ( .35 -x )² = 1.5 / x²

( .35 -x )² / x² = 4.75 / 1.5

( .35 -x )² / x² = 3.167

( .35 -x ) / x =  1.78

.35 -x = 1.78 x

2.78 x = .35

x = .126 m  ( - ve ) to the right of 1.50  µC

Answer:

(a) linear momentum = 8.75.  angular momentum =0. (b) 2.5m/s

Explanation:

A and B are on horizontal and so is the velocity given to A = 3.5m/s i (in i direction which is x direction ), this means that there would be motion only in x direction.

let's first calculate linear momentum.

(A)

p(total)= P_A+P_B = (2.5Kgx3.5m/s)+(1kgx0m/s) = 8.75m/s.

Angular momentum.

L=m*v x r. as the motion is only in x direction and there is no rotation in the system, there fore m*v x r = 0

(B)

Velocity of A and B come from the fact that total linear momentum is conserved.

p_final = P_A+P_B = (mA=mB)v_new = mA*V1+mB*vB.

second term on right is = 0 because B has 0 velocity.

solving for V_new and with the values of all unknown substituted in gives

V_new = (mA*VA)/(mA+mB)= 8.75/3.5= 2.5m/s

The diagram showing the 2 spheres is missing, so i have attached it.

Answer:

A) Linear momentum =(8.75 kg/m.s)i

Angular momentum =  (-0.5kg.m²/s)k

B) Va' = (1.5m/s)i and Vb' = (5 m/s)j

Explanation:

First of all, let's find the position of the mass centre;

y' = Σ(mi.yi)/mi = [2.5(0) + 1(0.2)]/(2.5+1) = 0.2/3.5 =0.057143 m

A) Linear momentum is given as;

L = m(a) x v(o) = (2.5 x 3.5)i = (8.75 kg/m.s)i

Angular momentum is given as;

HG = Vector GA  x m(a) x v(o)

Where vector GA is the position of the mass centre;

Thus;

HG = 0.057143j x (2.5 x 3.5)i = (-0.5 kg.m²/s) k

B) from conservation of linear momentum;

MaVo = Ma Va' + Mb Vb

2.5 x 3.5 = 2.5Va' + 1 Vb'

8.75 = 2.5Va' + 1 Vb' - - - - eq(1)

Also, for conservation of angular momentum ;

raMaVo = - raMa Va' + rbMb Vb'

from the diagram attached, ra + rb = 0.2

Now, ra is the same as  value as that of the centre of the mass.

Thus, ra = 0.057143.

rb = 0.2 - ra = 0.2 - 0.057143 = 0.14286

Thus;

0.057143 x 8.75 = -(0.057143 x 2.5)Va' + (0.14286 x 1) Vb'

0.5 = - 0.14286Va' + 0.14286Vb' - - - eq 2

Solving eq 1 and 2 simultaneously, we get; Va' = 1.5m/s and Vb' = 5 m/s

Answer:

[tex]\mu_s^{min}=0.885[/tex]

Explanation:

The centripetal force is provided by the static friction between the car and the road, and always have to comply with [tex]f\leq\mu_sN[/tex], so we have:

[tex]ma_{cp}=f\leq\mu_sN=\mu_smg[/tex]

Which means:

[tex]a_{cp}=\frac{v^2}{r}\leq\mu_sg[/tex]

So we have:

[tex]\frac{v^2}{gr}\leq\mu_s[/tex]

Which means that [tex]\frac{v^2}{gr}[/tex] is the minimum value the coefficient of static friction can have, which for our values is:

[tex]\mu_s^{min}=\frac{v^2}{gr}=\frac{(25m/s)^2}{(9.81m/s^2)(72m)}=0.885[/tex]

To keep the car from skidding, the frictional force between the car and the road must provide the necessary centripetal force to keep the car moving in a circle. The minimum coefficient of friction required would be about 0.86.

To determine this, we must understand that the friction force between the car and the road surface is what keeps the car in a curved path. If the speed of the car or the radius of the curve is too large, a greater force is needed to keep the car from skidding. This force must be supplied as an increased friction force, which implies a larger coefficient of friction.

The centripetal force needed to keep a car on the road as it rounds a turn is provided by the frictional force between the road and the car's tires. We can use the formula for centripetal force where Fc=m*v^2/r. Here, Fc is the centripetal force, m is the mass of the car, v is the speed of the car, and r is the radius of the curve. In this case, friction provides the centripetal force, so Fc is also equal to the force of static friction, which is less than or equal to the coefficient of static friction times the normal force (µ*m*g).

Setting these equal and solving for µ gives µ=v^2/(g*r). Plugging in the given values (v=25 m/s, g=9.8 m^2/s, and r=72 m), we find that the minimum coefficient of friction required is about 0.86. This is reasonable; a typical car with good tires on dry concrete requires a minimum coefficient of friction of about 0.7 to keep from skidding in a curve.

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Answer: In a contained incompressible fluid, any external pressure applied at one point will raise pressure equally at every point.

Explanation: This law was put forward by Blaise Pascal, a French mathematician in 1648. Pascal's Law states that in a contained incompressible fluid, any external pressure applied at one point will raise pressure equally at every point. Pascal's law has been used in fluid mechanics for different applications these includes:

- the hydraulic jack used in automobile listings,

- most automobile braking systems,

-water towers, and dams.

Answer:

the final temperature of the system of the system is 25.32°C

Explanation:

We are not given specific capacity of water and aluminium, so we use their standard values, also we are not given the density if water so we assume the standard vale of density of water

The aluminium calorimeter has a mass Mc= 120g

Volume of water in calorimeter = 100ml at θc =20°C

Density of water is

1000Kg/m³ = 1g/mL

Then, density = mass/ volume

Mass=density ×volume

Mass=1g/mL×100mL

Mass=100gram

Then, the mass of water is

Mw = 100gram

Mass of brass is Mb = 100gram

The temperature of brass is θb=100°C

The specific heat capacity of water is Cw= 1cal/g°C

The specific heat capacity of aluminum Ca=0.22cal/g°C

We are looking for final temperature θf=?

Given that the specific heat capacity of brass is Cb=0.09Cal/g°C

Using the principle of calorimeter;

The principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.

So, the calorimeter gained heat and the liquid in the calorimeter gain heat too

Heat gain by calorimeter(Hc) = Mc•Ca•∆θ

Where Mc is mass of calorimeter,

Ca is Specific Heat capacity of Calorimeter

∆θ=(θf-θc)

Hc=Mc•Ca•∆θ

Hc=120•0.22•(θf-20)

Hc=26.4(θf-20)

Hc=26.4θf-528

Also, heat gain by the water

Heat gain by wayer(Hw) = Mw•Cw•∆θ

Where Mw is mass of water,

Cw is Specific Heat capacity of water

∆θ=(θf-θw),

Note that the temperature of the water and the calorimeter are the same at the beginning i.e. θc=θw=20°C

Hw=Mw•Cw•∆θ

Hw=100•1•(θf-20)

Hw=100(θf-20)

Hw=100θf-2000

Also heat loss by the brass is given by

heat loss by brass

Heat loss by brass(Hb)= Mb•Cb•∆θ

Where Mb is mass of brass,

Cb is Specific Heat capacity of brass

∆θ=(θb-θf)

Therefore,

Hb=Mb•Cb•∆θ

Hb=100•0.09•(100-θf)

Hb=9(100-θf)

Hb=900-9θf

Applying the principle of calorimeter

Heat gain = Heat loss

Hc+Hw=Hb

26.4θf-528 + 100θf-2000=900-9θf

26.4θf+100θf+9θf=900+2000+528

135.4θf=3428

Then, θf=3428/133.4

θf=25.32°C

The final temperature of the system is 25.32 degree Celsius.

Given data:

The mass of aluminum cup is, m = 120 g .

The mass of brass piece is, m' = 100 g.

The volume of water in aluminum cup is, V = 100 ml.

The temperature of water is, T = 20 degree Celsius.

The specific heat of brass is, [tex]c''=0.09 \;\rm cal/g ^\circ C[/tex].

The temperature of brass is, T'' = 100 degree Celsius.

The principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.

So, first calculate heat gain by calorimeter,

H = mc (T' - T)

Here, c is the specific heat of aluminum and its value is, [tex]0.22 \;\rm cal/g^\circ C[/tex]. Solving as,

[tex]H = 120 \times 0.22 (T' - 20)\\H = 26.4T' - 528[/tex]

And, heat gain by water is,

[tex]H'=m'c'(T'-T)[/tex]

Here, c' is the specific heat of water. ( c' =1 )

Solving as,

[tex]H'=100 \times 1 \times (T'-20)\\H' = 100T' -2000[/tex]

Now, heat loss by brass is,

[tex]H'' = m'c'' (T''-T')\\\\H'' = 100 \times 0.09 \times (100-T')\\\\H'' = 900-9T'[/tex]

Applying the principle of calorimeter

Heat gain = Heat loss

H + H' = H''

(26.4T' - 528) + (100T' - 2000) = (900 - 9T')

135.4 T' = 3428

T ' = 25.32 degree Celsius

Thus, we can conclude that the final temperature of the system is 25.32 degree Celsius.

Learn more about the calorimetry here:

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Answer:

The minimum slope angle for which the skier can safely traverse the snow bridge is 47.82°

Explanation:

Given that,

Weight = 700 N

Normal force = 470 N

Suppose, Find the minimum slope angle for which the skier can safely traverse the snow bridge.

We need to calculate the minimum slope angle

Using balance equation

[tex]F_{N}=mg\cos\theta[/tex]

[tex]\theta=\cos^{-1}(\dfrac{F_{N}}{mg})[/tex]

Put the value into the formula

[tex]\theta=\cos^{-1}(\dfrac{470}{700})[/tex]

[tex]\theta=47.82^{\circ}[/tex]

Hence, The minimum slope angle for which the skier can safely traverse the snow bridge is 47.82°

When a 700 N skier crosses a snow bridge that can only support 470 N, the bridge will collapse due to the skier's weight exceeding the bridge's maximum normal force. Newton's second law can be used to understand this situation, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. The relationship between weight, mass, and the gravitational force can also be used to determine that the normal force should be equal to the skier's weight for the bridge to be safe.

The maximum normal force that the snow bridge can sustain is 470 N. The backcountry skier weighs 700 N, so the bridge will collapse under their weight.

To understand why, we can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. The skier's weight is acting downwards, while the normal force of the snow bridge is acting upwards. When the skier crosses the bridge, the normal force should be equal to their weight for the bridge to be in equilibrium. However, since the normal force is less than the skier's weight, the bridge cannot sustain the weight and will collapse.

This situation can be described using the formula: Weight = mass x gravitational force. So, we can calculate the mass of the skier by dividing their weight by the gravitational force. With a weight of 700 N and a gravitational force of 9.8 m/s², the mass is approximately 71.4 kg. Therefore, the normal force supporting the skier on the bridge should be equal to 700 N for it to be safe.

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Answer

[tex]3.9*10^{5}N/C[/tex]

Explanation:

from the expression for determing the magnetic field strength

[tex]E=\frac{q}{4\pi e_{0}d^2 }\\[/tex]

since the charge is given as

[tex]q=1.56*10^{-5}c\\[/tex]

and the distance is

d=60cm=0.6m

We can calculate the constant k

[tex]K=\frac{1}{4\pi e_{0}}\\ K=\frac{1}{4\pi *8.8542*10^{-12}}\\k=8.98*10^{9}\\[/tex]

if we substitute values, we arrive at

[tex]E=\frac{8.98*10^9 *1.56*10^{-5}}{0.6^2} \\E=389133.33\\E=3.9*10^{5}N/C[/tex]

Explanation:

Below is an attachment containing the solution.