The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Express your answer in terms of any or all of m, F, and t.

Answer:

a) 2 m/s

b) i) [tex]K.E = 50 (1.5t^2 + 2) ^2\\[/tex]

ii) [tex]F = 3tm[/tex]

Explanation:

The function for distance is [tex]x = 0.5t ^3 + 2t[/tex]

We know that:

Velocity = [tex]v= \frac{d}{dt} x[/tex]

Acceleration = [tex]a= \frac{d}{dt}v[/tex]

To find speed at time t = 0, we derivate the distance function:

[tex]x = 0.5 t^3 + 2t\\v= x' = 1.5t^2 + 2[/tex]

Substitute t = 0 in velocity function:

[tex]v = 1.5t^2 + 2\\v(0) = 1.5 (0) + 2\\v(0) = 2[/tex]

Velocity at t = 0 will be 2 m/s.

To find the function for Kinetic Energy of the box at any time, t.

[tex]Kinetic \ Energy = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times 100 \times (1.5t^2 + 2) ^2\\\\K.E = 50 (1.5t^2 + 2) ^2\\[/tex]

We know that [tex]Force = mass \times acceleration[/tex]

[tex]a = v'(t) = 1.5t^2 + 2\\a = 3t[/tex]

[tex]F = m \times a\\F= m \times 3t\\F = 3tm[/tex]

The speed of the box at time t=0 is 0 m/s. The acceleration, displacement, and velocity functions of the box as a function of time are a(t) = 3t, x(t) = .5t^3 + 2t and v(t) = 1.5t^2 + 2 respectively.

For part (a), the speed of the box at time t=0 can be found by taking the derivative of the position function x(t), which gives us the velocity function v(t). Therefore, v(t) = 1.5t^2 + 2 and v(0) = 0. Thus, the speed of the box at t=0 is 0 m/s.

For part (b), the box's acceleration at any time t can be found by taking the derivative of the velocity function v(t), which gives us the acceleration function a(t). Therefore, a(t) = 3t, the displacement function as a function of time is the original function x(t) = .5t^3 + 2t and the velocity function is as mentioned above, v(t) = 1.5t^2 + 2.

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Answer:

e. it is anywhere inside (the force is zero everywhere)

Explanation:

The relation between the force and electric field is given as follows

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

where q is the charge inside, and E is the electric field inside the balloon created by the charge on the surface.

By Gauss' Law, the electric field inside the balloon created by the charges on the surface (excluding the charge q, since the electric field of the same charge cannot apply a force on the same charge) is zero.

[tex]\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E4\pi r^2 = 0\\E = 0[/tex]

Since the external electric field inside the sphere is zero, then the force on the point charge is zero everywhere.

It is anywhere inside (the force is zero everywhere).

The electrical field on the surface of the charged sphere is calculated by applying Coulomb law;

[tex]E = \frac{Qr}{4\pi \varepsilon _0R^3}[/tex]

Electric force is calculated as follows;

F = Eq

The electric field inside the charged sphere is zero.

F = 0 x q = 0

Thus, we can conclude that, it is anywhere inside (the force is zero everywhere).

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Since there is a meeting of the cables at their midpoints, it is therefore understood that the force on them is the same but in the opposite direction, this to maintain the static balance between the two.

This can also be corroborated by applying the right hand rule for the force, at which depends of the magnetic field. The net force is zero because the cable segment to the left of the vertical cable feels an opposite force in the direction of the cable segment to the right.

Then the forces cancel.

Therefore the correct answer is C. Therefore the net force on each wire is zero

Answer:

1.42×10⁻¹⁶ N.

Explanation:

The force on a charge moving in a magnetic field is given as

F = qBvsin∅..................... Equation 1

Where F = Force on the charge, q = charge, B = Magnetic Field, v = speed, ∅ = angle between the magnetic field and the speed

Given: B = 0.78 T,  v = 1140 m/s, ∅ = 90° ( Perpendicular) q = 1.60 × 10⁻¹⁹ C.

Substitute into equation 1

F = 0.78(1140)(1.60 × 10⁻¹⁹)sin90°

F = 1.42×10⁻¹⁶ N.

Hence the force on the charge = 1.42×10⁻¹⁶ N.

4.45 * 10¹⁴ J is transferred to get this mass of aluminum from 20°C to its melting point, 660⁰C.

The quantity of heat required to change the temperature of a substance is given by:

Q = mcΔT

Where Q is the heat, m is the mass of the substance, ΔT is the temperature change = final temperature - initial temperature. c is the specific heat capacity

m = 1.7 billion pounds = 77 * 10⁷ kg, ΔT = 660 - 20 = 640°C, c = 903 J/kg•K

Hence:

Q = 77 * 10⁷ kg *  903 J/kg•K * 640°C

Q = 4.45 * 10¹⁴ J

4.45 * 10¹⁴ J is transferred to get this mass of aluminum from 20°C to its melting point, 660⁰C.

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The energy required to heat 1.7 billion pounds of aluminum from 20 degrees Celsius to 660 degrees Celsius is approximately 4.398 × 10^17 Joules.

The thermal energy transferred, or heat, to raise the temperature of a substance is given by the formula q=mcΔT where 'm' is the mass, 'c' is the specific heat capacity, and 'ΔT' is the change in temperature. For aluminum, the specific heat capacity is 0.897 Joules per gram per degree Celsius (J/g°C).

First, we need to convert 1.7 billion pounds of aluminum into grams since the specific heat value is in grams. There are about 453,592.37 grams in a pound, so this gives us about 7.711 × 10^14 grams of aluminum.

The change in temperature (ΔT) is the final temperature minus the initial temperature, or 660 degrees Celsius - 20 degrees Celsius, which equals 640 degrees Celsius.

So, to find the total energy required, we use the formula and substitute the known values: q=(7.711 × 10^14 g)*(0.897 J/g°C)*(640°C), which equals approximately 4.398 × 10^17 Joules.

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Answer: The correct option is B (The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain)

Explanation:

Anterior cruciate ligament (ACL) is one of the important ligaments found at the knee joint which helps to stabilise the joint. It connects the femur to the tibia bone at the knee joint.

Anterior cruciate ligament tear is one of the common knee joint injury which is seen in individuals( especially females) involved in sports( example soccer and basketball which involves sudden change in direction causing the knee to rotate inwards)

ACL tear occurs through both contact and non contact mechanisms. The contact mechanism of ACL injury occurs when force is directly applied at the lateral part of the knee while in non contact mechanism,tear occurs when the tibia is externally rotated on the planted foot.

Research has proven that women are prone to have ACL tear than men when competing in similar sports. This disparity exists due to structural differences that pose as risk factors. These includes

- the female ACL size is smaller than the male.

- the ACL of female has a lower modulus if elasticity( that is, less stiff) than in males leading to greater joint mobility than in the male.. therefore the option, (The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain) is correct.

Answer a) impulse is 1360 kg.m/s

Explanation: impulse is impact force times the time of impact

I = 3400 * 0.4 = 1360kg.m/s

Answer b) final velocity is 10.1m/s

Explanation: impulse is the change of momentum

I = m(v-u)

V and u are final and initial velocities respectively.

1360 = 200 (v - 3.3)

1360 = 200v - 660

V = (1360+660)÷200

V = 10.1m/s

Q2 answer is: astronaut has KE approximately 208 times that of the truck

Explanation :

KE = 0.5mv^2

For truck

KE = 11000 * 33.3^2 = 12197790J

For astronaut

KE = 42 * 7777.7^2 = 2540689926J

Comparing

2540689926/12197790 = 208.2

NB: speed has been converted to m/s by multiplying with 0.28 I.e 1000/3600

Answer:

[tex]c. 5cm \times 8cm[/tex]

Explanation:

The dimensions [tex]5cm\times 8cm[/tex] have the highest cross-sectional area combination of [tex]40cm^2[/tex].

-Resistance reduces with an increase in cross sectional area.

-[tex]Reason-[/tex]Electrons have alarger area to flow through.

Answer:

[tex]v=177.95m/s[/tex]

Explanation:

First, determine circle's radius  between Earth's pole and the location. This can be calculated as:

[tex]r=R_e_a_r_t_hCos(90\frac{3}{4})\\R_e_a_r_t_h=6.37\times10^6m\\r=6.37\times10^6\times Cos67.5\textdegree\\r=2,437,693.46m\\[/tex]

The angular speed of earth is constant and is :

[tex]w=\frac{2\pi}{T}=\frac{2\pi}{24\times 3600}\\=7.3\times10^{-5}rad/s[/tex]

Velocity is:

[tex]v=wr\\=7.3\times10^{-5}\times 2,437,693.46\\v=177.95m/s[/tex]

Explanation:

Below is an attachment containing the solution.

The magnitude of the magnetic force experienced by the bottom side of the rectangular coil, placed in a uniform magnetic field and carrying a current, is 21.16 N.

The subject of this question relates to the interaction of a current-carrying wire in a magnetic field, a fundamental concept in physics. In this particular setup, the magnetic force on each segment of the rectangular coil can be determined by the formula: F = I (L × B), where F is the magnetic force, I is the current, L is the length of the wire and B is the magnetic field. But in this instance, we're specifically interested in the force exerted on the bottom side of the loop, for which the magnetic field and current are perpendicular to each other.

Therefore, the force is given by F = ILB. By substituting the given values into the equation—the length of the bottom side (b = 0.46 m), the magnitude of the current (I = 10 A), and the strength of the magnetic field (B = 4.60 T)—we obtain: F = (10 A)(0.46 m)(4.60 T) = 21.16 N.

So, the magnitude of the magnetic force on the bottom side of the loop is 21.16 N.

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Answer:

Force is needed to begin to move the book and  force needed to keep the book moving at a constant velocity is 10.78 N and 5.85 N.

Explanation:

Given :

Mass of book , M = 2.03 kg.

Coefficient of static friction , [tex]\mu_s=0.542 \ .[/tex]

Coefficient of kinetic friction , [tex]\mu_k=0.294\ .[/tex]

Force, required needed to begin to move the book ,

[tex]F=\mu_sN=\mu_s(mg)=0.542\times 2.03\times 9.8=10.78\ N.[/tex]

Now, We know kinetic friction acts when object is in motion .

Therefore , Force, required o keep the book moving at a constant velocity

[tex]F=\mu_kN=\mu_k(mg)=0.294\times 2.03\times 9.8=5.85\ N.[/tex]

Hence, this is the required solution.

Final answer:

During the elevator's deceleration, the scale reading would increase to approximately 96.3 N as the passenger experiences an upward acceleration.

Explanation:

During the stopping process of the elevator that is descending at 4 m/s and comes to a rest in 2 seconds, the reading on the scale will change due to the acceleration the passenger experiences while the elevator decelerates. The initial velocity v0 is -4 m/s (downwards) and the final velocity v is 0 m/s, as the elevator comes to a stop. The time t taken to stop is 2 seconds, so to find the acceleration, we use the formula a = (v - v0)/t.

This gives us an acceleration of (0 - (-4))/2 = 2 m/s2 upwards, since deceleration is a positive acceleration in the opposite direction of movement. A person's apparent weight on a scale is the normal force exerted by the scale, which can be calculated from Newton's second law, F = ma, where m is mass and a is the acceleration. In this case, the person's actual mass (m) can be deduced from their weight (W = mg), where g is the acceleration due to gravity (approximately 9.81 m/s2).

By rearranging the weight formula, m = W/g, we find m = 80 N / 9.81 m/s2 ≈ 8.16 kg. Now, we add the elevator's upward acceleration to the gravity, so the total acceleration the person feels is (g + a) = 9.81 m/s2 + 2 m/s2 = 11.81 m/s2. Applying F = ma gives us the scale reading: F = 8.16 kg × 11.81 m/s2 ≈ 96.3 N.

The correct answer is that the reading on the scale during the stopping process is 160 N.

When the elevator comes to a stop, it decelerates, which means there is an upward acceleration. According to Newton's second law, the net force acting on an object is equal to the mass of the object times its acceleration ([tex]F_net = m * a[/tex]). In this case, the net force is the difference between the normal force and the gravitational force [tex](F_net = N - mg)[/tex], where mg is the weight of the person.

Given that the elevator decelerates at a rate of 4 m/s in 2 seconds, we can calculate the deceleration (a) as follows:

[tex]\[ a = \frac{\Delta v}{\Delta t} = \frac{4 \text{ m/s}}{2 \text{ s}} = 2 \text{ m/s}^2 \][/tex]

Now, we can set up the equation for the net force during deceleration:

[tex]\[ N - mg = m * a \][/tex]

Since the weight of the person (mg) is 80 N, we can substitute this value into the equation:

[tex]\[ N - 80 \text{ N} = m * 2 \text{ m/s}^2 \][/tex]

To find the normal force (N) during deceleration, we need to solve for N:

[tex]\[ N = m * 2 \text{ m/s}^2 + 80 \text{ N} \][/tex]

We know that when the elevator was moving at a constant rate, the normal force was equal to the weight of the person [tex](N_constant = 80 N).[/tex] During deceleration, the normal force must be twice the weight of the person to provide the additional force required to decelerate the person:

[tex]\[ N = 2 * 80 \text{ N} = 160 \text{ N} \][/tex]

Therefore, the reading on the scale during the stopping process is 160 N.

Explanation:

Below is an attachment containing the solution .

Answer:

The current in the primary coil would be [tex]0.5\ A[/tex].

Explanation:

Given the power supplied by a transformer is 60 watts.

And the voltage in the primary coil is 120 Volts.

We need to find the current supply in the primary coil.

We will use the formula

[tex]P=V\times I[/tex]

Where,

[tex]P[/tex] is the power in Watts.

[tex]V[/tex] is the voltage in Volts.

[tex]I[/tex] is the current supply in Ampere.

[tex]I=\frac{P}{V}\\\\I=\frac{60}{120}\\ \\I=0.5\ A[/tex]

So, the current in the primary coil would be [tex]0.5\ A[/tex].

Answer:

V= 3.55 V

Explanation:

         [tex]V_{rint} = I* r_{int}[/tex]

        [tex]V = V_{b} - V_{rint} = 6.00 V - 0.207A* r_{int}[/tex]

        [tex]r_{int} = \frac{V_{b} - V}{I} = \frac{6.00 V - 5.03V}{0.207A} = 4.7 \Omega[/tex]

         [tex]V = V_{b} - V_{rint} = 6.00 V - 0.523A* 4.7 \Omega = 3.55 V[/tex]

Answer: 5cm

Explanation: Since the radius of curvature is 10cm, the focal length of the mirror (f) is

f = r/2

Where r is the radius of curvature.

Since r = 10cm, f = 10/2 = 5cm.

To produce a parallel light rays of a flash light, it means the the image will be at infinity this making the image distance to be infinite.

From the mirror formulae

1/u + 1/v = 1/f

Where u = object distance =?, v = image distance = infinity and f = focal length = 5cm.

Let us substitute the parameters, we have that

1/u + 1/∞ = 1/5

1/∞ = 0

Hence 1/u = 1/5

u = 5cm.

Hence the object needs to be placed nearly 5cm to the mirror

Final answer:

The light bulb should be placed at the focal length of the concave mirror, which is 5 cm, to produce near-parallel light rays in a flashlight.

Explanation:

The distance between the light bulb and a concave mirror to produce a beam of near-parallel light rays in a flashlight is most nearly the focal length of the mirror. Since the mirror's radius of curvature (R) is 10 cm, using the formula R = 2f, we can calculate the focal length (f). Dividing the radius of curvature by 2, we get f = R/2 = 10 cm / 2 = 5 cm. Therefore, the light bulb should be placed approximately 5 cm from the mirror to achieve near-parallel rays.

Answer:

8.46 N/C

Explanation:

Using Gauss law

[tex]E=\frac {kQ}{r^{2}}[/tex]

Gauss's Law states that the electric flux through a surface is proportional to the net charge in the surface, and that the electric field E of a point charge Q at a distance r from the charge

Here, K is Coulomb's constant whose value is [tex]9\times 10^{9} Nm^{2}/C^{2}[/tex]

r = 0.43 + 0.106 = 0.536 m

[tex]E=\frac {9\times 10^{9}\times 0.270\times 10^{-9}}{0.536^{2}}=8.4581755402094007\approx 8.46 N/C[/tex]

The magnitude of the electric field at a point 0.106 m outside a solid metal sphere with a radius of 0.430 m and a net charge of 0.270 nC is approximately 892 N/C, to three significant figures.

The subject of your question is Physics, specifically focussing on electrostatics and the calculation of the electric field outside a charged sphere. The formula for the electric field (E) due to a point charge is given by Coulomb's Law, which is E = kQ/r^2, where 'Q' is the charge, 'r' is the distance from the charge, and 'k' is Coulomb's constant, approximately 8.99 x 10^9 N.m^2/C^2. Since the electric field due to a uniformly distributed spherical charge behaves as if all the charge is concentrated at the center, you can use this formula for the magnitude of the electric field outside the sphere.

Substituting the provided values (converted to appropriate units), we find E = (8.99 x 10^9 N.m^2/C^2 x 0.270 x 10^-9 C)/(0.536 m)^2, which gives E approximately equal to 892 N/C to three significant figures.

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Answer:

Please find attached file for complete answer solution and explanation of same question.

Explanation:

Answer:

The speed of the speeder is [tex]8.348x10^6m/s[/tex].

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).

That shift can be used to find the velocity of the object (in this case the speeder) by means of the Doppler velocity.

[tex]v = c\frac{\Delta \lambda}{\lambda_{0}}[/tex]   (1)

Where [tex]\Delta \lambda[/tex] is the wavelength shift, [tex]\lambda_{0}[/tex] is the wavelength at rest, v is the velocity of the source and c is the speed of light.

[tex]v = c(\frac{\lambda_{0}-\lambda_{measured}}{\lambda_{0}})[/tex]

For this case [tex]\lambda_{measured}[/tex] is equal to 564.2 nm and [tex]\lambda_{0}[/tex] is equal to 575.9 nm.

[tex]v = (3x10^8m/s)(\frac{575.9 nm - 564.2 nm}{564.2 nm)})[/tex]

[tex]v = 8.348x10^6m/s[/tex]

Hence, the speed of the speeder is [tex]8.348x10^6m/s[/tex].

The question involves calculating the speed of a car based on the Doppler shift of light from yellow to green. The formula for Doppler shift of light is utilized to determine the relative velocity, but the practicality of such an effect for a moving car is negligible.

The question relates to the concept of the Doppler Effect for light, where the observed wavelength of light emitted by a source changes due to relative motion between the source and the observer. In this case, to calculate how fast the speeder would have to be traveling for the Doppler shift to change the color of a yellow light (575.9 nm) to green (564.2 nm), we can use the Doppler shift formula for light:

f' = f (c + v) / (c - v), where:

Since frequency is inversely proportional to wavelength (f = c / λ), the formula can be rearranged in terms of wavelength. Solving for v when the source is moving towards the observer gives us the expression:

v = c (λ0 - λ) / λ, where λ0 is the original wavelength and λ is the observed wavelength.

The speed of the car can thus be calculated accordingly. However, the effect of Doppler shift at such speeds is so small that it would not account for the perceptual change from yellow to green in a real-world scenario.

Answer:

a) [tex]F_H=776.952\ N[/tex]

b) [tex]F_g=706.32\ N[/tex]

c) [tex]v=5.4249\ m.s^{-1}[/tex]

d) [tex]KE=1059.48\ J[/tex]

Explanation:

Given:

a)

Now the force of lift by the helicopter:

Here the lift force is the resultant of the force of gravity being overcome by the force of helicopter.

[tex]F_H-F_g=m.a[/tex]

where:

[tex]F_H=72\times 0.981+72\times9.81[/tex]

[tex]F_H=776.952\ N[/tex]

b)

The gravitational force on the astronaut:

[tex]F_g=m.g[/tex]

[tex]F_g=72\times 9.81[/tex]

[tex]F_g=706.32\ N[/tex]

d)

Since the astronaut has been picked from an ocean we assume her initial velocity to be zero, [tex]u=0\ m.s^{-1}[/tex]

using equation of motion:

[tex]v^2=u^2+2a.h[/tex]

[tex]v^2=0^2+2\times 0.981\times 15[/tex]

[tex]v=5.4249\ m.s^{-1}[/tex]

c)

Hence the kinetic energy:

[tex]KE=\frac{1}{2} m.v^2[/tex]

[tex]KE=0.5\times 72\times 5.4249^2[/tex]

[tex]KE=1059.48\ J[/tex]

Answer:

Explanation:

mass of helicopter, m = 72 kg

height, h = 15 m

acceleration, a = g/10

(a) Work done by the force

Work, W = force due to helicopter x distance

W = m x ( g + a) x h

W = 72 ( 9.8 + 0.98) x 15

W = 11642.4 J

(b) Work done by the gravitational force

W = - m x g x h

W = - 72 x 9.8 x 15

W = - 10584 J

(c) Kinetic energy = total Work done

K = 11642.4 - 10584

K = 1058.4 J

(d) Let the speed is v.

K = 0.5 x m v²

1058.4 = 0.5 x 72 x v²

v = 5.42 m/s

Now the force of lift by the helicopter:

where:

force by the helicopter

force of gravity

b)

The gravitational force on the astronaut:

d)

Since the astronaut has been picked from an ocean we assume her initial velocity to be zero,

using equation of motion:

c)

Hence the kinetic energy:

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Answer:

 v = 0.0147 m / s

Explanation:

For this exercise let's use energy conservation

Starting point. Fully stretched spring

            Em₀ = Ke = ½ k (x-x₀)²

Final point. Unstretched position

          Emf = K = ½ m v²

          Emo = Emf

         ½ k (x- x₀)² = ½ m v²

           v = √m/k    (x-x₀)

Let's calculate

            v = √(0.022 / 0.5)      (0.26-0.19)

            v = 0.0147 m / s

The speed of the mass at the mean position is 0.333 m/s

The potential energy stored in a fully stretched spring

PE = ½ kx²

where x is the stretch of the spring  = 26 -19 = 7 cm = 0.07 m

At the mean position, where x = 0, the PE stored in sprig is zero,

So according to the law of conservation of energy total energy must remain conserved so all the energy is converted into kinetic energy KE of the mass

KE = ½ mv²

where m is the mass and v is the velocity

½ kx² = ½ mv²

where k is the spring constant = 0.5 N/m

and m is the mass = 0.022 kg

[tex]v=\sqrt{\frac{k}{m} } x[/tex]

[tex]v=\sqrt{\frac{0.5}{0.022} } 0.07[/tex]

v = 0.333 m/s

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Answer:

Part (a) current in the wire is 1.144 A

Part (b) the drift velocity of electrons in the wire is 1.028 x 10⁻⁴ m/s

Explanation:

Given;

diameter d  = 1.02 mm

current density J = 1.40×10⁶ A/m²

number of electron = 8.5×10²⁸ electrons

Part (a) Current in the wire

I = J×A

Where A is area of the wire;

[tex]A = \frac{\pi d^2}{4} \\\\A = \frac{\pi (1.02X10^{-3})^2}{4} = 8.1723 X10^{-7} m^2[/tex]

I = 1.40 x 10⁶ x 8.1723 x 10⁻⁷

I = 1.144 A

Part (b) the drift velocity of electrons in the wire

[tex]V = \frac{J}{nq} = \frac{1.4X10^6}{8.5X10^{28} X 1.602X10^{-19}} = 1.028 X10^{-4} m/s[/tex]

We were given the

diameter = 1.02 mm

current density = 1.40×10⁶ A/m²

number of electron = 8.5×10²⁸ electrons

We can use the formula:

I = J×A

where I is current, J is density and A is area.

A = π d²

        4

  = π (1.02ₓ 10⁻³)² = 8.1723 x 10⁻⁷

              4

I = J×A

I = 1.40 x 10⁶ x 8.1723 x 10⁻⁷

I = 1.144 A

V = J/ nq

    =   1.4 ₓ 10⁶ / (8.5ₓ 10²⁸ₓ 1.602ₓ 10⁻¹⁹)

   = 1.028ₓ 10⁻⁴ m/s

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Answer:

[tex]d=2.4\ miles[/tex]

Explanation:

Given:

According to given condition:

[tex]t_w=t_b+\frac{36}{60}[/tex]

where:

[tex]t_w=[/tex] time taken to reach the building by walking

[tex]t_b=[/tex] time taken to reach the building by biking

We know that,

[tex]\rm time=\frac{distance}{speed}[/tex]

so,

[tex]\frac{d}{v_w} =\frac{d}{v_b} +\frac{36}{60}[/tex]

[tex]\frac{d}{3}=\frac{d}{12} +\frac{3}{5}[/tex]

[tex]d=2.4\ miles[/tex]

Answer:

The distance from her apartment to the classroom building is 2.4 miles.

Explanation:

Given that,

Time  = 36 min

Walking average speed of her = 3 m/h

biking average speed of her = 12 m/h

If she takes n minutes to ride, then if the distance is d miles,

We need to calculate the distance

Using formula of time

[tex]t=t_{1}+t_{2}[/tex]

[tex]t=\dfrac{d}{v_{1}}+\dfrac{d}{v_{2}}[/tex]

Put the value into the formula

[tex]\dfrac{36}{60}=\dfrac{d}{3}-\dfrac{d}{12}[/tex]

[tex]d=\dfrac{36\times12}{60\times3}[/tex]

[tex]d=2.4\ miles[/tex]

Hence, The distance from her apartment to the classroom building is 2.4 miles.

To calculate the radial acceleration of a ball in circular motion, derive its velocity using the horizontal distance traveled upon string break and use the centripetal acceleration formula.

The question provided involves calculating the radial acceleration of a ball in circular motion before the string breaks. To solve this, we must understand that the radial or centripetal acceleration formula is a = v^2 / r, where a is the centripetal acceleration, v is the velocity of the object in circular motion, and r is the radius of the circle.

However, the information given directly does not include the velocity v. We can derive the velocity using the horizontal distance the ball traveled after the string broke. Since the only force acting on the ball after the string breaks is gravity, the horizontal motion can be considered uniform. The formula distance = velocity x time (d = vt) can be rearranged to find the velocity (v = d/t). Using the principle of projectile motion, the time (t) it takes for the ball to hit the ground can be found using the formula derived from the vertical motion due to gravity: t = sqrt(2h/g), where h is the height above the ground and g is the acceleration due to gravity (9.81 m/s2).

Coming back to finding the radial acceleration, once we have the velocity, we simply substitute values into the centripetal acceleration formula. Note that the actual calculations were not performed as the goal here is to elucidate the strategy for solving the problem.

Final answer:

The radial acceleration of the ball during circular motion was calculated to be approximately 47.7 m/s².

Explanation:

To find the radial acceleration of the ball during its circular motion, we'll use the information provided about the ball's horizontal displacement after the string breaks.

In a situation where the string breaks and an object follows a projectile motion, the horizontal component of its initial velocity (vx) can be calculated using the horizontal displacement (d) and the time of flight (t). The formula for horizontal displacement is d = vx * t.

The time of flight (t) can be found using the vertical motion equations, considering that the ball drops 1.50 m. Since the ball is released from rest in the vertical direction, we have:

Vertical displacement (y): 1.50 m

Acceleration due to gravity (g): 9.81 m/s2

Initial vertical velocity (vy0): 0 m/s

Using the equation y = vy0 * t + 0.5 * g * [tex]t^2[/tex], we can solve for the time of flight (t), which will also be the same time the ball is moving horizontally since horizontal and vertical motions are independent.

Applying this equation to calculate the time (t):

1.50 = 0 + 0.5 * 9.81 * [tex]t^2[/tex]
t = sqrt(2 * 1.50 / 9.81)
t ≈ 0.553 s (rounded to three significant figures)

Now we can find the horizontal velocity (vx) using the horizontal distance (d):

2.10 = vx * 0.553
vx ≈ 3.80 m/s

The horizontal velocity of the ball at the instant the string breaks is the same as the tangential velocity of the ball while it was in circular motion. Therefore, we can calculate the radial or centripetal acceleration (ar) using the formula for centripetal acceleration:

ar = v2 / r
where v is the tangential velocity and r is the radius.

So:

ar = [tex](3.80 m/s)^2[/tex] / 0.300 m
ar ≈ 47.7 m/s2

The radial acceleration of the ball during its circular motion was approximately 47.7 [tex]m/s^2[/tex].

Answer:

The scale reading at the top = 565.8N

The scale reading at the bottom = 610.44N

Explanation:

We are given:

t = 32seconds

R = 9.7meters

mass (m) = 60Kg

We take g as gravitational field = 9.8

Therefore, to find the scale reading (N) at the top, let's use the formula:

[tex] N_t = m (g - w^2 R) [/tex]

Where w = 2π/t

w = 2π/32 = 0.197

Substituting the figures into the equation, we have

[tex] N = 60 (9.8 - 0.197^2 * 9.7) [/tex]

N = 60 (9.8 - 0.37)

N = 60 * 9.43

Na = 565.8N

To find scalar reading at the bottom, we use;

[tex] N_b = m(g + w^2 R) [/tex]

= 60 (9.8 + 0.37)

= 60 * 10.17

[tex] N_b = 610.44 [/tex]

Answer:

The scale reading at the top is [tex]z_{top} = 565.6\mu N[/tex]

The scale reading at the bottom is [tex]z_{bottom} = 610.356\ N[/tex]

Explanation:

From the question

       The radius is [tex]r = 9.7 m[/tex]

       The period is [tex]T = 32sec[/tex]

      The mass is [tex]m =60kg[/tex]

Generally the mathematical representation for angular velocity of the wheel is

                  [tex]\omega = \frac{2 \pi}{T}[/tex]

                    [tex]= \frac{2*3.142}{32}[/tex]

                    [tex]= 0.196 \ rad/sec[/tex]

The velocity at which the point scale move can be obtained as

                     [tex]v = r\omega[/tex]

                        [tex]= 9.7* 0.196[/tex]

                       [tex]= 1.9 m/s[/tex]

Considering the motion of the 60kg mass as shown on the first and second uploaded image

Let the z represent the reading on the scale which is equivalent to the normal force acting on the mass.

          Now at the topmost the reading of the scale would be

                       [tex]mg - z_{top} =\frac{mv^2}{r} =m\omega^2r[/tex]

Where mg is the gravitational force acting on the mass and [tex]\frac{mv^2}{r}[/tex] is the centripetal force keeping the mass from spiraling out of the circle

Now making z the subject of the formula

                    [tex]z_{top} = mg - \frac{mv^2}{r}[/tex]

                       [tex]= m(g- \frac{v^2}{r})[/tex]

                       [tex]= 60(9.8 - \frac{1.9^2}{9.7} )[/tex]

                      [tex]= 565.6\mu N[/tex]

Now at the bottom the scale would be

                      [tex]z_{bottom}-mg = \frac{mv^2}{r} = m \omega^2r[/tex]

This is because in order for the net force to be in the positive y-axis (i.e for the mass to keep moving in the Ferris wheel) the Normal force must be greater than the gravitational force.  

Making  z the subject

                  [tex]z_{bottom}= mg +m\omega^2r[/tex]

                  [tex]z =m(g+\omega^2r)[/tex]

                  [tex]z= 60(9.8 + (0.196)^2 *(9.7))[/tex]

                     [tex]= 610.356\ N[/tex]

Answer:

The distance is 709.5 m.

Explanation:

Given that,

Speed = 150 m/s

Distance = 110 m

Suppose, How far short of the target should it drop the package?

We need to calculate the time

Using equation of motion

[tex]s=ut+\dfrac{1}{2}gt^2[/tex]

[tex]t^2=\dfrac{2s}{g}[/tex]

Where, g = acceleration due to gravity

t = time

Put the value into the formula

[tex]t=\sqrt{\dfrac{2\times110}{9.8}}[/tex]

[tex]t=4.73\ sec[/tex]

We need to calculate the distance

Using formula of distance

[tex]d= vt[/tex]

Put the value into the formula

[tex]d=150\times4.73[/tex]

[tex]d=709.5\ m[/tex]

Hence, The distance is 709.5 m.

Answer:

a) 201

b) 0

Explanation:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

For the time interval from 0 to 2, sine wave reaches to zero for 201 times. Hence the current equals to zero for 201 times.

No charge transported through the light in the first second.

Given that, the current flowing through the bulb is [tex]I(t)=114 sin (100\pi t) \;\rm A[/tex].

The general equation of the current is,

[tex]I(t) =Asin(2\pi ft)[/tex]

So the frequency can be calculated as,

[tex]2\times \pi\times f\times t = 100\times \pi[/tex]

[tex]f =50 \;\rm Hz[/tex]

Hence the frequency of the sine wave is 50 Hz.

For the time interval from 0 to 2, the number of zero for the sine wave is,

[tex]No.\;of\; zero = (2\times50\times2 )+1[/tex]

[tex]No.\;of\;zero=201[/tex]

So, For the time interval from 0 to 2, sine wave reaches to zero for 201 times. Hence the current equals to zero for 201 times.

The charge can be calculated  by the formula given below.

[tex]Q(t) = \int\limits^{t_1}_{t_2} {I(t)} \ dt[/tex]

[tex]Q(t)=\int\limits^1_0 {114sin(100\pi t)} \ dt[/tex]

[tex]Q(t) = \dfrac {-114cos(100\pi t)}{100\pi}[/tex]

[tex]Q(t) = \dfrac {-114cos(100\pi \times 1)}{100\pi}-\dfrac {-114cos(100\pi \times0)}{100\pi}[/tex]

[tex]Q(t)=0[/tex]

Hence, no charge transported through the light in the first second.

For more details, follow the link given below.

https://brainly.com/question/1345174.

Answer:

B = (4.76 × 10⁻⁷) T

Explanation:

From Biot Savart's law, the magnetic field formula is given as

B = (μ₀I)/(2πr)

B = magnetic field = ?

I = current = 238 mA = 0.238 A

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

r = 10 cm = 0.1 m

B = [4π × 10⁻⁷ × 0.238)/(2π×0.1)]

B = (4.76 × 10⁻⁷) T

The direction of the magnetic field is in the clockwise direction wrapped around the current-carrying wire.

Hope this Helps!!!

Answer:

ΔS = - 3.74 × 10⁻²³ J/K = - 3.7 × 10⁻²³ to 2 s.f

Explanation:

The change in entropy for a system with changing allowable Microsystems is given as

ΔS = K In (W/W₀)

K = Boltzmann's constant = 1.381 × 10⁻²³ J/K

W₀ = initial number of microstates

To calculate this, how many ways can 2 atoms occupy 16 available microstates with order important?

That is, ¹⁶P₂ = 16!/(16 - 2)! = 16 × 15 = 240 microstates.

W = the number of microstates for Chlorine atom when Bromine atom desorps = 16 microstates.

ΔS = K In (W/W₀)

ΔS = (1.381 × 10⁻²³) In (16/240)

ΔS = (1.381 × 10⁻²³) × 2.7081

ΔS = - 3.74 × 10⁻²³ J/K

Entropy change is positive when an atom desorbs from a surface and drifts away, allowing for more freedom of movement and increasing the system's entropy.

Entropy change is a measure of disorder in a system. When an atom desorbs from a surface and drifts away, the change in entropy is positive as it increases the freedom of movement of the atom. This occurs because there are more possible locations for the atom to occupy, leading to an increase in entropy.

Answer:

The direction of the electric field vector is horizontal and pointing north.

Option (C) is correct option.

Explanation:

Given :

The direction of wave propagation is toward the west.

The direction of magnetic field vector is vertically upward.

According to the theory of electromagnetic wave propagation, the electric field and magnetic field is perpendicular to each other and the direction of propagation is also perpendicular to both electric and magnetic field vector.

⇒   [tex]\vec{E} + \vec{B} = \vec {k}[/tex]

From right hand rule, the fingers goes towards horizontal and pointing north and curl the finger goes towards  vertically upward and thumb will give you the direction of wave propagation toward west.

Hence, the direction of electric field vector is horizontal and pointing north.

The correct direction of the electric field vector in an electromagnetic wave propagating to the west with an upward magnetic field is horizontal and pointing north, following the right-hand rule for perpendicularity and direction of electromagnetic waves.

The subject of this question is the direction of the electric field vector in an electromagnetic wave that is propagating towards the west, with a magnetic field vector that points vertically upward. According to the properties of electromagnetic waves, the electric field (E) and magnetic field (B) are perpendicular to each other and to the direction of wave propagation. Therefore, if the magnetic field is pointing upward and the wave is moving west, the electric field must be pointing horizontally. Since the right-hand rule dictates that E x B gives the direction of wave propagation, and the wave is moving west, the electric field cannot be pointing east or west as it would not satisfy the right-hand rule. Thus, the electric field is either pointing north or south. To determine the correct direction between north and south, we rely on the right-hand rule: the fingers of the right hand point in the direction of E, the curled fingers point towards B, and the thumb points in the direction of the wave propagation (west in this case). If the magnetic field points up, and propagation is to the west, the electric field must be directed to the north. Therefore, the correct answer is C. horizontal and pointing north.