A 0.37 kg object is attached to a spring with a spring constant 175 N/m so that the object is allowed to move on a horizontal frictionless surface. The object is released from rest when the spring is compressed 0.19 m. Find the force on the object when it is released. Answer in units of N. What is the acceleration at this instant? Answer in units of m/s^2.​

Answer:

string stretched is 1.02 cm

Explanation:

given data

length = 80-cm

diameter = 1.0-mm

tension = 2000 N

solution

we get here string stretched that will be as and here  

we know that young modulus for steel = 200 × [tex]10^{9}[/tex]

so here stress will be

stress = y × strain  .............1

that is express as

[tex]\frac{force}{area} = \frac{Y \Delta L}{L}[/tex]

ΔL = [tex]\frac{0.80*2000}{\pi * 0.0005^2*200*10^9}[/tex]

ΔL = 0.0102 m

ΔL = 1.02 cm

so string stretched is 1.02 cm

Answer: 0.4469 m/s

Explanation:

First, convert mile into meters.

1 mile = 1609 m

Secondly, convert hour into seconds

1 hour = 3600 s

Finally, convert 1.000 mph to meters per second.

1.000 mph is same as 1 mile / 1 hour

Put SI units we converted earlier.

1 mile / 1 hour = 1609 m / 3600 s

1 mile / 1 hour = 0.4469 m/s

The appropriate solution is "0.4469 m/s". A further explanation is provided below.

According to the question,

Given facts:

The speed of the car will be:

= [tex]1 \ mile/hour[/tex]

By substituting the above given values, we get

= [tex]1.000\times \frac{1609 \ m}{3600 \ s}[/tex]

= [tex]1.000\times 0.447[/tex]

= [tex]0.4469 \ m/s[/tex]

Thus the above solution is correct.

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Answer:

the marble has lost the same number of electrons acquired by the piece of silk.  

the marble acquires a positive charge and attracts the piece of silk.

Explanation:

When an object or a material acquires extra electrons it becomes negatively charged while when it losses electrons it becomes positively charged. The glass marble was rubbed against the silk cloth, and the silk cloth acquires extra electron thereby becoming negatively charged, the same amount of charge the silk cloth acquires was lost from the glass marble thereby causing the marble to become positively charged. Also opposite charges attracts while likes charges repel and due to equal and opposite charge on silk and glass marble, they both will attract each other. thus option A and E is correct.

The question is not complete. Kindly find the complete question below:

Host A converts analog to digital at a = 58 Kbps  Link transmission rate R = 1.9 Mbps  Host A groups data into packets of length L = 112 bytes  Distance to travel d = 931.9 km  Propagation speed s = 2.5 x 108 m/s  Host A sends each packet to Host B as soon as it gathers a whole packet.  Host B converts back from digital to analog as soon as it receives a whole packet.  How much time elapses from when the first bit starts to be created until the conversion back to analog begins? Give answer in milliseconds (ms) to two decimal places, normal rounding, without units (e.g. 1.5623 ms would be entered as "1.56" without the quotes)

Answer / Explanation

The answer is 19.65

Answer:

The velocity of the skateboard is 0.774 m/s.

Explanation:

Given that,

The spring constant of the spring, k = 3086 N/m

The spring is stretched 18 cm or 0.18 m

Mass of the student, m = 100 kg

Potential energy of the spring, [tex]P_f=20\ J[/tex]

To find,

The velocity of the car.

Solution,

It is a case of conservation of energy. The total energy of the system remains conserved. So,

[tex]P_i=K_f+P_f[/tex]

[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2+20[/tex]

[tex]\dfrac{1}{2}\times 3086\times (0.18)^2=\dfrac{1}{2}mv^2+20[/tex]

[tex]50-20=\dfrac{1}{2}mv^2[/tex]

[tex]30=\dfrac{1}{2}mv^2[/tex]

[tex]v=\sqrt{\dfrac{60}{100}}[/tex]

v = 0.774 m/s

So, the velocity of the skateboard is 0.774 m/s.

Answer:

(a) The speed of proton decreases as it moves from A to B.

(b) Plate B is at a higher potential.

(c) Plate B is positive, plate A is negative.

(d) Parallel lines parallel to the two plates.

    Parallel lines equally spaced.

Explanation:

The electric potential energy is given by the following formula:

[tex]U = \frac{1}{4\pi \epsilon_0}\frac{qq_0}{r}[/tex]

Alternatively, potential energy in a uniform electric field is

[tex]U = qEr[/tex]

where 'r' is the distance from negative to positive plates. This definition is analogues to that of gravitational potential energy, U = mgh.

If the positively charged proton is gaining potential energy as it gets closer to plate B, then plate B is charged positively.

(a) According to this information, the speed of proton decreases as it moves from A to B. This is similar to the speed of an object which is gaining potential energy by moving higher.

(b) By the same gravitational analogy, plate B is at a higher potential.

(c) As explained before, Plate A is negative and Plate B is positive.

(d) The equipotential lines are parallel to electric field lines which are perpendicular to the plates. So, the equipotential lines are parallel to the plates. Since the electric field between the plates is uniform, then the equipotential lines are equally seperated.

Answer:

a. [tex]\bar{d}=4.34 cm[/tex]

b. [tex]\sigma=0.023 cm[/tex]

c. [tex]\rho=(0.0089\pm 0.00058) kg/cm^{3}[/tex]

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

[tex]\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}[/tex]

[tex]\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}[/tex]

[tex]\bar{d}=4.34 cm[/tex]

b) The standard deviation equations is:

[tex]\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}[/tex]

If we put all this values in that equation we will get:

[tex]\sigma=0.023 cm[/tex]

Then the mean diameter will be:

[tex]\bar{d}=(4.34\pm 0.023)cm[/tex]

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: [tex]V=\pi R^{2}h[/tex]

Then:

[tex]\rho=\frac{m}{\pi R^{2}h}[/tex]

Using the values that we have, we can calculate the value of density:

[tex]\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}[/tex]

We need to use propagation of error to find the error of the density.

[tex]\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}[/tex]  

Let's find each partial derivative:

1. [tex]\frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089[/tex]

2.  [tex]\frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004[/tex]

3. [tex]\frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071[/tex]

Therefore:

[tex]\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}[/tex]

[tex]\delta\rho=0.00058[/tex]

So the density is:

[tex]\rho=(0.0089\pm 0.00058) kg/cm^{3}[/tex]

I hope it helps you!

Answer:

The boat moves away from the dock at 0.5 m/s.

Explanation:

Hi there!

Since no external forces are acting on the system boy-boat at the moment at which the boy lands on the boat, the momentum of the system is conserved (i.e. it remains constant).

The momentum of the system is calculated as the sum of the momentum of the boy plus the momentum of the boat. Before the boy lands on the boat, the momentum of the system is given by the momentum of the boy.

momentum of the system before the boy lands on the boat:

momentum of the boy + momentum of the boat

m1 · v1 + m2 · v2 = momentum of the system

Where:

m1 and v1: mass and velocity of the boy.

m2 and v2: mass and velocity of the boat.

Then:

50 kg · 2.0 m/s + 150 kg · 0 m/s = momentum of the system

momentum of the system = 100 kg m/s

After the boy lands on the boat, the momentum of the system will be equal to the momentum of the boat moving with the boy on it:

momentum of the system = (m1 + m2) · v (where v is the velocity of the boat).

100 kg m/s = (50 kg + 150 kg) · v

100 kg m/s / 200 kg = v

v = 0.5 m/s

The boat moves away from the dock at 0.5 m/s.

Final answer:

The child running off a dock and landing in a rowboat scenario involves applying the conservation of momentum principle to find the boat's final velocity. The speed of the rowboat is 0.67 m/s.

Explanation:

Given:

Child mass (m1) = 50 kg

Child velocity (v1) = 2.0 m/s

Boat mass (m2) = 150 kg

Let the final velocity of the boat be v2

Using the conservation of momentum:

m1v1 = (m1 + m2)v2

Substitute the values to find v2: 50 kg * 2.0 m/s = (50 kg + 150 kg) * v2

Solving for v2, we get v2 = 0.67 m/s

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Answer:

1) 13.377 m/s

2) 12.044 m/s

3) 8.893 m

4) 32.85 m

5) 19.05 m/s

6) 3.25 m

Explanation:

1)

V_o,x = V_o * cos (Q)

V_o,x = 18 * cos (42)

V_o,x = 13.377 m/s

2)

V_o,y = V_o * sin (Q)

V_o,y = 18 * sin (42)

V_o,y = 12.044 m/s

3)

Maximum height is reached when V,y = 0

V,y = V_o,y + a*t

0 = 12.044 - 9.81t

Solve above equation for t:

t = 1.228 s

Compute S_y @t = 1.228 s

S_y = S_o,y + V_o,y*t + 0.5*a*t^2

S_y = 1.5 + 12.044*1.228 - 4.905*1.228^2

S_y = 8.893 m

4)

Time taken for the ball to complete path:

S_y = S_o,y + V_o,y*t + 0.5*a*t^2 = 1.5

V_o,y*t + 0.5*a*t^2 = 0

12.044*t - 4.905*t^2 = 0

t = 0, t = 2.455 s

Total distance traveled in horizontal direction S_x @ t = 2.455 s

S_x = S_o,x + V_o,x*t

S_x = 0 + 13.377*2.455 = 32.85 m

5)

S_y = S_o,y + V_o,y*t + 0.5*a*t^2 = 10

10 = 1.5 + V_o,y*t -4.905*t^2   .... Eq 1

Maximum height is reached when V,y = 0

V,y = V_o,y + a*t

0 =  V_o,y - 9.81t  .... Eq2

Solve Eq 1 and Eq 2 simultaneously

V_o,y = 9.81*t

10 = 1.5 + 9.81*t^2 -4.905*t^2

8.5 = 4.905*t^2

t = 1.316 s

V_o,y = 12.914 m/s

Compute Velocity

V = sqrt (V_o,x^2 + V_o,y^2)

V = sqrt (14^2 + 12.914^2)

V = 19.05 m/s

6)

Total distance traveled in horizontal direction between players is 32.85m

S_x = S_o,x + V_o,x*t

S_x = 0 + 14*t = 32.85 m

t = 2.3464 s

Compute Sy @ t = 2.3464 s

S_y = S_o,y + V_o,y*t + 0.5*a*t^2

S_y = 10  - 4.905*(1.1732)^2

S_y = 3.25 m

Answer: wavelength λ = 2.9Å

Explanation:

Using the particle in a box model. The energy level level increases with n^2

En = (n^2h^2)/ 8mL^2 .....1

For the ground state, n = 1 to level n= 5, the energy level changes from E1 to E5

∆E = (5^2 - 1^2)h^2/8mL^2

but 5^2 - 1^2 = 24.

so,

∆E = 24h^2/8mL^2 .....2

And the wavelength of the radiation can be derived from the equation below:

E = hc/λ

λ = hc/E .......3

Substituting equation 2 to 3

λ = hc/[(24h^2)/ 8mL^2]

λ = 8mcL^2/(24h)

λ = 8mcL^2/24h .....4

Where,

n = energy state

h = Planck's constant = 6.626 × 10^-34 Js

m= mass of electron = 9.1 × 10^-31 kg

L = length = 45.7pm = 45.7×10^-12 m

E = energy

c= speed of light = 3.0 ×10^8 m/s

λ= wavelength

Substituting the values into equation 4 above

λ = [(8×9.1×3×45.7^2)/(24×6.626)] × 10^(-31+8-24+34)

λ = 2868.285390884 × 10^-13 m

λ = 2.9 × 10^-10 m

λ = 2.9Å

The wavelength of the electromagnetic radiation required to excite an electron can be calculated using the formula wavelength = (2 * box length) / n, where n is the energy level. The wavelength of the electromagnetic radiation required is 9.14 pm.

To calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with n = 5 in a one-dimensional box, we can use the formula:

wavelength = (2 * box length) / n

Given that the box length is 45.7 pm and n = 5, we can substitute the values into the formula:

wavelength = (2 * 45.7 pm) / 5

Simplifying the expression gives us:

wavelength = 9.14 pm

Therefore, the wavelength of the electromagnetic radiation required is 9.14 pm.

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Answer: d(5) = 115m

Position after 5 seconds is 115m

Explanation:

Given;

Initial position d(0) = 0

Time = 5 sec

Velocity function vx(t) = 10t - 2

To determine its position after 5 sec we need to calculate the position function.

d(t) = integral of vx(t)

d(t) = ∫10t - 2

d(t) = (10/2)t^2 - 2t + c

d(t) = 5t^2 - 2t + d(0)

c = d(0) = 0

d(t) = 5t^2 - 2t

So, at time t = 5

d(5) = 5(5^2) -2(5) = 125 - 10

d(5) = 115m

Answer:

[tex]2.96866\times 10^{17}\ C[/tex]

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]

r = Distance between the objects and particles

[tex]q_1=q_2[/tex] = Charges

M = Mass of Sun = [tex]1.989\times 10^{30}\ kg[/tex]

m = Mass of Earth = [tex]5.972\times 10^{24}\ kg[/tex]

Here, the Electric force will balance the gravitational force

[tex]\dfrac{GMm_2}{r^2}=\dfrac{kq_1q_2}{r^2}\\\Rightarrow q=\sqrt{\dfrac{GMm}{k}}\\\Rightarrow q=\sqrt{\dfrac{6.67\times 10^{-11}\times 5.972\times 10^{24}\times 1.989\times 10^{30}}{8.99\times 10^{9}}}\\\Rightarrow q=2.96866\times 10^{17}\ C[/tex]

Charge on each particle will be [tex]2.96866\times 10^{17}\ C[/tex]

To develop this problem we will apply the concepts related to angular kinematic movement, related to linear kinematic movement. Linear velocity can be described in terms of angular velocity as shown below,

[tex]v = r\omega \rightarrow \omega = \frac{v}{r}[/tex]

Here,

v = Lineal velocity

[tex]\omega[/tex]= Angular velocity

r = Radius

Our values are

[tex]v = 6/ms[/tex]

[tex]r = \frac{d}{2} = \frac{120*10^{-3}}{2} = 0.06m[/tex]

Replacing to find the angular velocity we have,

[tex]\omega = \frac{6m/s}{0.06m}[/tex]

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

Convert the units to RPM we have that

[tex]\omega = 100rad/s (\frac{1rev}{2\pi rad})(\frac{60s}{1m})[/tex]

[tex]\omega = 955.41rpm[/tex]

Therefore the angular speed of the wheels when the scooter is moving forward at 6.00 m/s is 955.41rpm

The angular speed of the scooter's wheels when the scooter is moving at 6.00 m/s is 100 rad/sec. This value is calculated using the formula for relating linear velocity, radius, and angular speed, and converting the wheel's diameter to a radius in meters.

To find the angular speed of the scooter's wheels, we need to use the equation that relates linear velocity (v), radius (r), and angular speed (w). This equation is v = r*w where v is the linear speed, r is the radius, and w is the angular speed which we are trying to find.

First, radius r needs to be calculated using the provided diameter as r = diameter / 2 = 120 mm / 2 = 60 mm. Since the linear speed is provided in m/s, we need to convert the radius from mm to m. So, r = 60 mm = 0.06 m.

Then, we can substitute the known values into the equation. 6 m/s = 0.06 m * w, and solve for w: w = (6 m/s) / 0.06 m = 100 rad/s. Therefore, the angular speed of the wheels when the scooter is moving forward at 6.00 m/s is 100 rad/sec.

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Answer:

I = 70.2mA

V1 = 3.09V

V2 = 1.19V

V3 = 7.72V

Explanation:

Total resistance for a series connection = R1 + R2 + R3 = 44ohm + 17ohm + 110ohm = 171ohm

From ohm's law

Voltage (V) = current (I) × resistance (R)

I = V/R = 12/171 = 0.0702A = 0.0702×1000mA = 70.2mA

V1 = I×R1 = 0.0702×44 = 3.09V

V2 = I×R2 = 0.0702×17 = 1.19V

V3 = I×R3 = 0.0702×110 = 7.72V

Answer:

a) I = 0.0702A = 70.2mA

b) V1 = 3.09V

c) V2 = 1.19V

d) V3 = 7.72V

Explanation:

Given that the circuit consist of three series resistors.

For resistors arranged in series, the total resistance R can be given as:

R = R1 + R2 + R3

R1 = 44 ohms

R2 = 17 ohms

R3 = 110 ohms.

R = 44 + 17 + 110 = 171 ohms

V = 12 V

a) The current of a circuit is given by;

Potential difference V = current × total resistance

V = IR

Making I the subject of formula,

I = V/R

I = 12/171 = 0.0702A

I = 7.02mA

b) the potential difference across any resistors is given by:

V = IR

Since the arrangement is parallel, the same current flows through each of the resistors.

V1 = IR1

V1 = 0.0702 × 44

V1 = 3.09V

c) applying the same rule as b above:

V2 = IR2

V2 = 0.0702 × 17

V2 = 1.19V

d) applying the same rule as b above.

V3 = IR3

V3 = 0.0702 × 110

V3 = 7.72V

Gauss' law states that the total charge contained within a closed surface immediately proportionately affects the electric flux through that surface. The flux across the spherical surface will quadruple if the charge is tripled.

The enclosed charge determines the electric flux through a closed surface, not the size or shape of the surface. The quantity of charge confined does not change when the sphere's volume is doubled, hence the flux does not change.

Similar to component (b), the flux is unaffected by changing the surface's shape as long as the enclosed charge stays constant.

The total charge contained by the surface is the only factor that affects the electric flux across a closed surface.

Thus, since there is no longer any charge contained by the closed surface if the charge is transported outside of it, there is no longer any electric flux through the surface.

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The total flux through a spherical surface surrounding a point charge depends on the charge, while the flux through the surface remains constant when the volume of the sphere or the shape of the surface is changed. If the charge is moved to another location inside the surface, the total flux remains constant, but if the charge is moved outside the surface, the total flux goes to zero.

(a) When the charge is tripled, the total flux through the surface also triples. This is because the total flux is directly proportional to the charge enclosed by the surface.

(b) When the volume of the sphere is doubled, the total flux through the surface remains constant. This is because the total flux is independent of the volume of the surface.

(c) When the surface is changed to a cube, the total flux through the surface remains constant. This is because the total flux is independent of the shape of the surface.

(d) When the charge is moved to another location inside the surface, the total flux through the surface remains constant. This is because the total flux is independent of the position of the charge within the surface.

(e) When the charge is moved outside the surface, the total flux through the surface goes to zero. This is because the total flux is zero when there are no charges enclosed by the surface.

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Answer:

 r = 22.78 m

Explanation:

For this exercise let's use Newton's second law

Axis y

                 N- W = 0

                 N = W

X axis

                F = m a

Where the acceleration is centripetal

               a = v² / r

The force is the friction that the formula has

               fr = μ N

               fr = μ mg

Let's replace

            μ m g = m v² / r

           r = v² / μ g

Let's reduce the speed to the SI system

         v = 45 km / h (1000 m / 1 km) (1h / 3600 s) = 12.5 m / s

          r = 12.5 2 / (0.7 9.8)

          r = 22.78 m

Answer:

-0.912 m/s

Explanation:

When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.

[tex] (m_c + m_b)v_b + m_pv_p = 0[/tex]

where [tex]m_c = 24.6 kg, m_b = 39 kg, m_p = 5.8 kg[/tex] are the mass of the child, the boat and the package, respectively. [tex], v_p = 10m/s, v_b[/tex] are the velocity of the package and the boat after throwing.

[tex] (24.6 + 39)v_b + 5.8*10 = 0[/tex]

[tex]63.6v_b + 58 = 0[/tex]

[tex]v_b = -58/63.6 = -0.912 m/s[/tex]

The position and velocity of the particle in equilibrium position at the given parameters are;

A) x = -2.34 m

B) v_x = -1.3 m/s

C) x(4.5) = -0.076 m

D) v = 0.314 m/s

We are given;

Initial distance; x_i = 0.27 m

Initial velocity; v_xi = 0.14 m/s

Acceleration; a_x = - 0.32 m/s²

Time; t = 4.5 s

A) Formula for the particle's position as a function of time under constant acceleration is;

x = x_i + v_xi•t + ½a_x•t²

x = 0.27 + (0.14 × 4.5) + ½(-0.32 × 4.5²)

x = -2.34 m

B) Formula for it's velocity at the end of the time interval is;

v_x = v_xi + a_x•t

v_x = 0.14 + (-0.32 × 4.5)

v_x = -1.3 m/s

C) Formula for position in simple harmonic motion is;

x(t) = A cos(ωt + ϕ)

We know that acceleration is;

a = -ω²x

Thus;

-0.32 = -ω²(0.27)

ω = √(0.32/0.27)

ω = 1.089 rad/s

Now, velocity is the derivative of x(t). Thus;

v(t) = x'(t) = -Aω sin (ωt + ϕ)

At t = 0, we have;

0.14 = -A(1.089) × sin ϕ  - - -(1)

Also, at t = 0,

0.27 = A cos ϕ  - - - (2)

Divide equation 1 by equation 2 to get;

0.14/0.27 = -1.089 tan ϕ

ϕ = tan^(-1) (0.14/(0.27 × -1.089))

ϕ = -25.46°

Thus, putting -25.46° for ϕ in eq 2 gives;

0.27 = A cos (-25.46)

0.27 = A × 0.90183

A = 0.27/0.90183

A = 0.2994

Thus,

x(4.5) = 0.2994 cos((1.089 × 4.5) + (-25.46))

x(4.5) = -0.076 m

D) v = -0.2994 × 1.089 × sin 254.6

v = 0.314 m/s

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The position of the particle after 4.50 s of constant acceleration is 207.360 m, and its velocity is 91.580 m/s. In simple harmonic motion, the position of the particle at the end of 4.50 s is 0 m, and its velocity is -1.080 m/s.

(a) Position:

To find the position of the particle after 4.50 s, we can use the equation x = x0 + v0t + 0.5at^2, where x is the final position, x0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time interval.

Plugging in the given values:

x = 0.270 m + (0.140 m/s)(4.50 s) + 0.5(20.320 m/s^2)(4.50 s)^2

x = 0.270 m + 0.630 m + 206.460 m

x = 207.360 m

Therefore, the position of the particle at the end of 4.50 s is 207.360 m.

(b) Velocity:

To find the velocity of the particle at the end of 4.50 s, we can use the equation v = v0 + at, where v is the final velocity.

Plugging in the given values:

v = 0.140 m/s + (20.320 m/s^2)(4.50 s)

v = 0.140 m/s + 91.440 m/s

v = 91.580 m/s

Therefore, the velocity of the particle at the end of 4.50 s is 91.580 m/s.

(c) Position:

Since the particle is in simple harmonic motion, its position can be described by the equation x = x0 + A* sin(ωt + φ), where x0 is the equilibrium position, A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant.

Plugging in the given values:

x = 0 + 0.270*sin(4.00 rad/s * 4.50 s + 0)

x ≈ 0.270*sin(18.000 rad)

x ≈ 0.270*sin(π rad)

x ≈ 0 m

Therefore, the position of the particle at the end of 4.50 s in simple harmonic motion is 0 m.

(d) Velocity:

The velocity of the particle in simple harmonic motion can be described by the equation v = A*ω*cos(ωt + φ).

Plugging in the given values:

v = 0.270*4.00*cos(4.00 rad/s * 4.50 s + 0)

v ≈ 1.080*cos(18.000 rad)

v ≈ 1.080*cos(π rad)

v ≈ -1.080 m/s

Therefore, the velocity of the particle at the end of 4.50 s in simple harmonic motion is -1.080 m/s.

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Answer:

(a). The critical angle for the liquid when surrounded by air is 44.37°

(b). The angle of refraction is 26.17°.

Explanation:

Given that,

Incidence angle = 26.7°

Refraction angle = 18.3°

(a). We need to calculate the refraction of liquid

Using Snell's law

[tex]n=\dfrac{\sin i}{\sin r}[/tex]

Put the value into the formula

[tex]n=\dfrac{\sin 26.7}{\sin 18.3}[/tex]

[tex]n=1.43[/tex]

We need to critical angle for the liquid when surrounded by air

Using formula of critical angle

[tex]C=\sin^{-1}(\dfrac{1}{n})[/tex]

Put the value into the formula

[tex]C=\sin^{-1}(\dfrac{1}{1.43})[/tex]

[tex]C=44.37^{\circ}[/tex]

(b). Given that,

Incidence angle = 37.5°

Speed of light in mineral [tex]v=2.17\times10^{8}\ m/s[/tex]

We need to calculate the index of refraction

Using formula of index of refraction

[tex]n=\dfrac{c}{v}[/tex]

Put the value into the formula

[tex]n=\dfrac{3\times10^{8}}{2.17\times10^{8}}[/tex]

[tex]n=1.38[/tex]

We need to calculate the angle of refraction

Using Snell's law

[tex]n=\dfrac{\sin i}{\sin r}[/tex]

[tex]\sin r=\dfrac{\sin i}{n}[/tex]

Put the value into the formula

[tex]\sin r=\dfrac{\sin 37.5}{1.38}[/tex]

[tex]r=\sin^{-1}(\dfrac{\sin 37.5}{1.38})[/tex]

[tex]r=26.17^{\circ}[/tex]

Hence, (a). The critical angle for the liquid when surrounded by air is 44.37°

(b). The angle of refraction is 26.17°.

Answer

a) Angle of incidence i  = 26.7°

Angle of refraction r = 18.3°

From Snell’s law index of refraction of the liquid

[tex]n = \dfrac{sin\ i}{sin\ r}[/tex]

[tex]n = \dfrac{sin\ 26.7^0}{sin\ 18.3^0}[/tex]

      n = 1.43

So, critical angle

[tex]C= sin^{-1}(\dfrac{1}{n})[/tex]

[tex]C= sin^{-1}(\dfrac{1}{1.43})[/tex]

       C = 44.33°

b) Angle of incidence, i = 37.5°

  speed of light in mineral oil , v = 2.17 x 10⁸ m/s

  speed of light in air, c = 3 x 10⁸ m/s

refractive index of the oil

 [tex]n = \dfrac{c}{v}[/tex]

 [tex]n = \dfrac{3\times 10^8}{2.17\times 10^8}[/tex]

  n = 1.38

again using Snell's law

[tex]n = \dfrac{sin\ i}{sin\ r}[/tex]

[tex]sin\ r = \dfrac{sin\ i}{n}[/tex]

[tex]sin\ r = \dfrac{sin\ 37.5^0}{1.38}[/tex]

[tex] r = sin^{-1}(0.441)[/tex]

       r = 26.18°

hence, the angle of refraction is equal to r = 26.18°

Answer:

Explanation:

When  a positively charged glass rod is brought near the uncharged sphere (which contains equal amount of positive and negative charge) then it started attracting metal sphere. This occurs because glass rod polarize metal sphere such that negative charge is induced at end near to the rod.

But when rod is touched to sphere it started to repel because some of the rod positive charge goes into metal sphere and thus similar charges repel each other.

A light bulb acts as a non-ohmic resistor due to its changing resistance with temperature changes, primarily caused by increased atomic vibrations impeding electron flow. This contrasts with Ohm's Law's assumption of constant resistance, illustrating the complex atomic level interactions that govern a light bulb's resistance and thereby its non-ohmic behavior.

The reason a light bulb is considered a non-ohmic resistor primarily lies in its changing resistance with variations in temperature, a characteristic that defies the principle of Ohm's Law which presumes constant resistance. In a light bulb, particularly an incandescent one, the filament's resistance increases significantly as it heats up from room temperature to its operating temperature. This increase in temperature, and consequently resistance, is not simply due to thermal expansion but is rooted in atomic level interactions.

At the atomic level, as the filament's temperature rises, the atoms inside the metal filament vibrate more vigorously. This enhanced vibration creates more impediments for the free flow of electrons, which is the principal cause of electrical current. Hence, with more obstacles in their path, electrons face increased resistance. This change in resistance with temperature illustrates the non-ohmic behavior as it shows that the resistance isn't constant but varies with temperature. It's clear that thermal expansion plays a role in this scenario, but the key factor is the increased atomic vibrations that hinder electron flow.

The power dissipation in resistors, and by extension in light bulbs, can be described by the equations P = V^2/R and P = I^2R. These seemingly contradictory formulae actually complement each other in explaining how power dissipation can either increase or decrease with rising resistance, depending on whether the scenario is considered from the perspective of voltage or current, further illustrating the complex relationship between these variables in a non-ohmic conductor like a light bulb.

Answer:

q1= +0.75 nC

Explanation:

As the electrostatic force is linear, we can apply the superposition principle to calculate the total force on q₃ due to q₂ and q1, according to Coulomb's Law, as follows:

F₃₂ = k*q₃*q₂/r₃₂² = 9*10⁹ N*m²/C²*+5nC*(-3 nC) / (0.04m)² = -84.4*10⁻⁶ N

F₃₁ = k*q₃*q₁ / r₃₁² = 9*10⁹ N*m²/C²*q₁*(+5 nC) / (0.02m)²

The total force on q₃ is just the sum of F₃₂ and F₃₁, which must add to 0, as follows:

F₃ = F₃₂ + F₃₁ = 0

⇒ -84.4* 10⁻⁶ N = -9*10⁹ N*m²/C²*q₁*(+5 nC) / (0.02m)²

Solving for q₁, we get:

q₁ = (84.4 / 11.25)*10⁻¹⁰ C = +0.75 nC

q₁ must be positive, in order to counteract the attractive force on q₃ due to q₂.

Final answer:

When two conducting spheres come into contact, they share their total charge evenly. If sphere A has a charge of -5 nC and sphere B has -3 nC, each sphere will have -4 nC after contact and separation.

Explanation:

When two identical conducting spheres are brought into contact, charge redistribution occurs. The charge will spread evenly across both spheres, because both have equal capacity to hold charge being identical conductors. To find the final charge on each sphere, one must add the initial charges of both spheres and then divide by two because the total charge will be shared equally.

In the scenario given where sphere A has a charge of −5 nC and sphere B has a charge of −3 nC:

Total charge before contact: -5 nC + (-3 nC) = -8 nC

After contact and separation, each sphere will have half the total charge: -8 nC / 2 = -4 nC

Therefore, both sphere A and sphere B will end up with a charge of −4 nC after being brought into contact and separated. The equivalent number of electrons is determined by the charge on each sphere divided by the elementary charge (approximately 1.6 x 10^-19 Coulombs).

The time it takes for the plane to circle the Earth at the equator is approximately 182588 seconds or about 50.72 hours.

In order to calculate the time it takes for the plane to circle the Earth at the equator, we need to determine the distance of the circumference of the Earth at the equator. The circumference of a circle is given by the formula C = 2πr, where r is the radius of the circle. In this case, the radius is approximately 6370 km. Plugging this value into the formula, we get C ≈ 2π(6370 km) ≈ 40030 km.

Next, we can calculate the time it takes for the plane to travel this distance by using the formula time = distance / speed. The speed of the plane is given as v = 219 m/s, which is equivalent to approximately 0.219 km/s. Therefore, the time it takes for the plane to circle the Earth at the equator is approximately 40030 km / 0.219 km/s ≈ 182588 seconds (or about 50.72 hours).

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Answer:

the direction that should be walked by Ricardo to go directly to Jane is 23.52 m, 24° east of south

Explanation:

given information:

Ricardo walks 27.0 m in a direction 60.0 ∘ west of north, thus

A= 27

Ax =  27 sin 60 = - 23.4

Ay = 27 cos 60 = 13.5

Jane walks 16.0 m in a direction 30.0 ∘ south of west, so

B = 16

Bx = 16 cos 30 = -13.9

By = 16  sin 30 = -8

the direction that should be walked by Ricardo to go directly to Jane

R = √A²+B² - (2ABcos60)

   = √27²+16² - (2(27)(16)(cos 60))

   = 23.52 m

now we can use the sines law to find the angle

tan θ = [tex]\frac{R_{y} }{R_{x} }[/tex]

         = By - Ay/Bx -Ax

         = (-8 - 13.5)/(-13.9 - (-23.4))

     θ  = 90 - (-8 - 13.5)/(-13.9 - (-23.4))

         = 24° east of south

Answer:

a) When the x-component of the position of the particle is 29 m, the y-component is 45 m.

b)When the x-component of the position of the particle is 29 m, its speed is 21.8 m/s.

Explanation:

I´ve found the complete question on the web:

A particle starts from the origin at t=0 with a velocity of 8.0 j m/s and moves in the x-y plane with a constant acceleration of (4i + 2j) m/s^2. At the instant the particle's x-coordinate is 29m. What are its y-coordinate and speed?

a) First, let´s find at which time the particle is located at 29 m along the x-axis. For that let´s use the equation of position of an object moving in a straight line at constant acceleration:

x = x0 + v0x · t + 1/2 · ax · t²

Where:

x = x-component of the position of the particle at time t

x0 = initial x-position.

v0x = initial x-component of the velocity.

t = time

ax = x-component of the acceleration.

We have the following data:

x = 29 m

x0 = 0 (because the particle starts from the origin, x = 0 and y = 0).

v0x = 0 (the initial velocity only has an y-component).

ax = 4 m/s²

Then, the equation of position gets reduced to this:

x = 1/2 · a · t²

29 m = 1/2 · 4 m/s² · t²

29 m/ 2 m/s² = t²

t = 3.8 s

Now, we can find the y-component of the position of the particle at that time:

y = y0 + v0y · t + 1/2 · ay · t²

Where:

y = y-component of the position of the particle at time t.

y0 = initial y-component of the position.

v0y = initial y-component of the velocity.

ay = y-component of the acceleration

t = time.

We have the following data:

y0 = 0

v0y = 8.0 m/s

ay = 2 m/s²

t = 3.8 s (calculated above)

Then, we can calculate "y" at t = 3.8 s

y = 0 + 8.0 m/s · 3.8 s + 1/2 · 2 m/s² · (3.8 s)²

y = 45 m

When the x-component of the position of the particle is 29 m, the y-component is 45 m.

b) To find the speed of the particle, let´s use the equation of velocity.

For the x-component of the velocity (vx):

vx = v0x + ax · t   (v0x = 0)

vx = 4 m/s² · 3.8 s

vx = 15.2 m/s

The y-component of the velocity will be:

vy = v0y + ay · t

vy = 8.0 m/s + 2 m/s² · 3.8 s

vy = 15.6 m/s

Then, the vector velocity will be:

v = (15.2, 15.6) m/s

To calculate the speed, we have to find the magnitude of the velocity vector:

[tex]|v| = \sqrt{(15.2 m/s)^{2}+(15.6 m/s)^{2}} = 21.8 m/s[/tex]

When the x-component of the position of the particle is 29 m, its speed is 21.8 m/s.

Final answer:

The work done by the motor in a CD player to make the CD spin can be calculated using the formula: Work = Torque * Angle. The final angular velocity can be calculated using the formula: v = r * ω. The moment of inertia can be calculated using the formula: Moment of Inertia = (1/2) * Mass * Radius^2.

Explanation:

The work done by the motor in a CD player to make the CD spin can be calculated using the formula:

Work = Torque * Angle

In this case, since the CD starts from rest, the initial angular velocity is zero. The final angular velocity can be calculated using the formula:

v = r * ω

where r is the radius of the CD and ω is the angular velocity.

The torque can be calculated using the formula:

Torque = Moment of Inertia * Angular Acceleration

Since we have the mass and diameter of the CD, the moment of inertia can be calculated using the formula:

Moment of Inertia = (1/2) * Mass * Radius^2

Once we have the torque and the angle, we can calculate the work done by the motor.

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

[tex]|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s[/tex]

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

To solve this problem we will apply the linear motion kinematic equations. With the information provided we will calculate the time it takes for the object to fall. From that time, considering that the ascent rate is constant, we will take the reference distance and calculate the distance traveled while the object hit the ground, that is,

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

[tex]-18 = 15*t + \frac{1}{2} 9.8*t^2[/tex]

[tex]t = 3.98s[/tex]

Then the total distance traveled would be

[tex]h = h_0 +v_0t[/tex]

[tex]h = 18+15*3.98[/tex]

[tex]h = 77.7m[/tex]

Therefore the railing will be at a height of 77.7m when it has touched the ground

The height  is the railing when the camera hits the ground should be considered as the 77.7 m.

Since we know that

[tex]h = vt - 1/2gt^2\\\\-18 = 15*1 + 1/2*9.8*t^2[/tex]

t = 3.98s

Now the total distance should be

[tex]= 18 + 15*3.98[/tex]

= 77.7 m

hence, The height  is the railing when the camera hits the ground should be considered as the 77.7 m.

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To develop this problem we will apply the linear motion kinematic equations. For this purpose we will define the change in speed as the product between acceleration and time.

[tex]v_f-v_i = at[/tex]

The relation between initial velocity final velocity and time is

[tex]v_f = v_i+at[/tex]

The acceleration is due to the acceleration due to gravity, then we have

[tex]v_f = v_i-gt[/tex]

At the maximum height the final velocity is zero. Then we have that

[tex]0 = v_i-gt[/tex]

[tex]t = \frac{v_i}{g}[/tex]

The time the player must wait before touching he ball is

[tex]t = \frac{4.41}{9.8}[/tex]

[tex]t = 0.45s[/tex]