Wilma I. Ball walks at a constant speed of 5.93 m/s along a straight line from point A to point B and then back from B to A at a constant speed of 3.15 m/s.
(a)

What is Wilma's average speed over the entire trip?Wilma I. Ball walks at a constant speed of 4.51 m/s along a straight line from point A to point B and then back from B to A at a constant speed of 3.15 m/s.

(a)

What is Wilma's average speed over the entire trip?

(b)

What is Wilma's average velocity over the entire trip?

Answer:

Explanation:

Given

Length of shell [tex]L=220\ m[/tex]

radius of cylindrical shell [tex]r=4\ cm[/tex]

surface charge density [tex]\sigma =14\ nC/m^2[/tex]

Total charge on the shell [tex]Q=surface\ area\times surface\ charge\ density[/tex]

surface area [tex]A=2\pi r\cdot L=2\pi \cdot 0.04\cdot 220=55.299\ m^2[/tex]

[tex]Q=\sigma \cdor A[/tex]

[tex]Q=14\times 10^{-9}\times 55.299[/tex]

[tex]Q=7.741\times 10^{-7}\ C[/tex]

The correct answer is  (a.) thousands of small images. Each hole in the aluminum foil acts as a tiny pinhole camera, allowing light to pass through and create an image on the other side. Since there are numerous holes, each one forms a separate image, resulting in a multitude of small, overlapping pictures. This phenomenon is known as a pinhole array or pinhole sieve.

If you punched thousands of holes in the aluminum foil of the scope (so there were more “holes” than “foil”), the resulting images in the viewer would be thousands of small images. Each hole acts as a pinhole camera, allowing light to pass through and create an image on the other side. Since there are numerous holes, each one would form a separate image, resulting in a multitude of small, overlapping pictures. This phenomenon is known as a pinhole array or pinhole sieve.

complete question;

If you punched thousands of holes in the aluminum foil of the scope (so there were more "holes" than "foil"), how many images would you see in the viewer?

Choices are:

a. thousands of small images

b. a few bright images

c. one large blurry image

d. no images at all, the light waves would cancel

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:

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.

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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The charge on the half of the rod farther from the sphere brought near to it, which had initially a neutral charge, is -4×104 units. This is due to a process known as charging by induction.

The charge on the half of the rod farther from the sphere is -4×104 units. This occurs due to a process called charging by induction. Basically, when a charged object is brought near to an initially neutral conductor, it polarizes the conductor. Negative charges are attracted towards the charged sphere, leaving the far side of the rod positively charged.

However, the problem statement tells us that there is a surplus of charge on the closer half of the rod, hence the further half must have a deficiency of charge by the same amount, resulting in -4×104 units of charge. Remember, in a neutral object, the total charge is zero. Thus, if we develop a surplus (+4×104) on one side, we must have an equal amount of deficit (-4×104) on the other side to maintain the total charge at zero.

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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:

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

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:

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

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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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:

63°

Explanation:

Draw a free body diagram of the ladder.  There are 4 forces:

Normal force N pushing up at the base of the ladder.

Friction force Nμ pushing right at the base of the ladder.

Weight force mg pushing down a distance x up the ladder.

Reaction force R pushing left at the top of the ladder.

Sum of forces in the x direction:

∑Fₓ = ma

Nμ − R = 0

R = Nμ

Sum of forces in the y direction:

∑Fᵧ = ma

N − mg = 0

N = mg

Sum of moments about the base of the ladder:

∑τ = Iα

R (L sin θ) − mg (x cos θ) = 0

R (L sin θ) = mg (x cos θ)

Substituting:

Nμ (L sin θ) = mg (x cos θ)

mgμ (L sin θ) = mg (x cos θ)

μ (L sin θ) = x cos θ

tan θ = x / (μL)

θ = atan(x / (μL))

Given x = 0.49 m, μ = 0.1, and L = 2.5 m:

θ = atan(0.49 m / (0.1 × 2.5 m))

θ ≈ 63°

To solve the problem we will first calculate the reaction and the normal force.

The angle of the ladder should be 63°.

Given to us

[tex]\sum F_y = 0\\N - mg = 0\\N = mg[/tex]

[tex]\sum F_x = 0\\N\mu - R = 0\\R = N\mu[/tex]

[tex]R (L\ sin\theta) - mg (x\ cos\theta) = 0\\R (L\ sin\theta) = mg (x\ cos\theta)[/tex]

Substituting the values of R and N,

[tex]N\mu (L\ sin \theta) = mg (x\ cos \theta)\\mg\mu (L\ sin \theta) = mg (x\ cos\theta)\\\mu (L\ sin \theta) = x\ cos \theta\\tan \theta = \dfrac{x}{\mu L}\\\theta = tan^{-1}( \dfrac{x}{\mu L})[/tex]

Substituting the values,

[tex]\theta = tan^{-1}(\dfrac{0.49\ m}{0.1\times 2.5 m})\\\\\theta = tan^{-1} (0.96)\\\\\theta = 62.969^o \approx 63^o[/tex]

Hence, the angle of the ladder should be 63°.

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Answer:

a) The number density is 3.623 × 10⁻³ [tex]\frac{mol}{m^{3} }[/tex]

The mass of the atmosphere is 1.3 × 10²²Kg

b) The pressure is 10⁻²⁰ Millimeter of mercury

c) The mass mixing ratio is 0.0107

The partial pressure of ethane is 0.01114 Pa

Yes it is condensable because it boiling point is -88.5 C  which is equivalent to 184.5 K i.e is adding 273 to -88.5C and the temperature of the atmosphere  is 37 K.

Explanation:

The explanation is on the first and second uploaded image

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:

Explanation:

Given

mass of bullet [tex]m=5.6\ gm[/tex]

velocity of bullet [tex]v=501.8\ m/s[/tex]

Depth of penetration [tex]d=4.59\ cm[/tex]

According to the work energy theorem work done by all the force will be equal to change in kinetic energy of Particle

Suppose F is the force which is opposing the bullet motion

change in kinetic Energy [tex]\Delta K=\frac{1}{2}mv^2-0[/tex]

[tex]\Delta K=\frac{1}{2}mv^2=\frac{1}{2}\times 5.6\times 10^{-3}\times (501.8)^2[/tex]

[tex]\Delta K=705.049\ J[/tex]

[tex]\Delta K=F\cdot d[/tex]

[tex]F=\frac{\Delta K}{d}[/tex]

[tex]F=\frac{705.049}{4.59\times 10^{-2}}[/tex]

[tex]F=15,360.54\ N[/tex]

[tex]F=1.536\times 10^4\ N[/tex]                      

Answer:

m will be equal to 1158.73 gram

Explanation:

We have given mass m when frequency is 0.83 Hz

So mass [tex]m_1=m[/tex] and frequency [tex]f_!=f[/tex] let spring constant of the spring is KK

Frequency of oscillation of spring is given by [tex]f=\frac{1}{2 \pi }\sqrt{\frac{k}{m}}[/tex]

From above relation we can say that [tex]{\frac{f_1}{f_2}}=\sqrt{\frac{m_2}{m_1}}[/tex]

It is given that when an additional 730 gram is added to m then frequency become 0.65 Hz , [tex]f_2=0.65Hz[/tex]

So [tex]m_2=m+730[/tex]

So  [tex]\frac{0.93}{0.65}=\sqrt{\frac{m+730}{m}}[/tex]

[tex]\frac{m+730}{m}=1.63[/tex]

[tex]0.63m=730[/tex]

m= 1158.73 gram

To find the value of m, we can set up a proportion using the formula for frequency and solve for m. The value of m is 0.93 kg.

To solve this problem, we can use the formula for the frequency of an object in simple harmonic motion:

f = 1 / T

Where f is the frequency and T is the period. Let's denote the mass of the object as m, and the original frequency as f1. When the additional mass is added, the frequency becomes f2. We can set up a proportion to solve for the value of m:

f1 / f2 = m / (m + 0.73)

Solving for m, we have:

m = (f1 / f2) * 0.73

Substituting the values f1 = 0.83 Hz and f2 = 0.65 Hz, we can find the value of m:

m = (0.83 / 0.65) * 0.73 = 0.93 kg

Therefore, the value of m is 0.93 kg.

Answer:

a.<-6500,0,3900>kgm/s

b.<-6500,0,3900>kgm/s

Explanation:

We are given that

Initial velocity of car,u=[tex]<19,0,23>[/tex]m/s

Final velocity of car=[tex]v=<14,0,26>m/s[/tex]

Mass of the car=m=1300 kg

a.We have to find the change in momentum during this manuver.

Change in momentum=[tex]\Delta P=m(v-u)[/tex]

Using the formula

[tex]\Delta P=1300(<14,0,26>-<19,0,23>)=1300(<-5,0,3>)=<-6500,0,3900>kgm/s[/tex]

Hence, the change in momentum during this maneuver=<-6500,0,3900>kgm/s

b.Impulse =Change in momentum of car

Impulse applied to the car=<-6500,0,3900>kgm/s

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

Answer:

Explanation:

Given

Lower Temperature [tex]T_L=294 \K[/tex]

Higher Temperature [tex]T_H=522 \K[/tex]

Maximum Possible efficiency is achieved when the engine works as carnot Engine

i.e. [tex]\eta _{max}=1-\frac{T_L}{T_H}[/tex]

[tex]\eta_{max}=1-\frac{294}{522}[/tex]

[tex]\eta _{max}=\frac{228}{522}=0.436[/tex]

[tex]\eta _{max}=43.64\ %[/tex]

The maximum possible efficiency of a heat engine can be determined using the temperatures of the hot and cold reservoirs in the formula Effc = 1 - Tc / Th. Applying this formula to the given temperatures of 294 K and 552 K results in a maximum efficiency of 46.8%.

The maximum possible efficiency of a heat engine (also known as the Carnot efficiency) can be calculated using the temperatures of the heat source (hot reservoir) and the heat sink (cold reservoir). This efficiency can be determined by using the formula Effc = 1 - Tc / Th, where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. Both these temperatures should be in Kelvin.

In the given problem, Tc is 294 K and Th is 552 K. Substituting these values into the formula gives the maximum possible efficiency:

Effc = 1 - 294 / 552 = 0.468. Thus, the maximum possible efficiency of this engine is approximately 46.8%.

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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]

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.

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:

98.9 kPa

Explanation:

given,

Pressure reading of barometer = 742 mm Hg

we know,

760 mm Hg = 1.013 x 10⁵ Pa

[tex]1\ mm\ Hg = \dfrac{1.013\times 10^5}{760}[/tex]

[tex]742\ mm\ Hg = \dfrac{1.013\times 10^5}{760}\times 742[/tex]

                           = 0.989 x 10⁵ Pa

                          = 98.9 x 10³ Pa

                          = 98.9 kPa

the reading of the barometer is equal to  98.9 kPa

The pressure reading from the barometer expressed in kilopascal is 98.9kPa.

Given that;

Pressure reading from the barometer; [tex]P = 742mmHg[/tex]

Pressure reading from the barometer in kilopascals; [tex]x = \ ?[/tex]

First we convert the units from Millimeter of Mercury (mmHg) to Pascal (Pa)

We are to use;

[tex]760 mm Hg = 1.013 * 10^5 Pa\\\\\frac{760mmHg}{760} = \frac{1.013 * 10^5 Pa}{760} \\\\1mmHg = 1.33289 * 10^2 Pa[/tex]

So,

Pressure reading is pascal

[tex]P = 742 * [1.33289*10^2Pa]\\\\P = 98900.438Pa[/tex]

Next we convert to kilopascal

We know that; [tex]1\ Pascal = 0.001\ kilopascal[/tex]

so

[tex]P = 98900.438 * 0.001 kPa\\\\P = 98.9kPa[/tex]

Therefore, the pressure reading from the barometer expressed in kilopascal is 98.9kPa.

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Final answer:

To find the particle's dynamics, we need to integrate the given acceleration function in terms of velocity and apply the initial conditions. The velocity function v(t) is √(400 - 2t), the position function s(t) would result from further integration, and the acceleration a(t) is -2√(400 - 2t).

Explanation:

A particle is moving in a straight line with its acceleration a(v) given as a(v) = (-2v) m/s², where v is the velocity in meters per second. To find the particle's position, velocity, and acceleration as functions of time, we'll need to integrate the acceleration function.

Finding the velocity as a function of time:

We have the acceleration in terms of velocity; thus, we can write:

'(v) = a(v) = -2v

(v) = -∫ 2v dv = -v² + C

When t=0, v = 20 m/s:

C = 20² = 400

v(t) = √(400 - 2t)

Finding the particle's position as a function of time:

s(t) = ∫ v(t) dt

s(t) = ∫ √(400 - 2t) dt

The integration will give us the position, s(t), in terms of t, which needs to be integrated carefully using appropriate techniques such as substitution.

Finding the acceleration as a function of time:

a(t) can be found by substituting the expression for v(t) into a(v), which gives us a(t) = -2√(400 - 2t).

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]

Answer:

t=320s

Explanation:

Given Data

Boat speed=20 m/s

River flows=5 m/s

Total trip of distance d=3.0km = 3000m

To find

Total time taken

Solution

As

[tex]Velocity=distance/time\\time=distance/velocity\\[/tex]

Here we have two conditions

First when boat moves upward and the river pushing back.then velocity is given as

velocity=20m/s-5m/s

velocity=15 m/s

Time for that velocity

[tex]t_{1} =distance/velocity\\t_{1}=\frac{3000m}{15m/s}\\ t_{1}=200s[/tex]

Now for second condition when river flows and boat speed on same direction

velocity=20m/s+5m/s

velocity=25 m/s

Time taken for that velocity

[tex]t_{2}=distance/velocity\\t_{2}=\frac{3000m}{25m/s}\\ t_{2}=120m/s[/tex]

Now the total time

[tex]t=t_{1}+t_{2}\\t=(200+120)s\\t=320s[/tex]

Answer:

F=1.4384×10⁻¹⁹N

Explanation:

Given Data

Charge q= -8.00×10⁻¹⁷C

Distance r=2.00 cm=0.02 m

To find

Electrostatic force

Solution

The electrostatic force between between them can be calculated from Coulombs law as

[tex]F=\frac{kq^{2} }{r^{2} }[/tex]

Substitute the given values we get

[tex]F=\frac{(8.99*10^{9} )*(-8.00*10^{-17} )^{2} }{(0.02)^{2} }\\ F=1.4384*10^{-19} N[/tex]

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

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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