PART ONE
A ladder rests against a vertical wall. There
is no friction between the wall and the ladder.
The coefficient of static friction between the
ladder and the ground is µ = 0.464 .
(USE THE PICTURE TO DETERMINE THE ANSWER)
Identify the set of equations which is correct.
ANSWER CHOICES:
1. A1, B2, C3
2. A2, B2, C1
3. A1, B1, C1
4. A1, B2, C2
5. A1, B1, C2
6. A2, B1, C3
7. A2, B1, C2
8. A1, B2, C1
9. A1, B1, C3
10. A2, B1, C1

PART TWO
Determine the smallest angle θ for which the
ladder remains stationary.
Answer in units of ◦

Answer:

0.47 m/s²

Explanation:

Assuming the string is inelastic, m₃ will accelerate downward at a rate of -a, and m₄ will accelerate upward at a rate of +a.

Draw a four free body diagrams, one for each hanging mass and one for each wheel.

For m₃, there are two forces: weight force m₃g pulling down, and tension force T₃ pulling up.  Sum of forces in the +y direction:

∑F = ma

T₃ − m₃g = m₃(-a)

For m₄, there are two forces: weight force m₄g pulling down, and tension force T₄ pulling up.  Sum of forces in the +y direction:

∑F = ma

T₄ − m₄g = m₄a

For m₁, there are two forces: tension force T₃ pulling down, and tension force T pulling right.  Sum of the torques in the counterclockwise direction:

∑τ = Iα

T₃r₃ − Tr₃ = (m₁r₃²) (a/r₃)

T₃ − T = m₁a

For m₂, there are two forces: tension force T₄ pulling down, and tension force T pulling left.  Sum of the torques in the counterclockwise direction:

∑τ = Iα

Tr₄ − T₄r₄ = (m₂r₄²) (a/r₄)

T − T₄ = m₂a

We now have 4 equations and 4 unknowns.  Let's add the third and fourth equations to eliminate T:

(T₃ − T) + (T − T₄) = m₁a + m₂a

T₃ − T₄ = (m₁ + m₂) a

Now let's subtract the second equation from the first:

(T₃ − m₃g) − (T₄ − m₄g) = m₃(-a) − m₄a

T₃ − m₃g − T₄ + m₄g = -(m₃ + m₄) a

T₃ − T₄ = (m₃ − m₄) g − (m₃ + m₄) a

Setting these two expressions equal:

(m₁ + m₂) a = (m₃ − m₄) g − (m₃ + m₄) a

(m₁ + m₂ + m₃ + m₄) a = (m₃ − m₄) g

a = (m₃ − m₄) g / (m₁ + m₂ + m₃ + m₄)

Plugging in values:

a = (1.64 kg − 1.27 kg) (9.8 m/s²) / (2.5 kg + 2.3 kg + 1.64 kg + 1.27 kg)

a = 0.47 m/s²

The difference in torque and mass applied determines the acceleration of

the system.

The acceleration is approximately 0.4703 m/s²

Reasons:

The mass of the left wheel = 2.5 kg

Radius of the left wheel = 24.03 cm

Mass of the right wheel = 2.3 kg

Radius of the right wheel = 31.38 cm

Mass on the left = 1.64 kg

Mass on the right = 1.27 kg

Acceleration due to gravity, g ≈ 9.8 m/s²

Required:

The acceleration of the system

Solution:

The given acceleration of the

T₃ - m₃·g = m₃·(-a)...(1)

T₄ - m₄·g = m₄·a...(2)

[tex]T_3 \cdot r_1 - T \cdot r_1 = I \cdot \alpha = m_1 \cdot r_1^2 \times \dfrac{a}{r_1} = \mathbf{m_1 \cdot r_1 \cdot a}[/tex]

T₃ - T = m₁·a...(3)

[tex]T \cdot r_2 - T_4 \cdot r_2 = \mathbf{ I \cdot \alpha} = m_2 \cdot r_2^2 \times \dfrac{a}{r_2} = m_2 \cdot r_2 \cdot a[/tex]

T - T₄ = m₂·a...(4)

Add equation (3) to equation (4) gives;

T₃ - T + (T - T₄) = T₃ - T₄ = m₁·a + m₂·a

Subtracting equation (2) from equation (1) gives;

(T₃ - m₃·g) - (T₄ - m₄·g) = m₃·a + m₄·a

T₃ - T₄ = -m₃·a - m₄·a - (m₄·g - m₃·g)

Which gives;

m₁·a + m₂·a = -m₃·a - m₄·a - (m₄·g - m₃·g)

a·(m₁ + m₂) = -a·(m₃ + m₄) - (m₄·g - m₃·g)

-(m₄·g - m₃·g) = (m₃·g - m₄·g) = a·(m₃ + m₄) + a·(m₁ + m₂) = a·(m₃ + m₄ + m₁ + m₂)

[tex]a = \mathbf{\dfrac{m_3 \cdot g - m_4 \cdot g}{(m_3 + m_4 + m_1 - m_2)}}[/tex]

[tex]a = \dfrac{1.64 \times 9.8 - 1.27\times 9.8}{(1.27 +1.64 + 2.5 + 2.3)} \approx 0.4703[/tex]

The acceleration is a ≈ 0.4703 m/s²

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Answer:

Tension must a rope be stretched is 36.8828 N

Explanation:

Wave speed in term of wavelength:

[tex]v=f. \lambda[/tex]

Where:

f is the frequency

[tex]\lambda[/tex] is the wavelength

Now:

[tex]v=36*0.790\\v=28.44 m/s[/tex]

Wave speed in term of tension force and mass per unit length

[tex]v=\sqrt{\frac{F}{Mass\ per\ unit\ length}}[/tex]                          Eq (1)

Where:

F is the tension force

[tex]Mass\ per\ unit\ length=\frac{0.105}{2.30} \\Mass\ per\ unit\ length=0.0456 Kg/m\\[/tex]

Since  [tex]v[/tex] is calculated above.On rearranging Eq (1) we will get:

[tex]F=v^2 *Mass\ per\ unit\ length\\F=(28.44)^2*0.0456\\F=36.8828 N[/tex]

Tension must a rope be stretched is 36.8828 N

Answer:

c. 1.2×10¹⁵ Hz

Explanation:

Work Function: This is the minimum amount of energy a photon requires to liberate an electron from the surface of a metal.

Mathematically, it can be represented as

E' = hf' ................................... Equation 1

Where E' = work function of gold, f' = minimum frequency ( threshold frequency), h = Planck's constant

Making f' the subject of the equation,

f' = E'/h................................. Equation 2

Given: E' = 4.8 ev = 4.8×1.6×10⁻¹⁹ J = 7.68×1.6×10⁻¹⁹ J, h = 6.626×10⁻³⁴Js.

Substituting into equation 2

f' = 7.68×10⁻¹⁹/ 6.626×10⁻³⁴

f' = 1.16×10¹⁵ Hz.

f' ≈ 1.2×10¹⁵ Hz

Thus the minimum frequency =  1.2×10¹⁵ Hz

The right option is c. 1.2×10¹⁵ Hz

Using Planck's equation, the minimum frequency of photons required to remove an electron from gold, given a work function of 4.8 eV, is approximately 1.15 * 10^15 Hz or 1.15 PHz.

The question is asking for the minimum frequency of photons capable of removing electrons from gold. This is related to the photoelectric effect, where electrons can be ejected from a metal surface by incident light. In this process, energy of the incoming light is absorbed by the electron, which can then be ejected if the light's energy is greater than the binding energy of the electron, which is also known as the work function.

The work function for gold is given as 4.8 eV. To calculate the minimum frequency, we need to employ Planck's equation, E = hf, which implies frequency, f = E/h. Given that E is the energy of the photon which must be equal to the work function (4.8 eV or 4.8 * 1.6 x 10^-19 J) and h is Planck’s constant (6.626 x 10^-34 J s), we can solve for f. The calculation gives approximately 1.15 * 10^15 Hz or 1.15 PHz, which is not listed among the provided answer choices.

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Answer:

Explanation:

(a). The line used to specify the orientation of the plane of paper is the line normal to the plane of sheet of paper

(b). The direction of the vector represents the normal to the  lat surface while the Magnitude represents the area of flat surface.

(c). Say the area of each smaller square is 1 square unit, then the area of graph paper is 64 square units. Direction of this area vector is given by a unit vector perpendicular to the graph sheet. If X and Y axes are in the plane of paper, then unit vector normal to the sheet of paper is K. Hence the complete vector is 64 K sq. units.

Area vector of each individual square is 1 squ. unit. where all these individual squares are parallel as vectors.

(d). Absolutely.

the entire sheet can be represented by a single vector. Its area vector is the sum of area vectors of three flat sides of triangular tube.

(e) NO.

Orientation of the individual squares is not the same for all squares. They cannot be represented by the same vector when compared to part C above, because they are in different directions even tough their magnitude are same.

(f) To represent a surface with a single area vector, divide the surface in to as many as possible flat pieces (if necessary infinitely large number of infinitesimally small pieces). Find the area vectors of all pieces. Add all the area vectors to obtain the single area vector resenting the complete surface.

But since the process can be done for any surface, any surface can be represented by a single area vector.

i hope this helps, cheers

The normal vector is perpendicular to the flat surface while the area vector is the direction in which the plane is embedded in 3 dimensions.

An area vector is an area (magnitude) with direction.

Therefore, the normal vector is perpendicular to the flat surface while the area vector is the direction in which the plane is embedded in 3 dimensions.

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To solve this problem we will apply the linear motion kinematic equations. Just as we will also find the relative speed of the body through the conservation of momentum. Our data is given as

[tex]M = 157kg[/tex]

[tex]m = 46kg[/tex]

[tex]v_1 = 1.48m/s[/tex]

PART A)

From the conservation of momentum,

[tex]\text{Momentum of Plank+girl}+\text{Mometum of girl} = 0[/tex]

[tex](M+m)v_2+mv_1 = 0[/tex]

[tex](M+m)v_2 = -m_v1[/tex]

[tex]v_2 = \frac{-Mv_1}{M+m}[/tex]

[tex]v_2 = \frac{-(46)(1.48)}{(157+46)}[/tex]

[tex]v_2 = -0.3353m/s[/tex]

Since the ice surface is frozen lake and girl is moving on it so the relative velocity will get added. Therefore the velocity of the girl relative to the ice surface is as,

[tex]v_1+v_2 = 1.48+(-0.33353)[/tex]

[tex]v_1+v_2 = 1.14647m/s[/tex]

The velocity of the girl relative to the ice surface is 1.14647m/s

PART B) The velocity of the plank plus girl is [tex]v_2 = -0.3353m/s[/tex]

Since the ice surface is frozen lake and plank is moving with girl on it so the relative velocity will get added. Therefore the velocity of the plank relative to the ice surface is as:

[tex]v_2 = -0.3353m/s[/tex]

"The correct answers are: (a) The girl's velocity relative to the surface of ice is 1.48m/s. (b) The velocity of the plank relative to the surface of ice is 0 m/s.

 (a) Since the girl is walking on the plank with a constant velocity of 1.48 m/s relative to the plank, and the plank itself is at rest on the frictionless ice surface, there are no external forces acting on the girl-plank system in the horizontal direction.

According to the principle of inertia, an object at rest will stay at rest, and an object in motion will stay in motion with a constant velocity unless acted upon by an external force. Therefore, the girl's velocity relative to the surface of ice is the same as her velocity relative to the plank, which is 1.48 m/s.

(b) The plank is initially at rest on the frictionless ice surface, and since there are no external forces acting on it in the horizontal direction, it will remain at rest relative to the ice surface. This means that the velocity of the plank relative to the surface of ice is 0 m/s. The girl walking on the plank does not affect the plank's velocity because her motion is internal to the girl-plank system, and there is no friction to transfer her momentum to the plank.

In summary, the girl's motion is relative to the plank, and since the plank remains stationary on the ice, her velocity relative to the ice is the same as her walking velocity on the plank. The plank itself does not move because of the lack of friction and external forces, thus its velocity relative to the ice remains 0 m/s."

Answer:

The electrostatic force between and electron and a proton is [tex]F=2.30\times 10^{-20}\ N[/tex]

Explanation:

It is given that, charge [tex]q_1[/tex] is placed at a distance [tex]r_o[/tex] from charge [tex]q_2[/tex]. The force acting between charges is given by :

[tex]F=\dfrac{kq_1q_2}{r_o^2}[/tex]

We need to find the force if the distance between them is reduced to [tex]r_o/4[/tex]. It is given by :

[tex]F'=\dfrac{kq_1q_2}{(r_o/4)^2}[/tex]

[tex]F'=16\times \dfrac{kq_1q_2}{r_o^2}[/tex]

[tex]F'=16\times F[/tex]

So, if the the distance between them is reduced to [tex]r_o/4[/tex], the new force becomes 16 times of the previous force.

The electrostatic force between and electron and a proton separated by 0.1 mm or [tex]10^{-4}\ m[/tex] is :

[tex]F=\dfrac{kq_1q_2}{r_o^2}[/tex]

[tex]F=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(10^{-4})^2}[/tex]

[tex]F=2.30\times 10^{-20}\ N[/tex]

So, the electrostatic force between and electron and a proton is [tex]F=2.30\times 10^{-20}\ N[/tex]. Hence, this is the required solution.

When the distance between two charges is reduced, the magnitude of the force on one charge due to the other increases. This is because the electrostatic force is inversely proportional to the square of the distance between the charges.

When the distance between two charges, q1 and q2, is reduced from r0 to r0/4, the magnitude of the force on q1 due to q2 increases. This is because the electrostatic force is inversely proportional to the square of the distance between the charges.

For example, if we have charges q1 = 2 C and q2 = 4 C, and the original distance r0 = 2 m, the force between them would be F = k × (q1 × q2) / r0^2 = k × (2 × 4) / (2^2) = k × 4, where k is the Coulomb's constant.

If we reduce the distance to r0/4 = 0.5 m, the force would become F' = k × q1 × q2) / (r0/4)^2 = k × 4) / (0.5^2) = k ×64, which is 16 times greater than the original force.

Answer:

544.68604 rad/s²

Explanation:

m = Mass of disk = 1.06 kg

R = Radius of disk = 0.433 m

T = Tension

[tex]T_2[/tex] = 167 N

[tex]T_1[/tex] = 42 N

Moment of inertia is given by

[tex]I=\dfrac{1}{2}mR^2\\\Rightarrow I=\dfrac{1}{2}\times 1.06\times 0.433^2[/tex]

The resultant torque of the system will be given by

[tex](T_2-T_1)R=\tau\\\Rightarrow (T_2-T_1)R=I\alpha\\\Rightarrow \alpha=\dfrac{(T_2-T_1)R}{I}\\\Rightarrow \alpha=\dfrac{(167-42)\times 0.433}{\dfrac{1}{2}\times 1.06\times 0.433^2}\\\Rightarrow \alpha=544.68604\ rad/s^2[/tex]

The angular acceleration of the disk is 544.68604 rad/s²

The change in the velocity with respect to time is called acceleration.

The acceleration depends on the following:-

According to the question, the data is as follows:-

m  = 1.06 kg

R = 0.433 m

T = Tension, T1 = 167 N , T2= 42 N

To calculate we will use the formula of the moment of inertia i.e

[tex]I =\frac{1}{2}mr^2[/tex]

After putting the value,

[tex]I = \frac{1}{2} *1.06*0.433^2[/tex]

The resultant torque of the system will be given by

[tex]torque = (T_2-T_1)R[/tex]

[tex]Ia = (T_2-T_1)R[/tex]

[tex]a =\frac{(T_2-T_1)R}{I}[/tex]

[tex]a= \frac{(167-42)*0.433}{\frac{1}{2}*1.06*0.433^2}[/tex]

After solving the equation, a is = 544.68604

Hence, The angular acceleration of the disk is 544.68604 rad/s²

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Answer:

Avg.velocity=(Δy/ Δt)  =(net displacement/ total time for journey)

Δy = 0

Δt = t

so avg. velocity = 0/t =0

Avg. speed =(total distance traveled/ total time for journey)

total distance = up +down = Ymax+Ymax= 2 Ymax

total time = t

avg. speed = 2 Ymax/t

Explanation:

Since there is no net displacement from the original position,velocity is zero. Claudia is right!

while it covered some distance in time t so its speed is not as qouted by Hossien

Answer:

To find the diameter of the wire, when the following are given:

Resistivity of the material (Rho), Current flowing in the conductor, I, Potential difference across the conductor ends, V, and length of the wire/conductor, L.

Using the ohm's law,

Resistance R = (rho*L)/A

R = V/I.

Crossectional area of the wire A = π*square of radius

Radius = sqrt(A/π)

Diameter = Radius/2 = [sqrt(A/π)]

Making A the subject of the formular

A = (rho* L* I)V.

From the result of A, Diameter can be determined using

Diameter = [sqrt(A/π)]/2. π is a constant with the value 22/7

Explanation:

Error and uncertainty can be measured varying the value of the parameters used and calculating different values of the diameters. Compare the values using standard deviation

The diameter of the wire can be derived from the resistivity, length, current, and voltage using Ohm's Law and the equation for the resistance of a wire. To find the uncertainty in the diameter, use error propagation techniques considering the uncertainties of the resistivity, length, and (voltage/current).

The diameter of the wire can be found using Ohm's Law and the formula for the resistance of a wire. The guiding formula is R = V/I (Ohm’s Law), where R is the resistance, V is the voltage, and I is the current.

The formula for resistance of a wire is R = ρL/A, where A (cross-sectional area) is π(d/2)^2 because the wire is cylindrical. Equating these two equations gives V/I = ρL/π(d/2)^2. Solving this for d (diameter) gives d = sqrt((4ρL(V/I))/π). To calculate the measurement error or uncertainty in the diameter of the wire, one would need to use error propagation techniques.

If the uncertainties in ρ, L, V/I are δρ, δL and δ(V/I) respectively, then the uncertainty in d, δd is given by δd = 1/2 [((4δρρ)/(ρL(V/I)))^2+ ((4δL/L)/(ρL(V/I)))^2+ ((4δ(V/I)/(V/I))/(ρL(V/I)))^2]^1/2.

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Answer:

[tex]\lambda_w=0.6509\ m[/tex]

Explanation:

Given:

Now the angular frequency of the spring oscillation:

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

[tex]\omega=\sqrt{\frac{9.3}{13.9} }[/tex]

[tex]\omega=0.81796\ rad.s^{-1}[/tex]

Now according to the question the wave is created after each cycle of the spring oscillation.

So the time period of oscillation:

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

[tex]T=\frac{0.81796}{2\pi}[/tex]

[tex]T=0.130182\ s[/tex]

Now the wave length of the water wave:

[tex]\lambda_w=v_w.T[/tex]

[tex]\lambda_w=5\times 0.130182[/tex]

[tex]\lambda_w=0.6509\ m[/tex]

Answer:

Wavelength will be 38.388 m

Explanation:

We have given mass m = 13.9 kg

Spring constant K= 9.3 N/m

Velocity v = 5 m /sec

Angular frequency is given by [tex]\omega =\sqrt{\frac{k}{m}}[/tex]

So [tex]\omega =\sqrt{\frac{9.3}{13.9}}=0.817[/tex]

Now we have to find frequency for further calculation

So frequency will be equal to [tex]f=\frac{\omega }{2\pi }=\frac{0.817}{2\times 3.14}=0.130Hz[/tex]

Now we have to find wavelength, it is ratio of velocity and frequency

There is a relation between frequency velocity and wavelength

[tex]v=f\lambda[/tex]

[tex]\lambda =\frac{v}{f}=\frac{5}{0.130}=38.388m[/tex]

Answer:

24.48 days

37.7 days

Explanation:

r = Radius

s denotes satellite

C denotes Charon

Time period is given by

[tex]T=\dfrac{2\pi r^{1.5}}{\sqrt{2GM}}[/tex]

So,

[tex]T\propto r^{1.5}[/tex]

[tex]\dfrac{T_C}{T_{s1}}=\dfrac{r_c^{1.5}}{r_{s1}^{1.5}}\\\Rightarrow T_{s1}=\dfrac{T_Cr_{s1}^{1.5}}{r_C^{1.5}}\\\Rightarrow T_{s1}=\dfrac{6.39\times 86400\times {48000000}^{1.5}}{19600000^{1.5}}\\\Rightarrow T_{s1}=2115886.41242\ s\\\Rightarrow T_{s1}=24.48\ days[/tex]

The time period of the first satellite is 24.48 days

[tex]T_{s2}=\dfrac{T_Cr_{s2}^{1.5}}{r_C^{1.5}}\\\Rightarrow T_{s2}=\dfrac{6.39\times 86400\times {64000000}^{1.5}}{19600000^{1.5}}\\\Rightarrow T_{s2}=3257620.23942\ s\\\Rightarrow T_{s2}=37.7\ days[/tex]

The time period of the second satellite is 37.7 days

Answer:

x = 150.0 m

Explanation:

If the acceleration is constant, we can find the value of the acceleration, starting from rest, applying the following kinematic equation:

x = 1/2*a*t² = 6.0 m

Solving for a:

a = 2*x / t² = 2*6.0 m / 1.0s² = 12 m/s²

Now, during the following 4.0 s, the car continues moving with this acceleration, but its initial velocity is not zero anymore, but the speed at 1.0 s, which is just 12 m/s (as it accelerates 12 m/s each second), so we can write again the same kinematic equation, taking into account that initial velocity for the second part, as follows:

x = x₀ + v₀*t + 1/2*a*t² = 6.0 m + 12m/s*4.0s + 1/2*12 m/s²*(4.0)s² = 150.0 m

⇒ x = 150.0 m

Answer:

L= 1 m,   ΔL = 0.0074 m

Explanation:

A clock is a simple pendulum with angular velocity

         w = √ g / L

Angular velocity is related to frequency and period.

         w = 2π f = 2π / T

We replace

        2π / T = √ g / L

        T = 2π √L / g

We will use the value of g = 9.8 m / s², the initial length of the pendulum, in general it is 1 m (L = 1m)

With this length the average time period is

           T = 2π √1 / 9.8

           T = 2.0 s

They indicate that the error accumulated in a day is 15 s, let's use a rule of proportions to find the error is a swing

           t = 1 day (24h / 1day) (3600s / 1h) = 86400 s

         e= Δt = 15 (2/86400) = 3.5 104 s

The time the clock measures is

           T ’= To - e

           T’= 2.0 -0.00035

           T’= 1.99965 s

Let's look for the length of the pendulum to challenge time (t ’)

           L’= T’² g / 4π²

           L’= 1.99965 2 9.8 / 4π²

           L ’= 0.9926 m

Therefore the amount that should adjust the length is

           ΔL = L - L’

           ΔL = 1.00 - 0.9926

           ΔL = 0.0074 m

Answer:

[tex]8.90392\times 10^{-9}\ W[/tex]

[tex]5.36518\times 10^{-9}\ W[/tex]

Explanation:

h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

t = Time taken = 3.8 ms

[tex]\lambda[/tex] = Wavelength

n = Number of protons = [tex]8\times 10^7[/tex]

Power is given by

[tex]P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{nh\dfrac{c}{\lambda}}{t}\\\Rightarrow P=\dfrac{8\times 10^7\times 6.626\times 10^{-34}\times \dfrac{3\times 10^8}{470\times 10^{-9}}}{3.8\times 10^{-3}}\\\Rightarrow P=8.90392\times 10^{-9}\ W[/tex]

The power is [tex]8.90392\times 10^{-9}\ W[/tex]

[tex]P=\dfrac{nh\dfrac{c}{\lambda}}{t}\\\Rightarrow P=\dfrac{8\times 10^7\times 6.626\times 10^{-34}\times \dfrac{3\times 10^8}{780\times 10^{-9}}}{3.8\times 10^{-3}}\\\Rightarrow P=5.36518\times 10^{-9}\ W[/tex]

The power is [tex]5.36518\times 10^{-9}\ W[/tex]

Answer:

Resultant velocity will be equal to 6.10 m/sec

Explanation:

We have given a motorbike is traveling with 5 m/sec in east

And a current is flowing at a rate of 3.5 m /sec in north

We know that east and north is perpendicular to each other

So resultant velocity will be vector sum of both velocity

So resultant velocity [tex]v=\sqrt{5^2+3.5^2}=6.10m/sec[/tex]

So resultant velocity will be equal to 6.10 m/sec

Answer:

a)   v_average = 11 m / s, b)  t = 0.0627 s

, c)    F = 7.37 10⁵ N

, d)   F / W = 35.8

Explanation:

a) truck speed can be found with kinematics

         v² = v₀² - 2 a x

The fine speed zeroes them

           a = v₀² / 2x

           a = 22²/2 0.69

           a = 350.72 m / s²

The average speed is

           v_average = (v + v₀) / 2

           v_average = (22 + 0) / 2

           v_average = 11 m / s

b) The average time

          v = v₀ - a t

          t = v₀ / a

          t = 22 / 350.72

          t = 0.0627 s

c) The force can be found with Newton's second law

             F = m a

             F = 2100 350.72

             F = 7.37 10⁵ N

.d) the ratio of this force to weight

             F / W = 7.37 10⁵ / (2100 9.8)

             F / W = 35.8

.e) Several approaches will be made:

- the resistance of air and tires is neglected

- It is despised that the force is not constant in time

- Depreciation of materials deformation during the crash

the magnitude of the charge of object A is[tex]\( 2.78 \times 10^{-10} \, \text{C} \).[/tex]

We can use the formula for electric field intensity ( E ) to calculate the charge of object A. The electric field intensity is given by:

[tex]\[ E = \frac{k \cdot |q|}{r^2} \][/tex]

Where:

- ( E ) is the electric field intensity (40.0 N/C),

- ( k) is Coulomb's constant [tex](\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),[/tex]

- [tex]\( |q| \)[/tex] is the magnitude of the charge on object A (what we're trying to find), and

- ( r ) is the distance from object A to point P (0.250 m).

First, rearrange the formula to solve for[tex]\( |q| \)[/tex]:

[tex]\[ |q| = \frac{E \cdot r^2}{k} \][/tex]

Now, substitute the given values into the formula:

[tex]\[ |q| = \frac{40.0 \, \text{N/C} \cdot (0.250 \, \text{m})^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2} \]\[ |q| = \frac{40.0 \times 0.0625}{8.99 \times 10^9} \, \text{C} \]\[ |q| = \frac{2.5}{8.99 \times 10^9} \, \text{C} \]\[ |q| = 2.78 \times 10^{-10} \, \text{C} \][/tex]

So, the magnitude of the charge of object A is[tex]\( 2.78 \times 10^{-10} \, \text{C} \).[/tex]

To solve this problem we will proceed to use Newton's second law for which the mass (m) multiplied by the acceleration (a) is defined as the Force (F) applied on a body, mathematically that is,

[tex]F = ma[/tex]

According to the statement for the first object, the acceleration is,

[tex]a = \frac{F}{m} \rightarrow 1^{st} Object[/tex]

For the second object the acceleration is,

[tex]8a = \frac{3F}{m_2} \rightarrow 2^{nd} Object[/tex]

Solving for the mass of the second object,

[tex]m_2 = \frac{3F}{8a}[/tex]

[tex]m_2 = \frac{3F}{8(F/m)}[/tex]

[tex]m_2 = \frac{3}{8}m[/tex]

Therefore the correct answer is D.

The question is about calculating the electric force on a point charge near an infinitely long wire using the concepts of electric fields and Coulomb's law within the domain of Physics.

The question involves calculating the magnitude of the electric force that an infinitely long, thin wire exerts on a point charge placed at a certain distance from it. This falls under the subject of Physics, specifically dealing with electrostatics and the concept of electric fields and forces.

First, we need to find the electric field due to the wire at the location of the point charge. The electric field E from an infinitely long wire is given by the formula E = (2kλ)/r, where k is Coulomb's constant (8.99 × 109 Nm2/C2), λ is the linear charge density of the wire, and r is the distance from the wire to the point where the electric field is being calculated.

After calculating E, we can find the force exerted on the charge q using F = qE. Since we're given q = -4.00 nC and λ = 3.00×10−9C/m, with r = 9.00 cm, we can substitute these values into the formulas to calculate the electric field and then the force acting on the point charge.

The magnitude of the electric force exerted by the wire on the point charge is 2.40 × 10⁻⁶ N.

To find the magnitude of the electric force that an infinitely long, thin wire with linear charge density λ = 3.00×10⁻⁹ C/m exerts on a point charge q = -4.00 nC placed a distance of 9.00 cm from the wire, we use the principle that an electric field E is created by a line of charge.

The electric field created by an infinitely long, thin wire is given by:

E =  (2kλ)/r

where  k = 8.99 × 10⁹ Nm²/C²

            λ = 3.00×10⁻⁹ C/m , charge density

             r = 9 × 10⁻² m

Substitute the given values:

E = (2 ×  8.99 × 10⁹ Nm²/C² × 3.00×10⁻⁹ C/m ) / 9 × 10⁻² m

Calculating this, we get:

E ≈ 600 N/C

The force F on the point charge q due to the electric field E is given by:

F = q × E

Since q = -4.00 nC = -4.00 × 10⁻⁹ C:

F = (-4.00 × 10⁻⁹ C) × 600 N/C ≈ -2.40 × 10⁻⁶ N

Therefore, the magnitude of the electric force is 2.40 × 10⁻⁶N.

Answer:

Electric field magnitude

E = K/qd

Where

K = kinetic energy of electron

d = electron distance

q = charge

Explanation:

Given the relationship between workdone and energy

Work-energy theorem:

Net workdone = Energy change

W = ∆E

In this case

W = ∆K.E

And,

∆K.E = K(final) - K(initial)

To stop the kinetic energy | K(final) = 0

K(initial) = K (given)

∆K.E = 0 - K = -K

Let the electric force on the electron has magnitude F.

And

W = -Fd = ∆K.E = -K

-Fd = -K

F = K/d .....1

The magnitude of the electric field E that can stop these electron in a distance d:

E = F/q ......2

Where q is the charge on electron.

substituting equation 1 to 2

E = (K/d)/q = K/qd

E = K/qd

Answer:

a)   [tex] = 9.30\times 10^{-3} N[/tex]

b) [tex]= 2.49\times 10^{-3} N[/tex]

Explanation:

Given data;

magnetic field [tex]= 0.60 \times 10^{-4}  T[/tex]

current I = 11 A

length of wire L = 14.6 m

Angle[tex] \theta = 75 A[/tex]

a) Magnetic force due to current

[tex]F = BIL sin \theta[/tex]

 [tex] = 0.60 \times 10^{-4} \times 11 \times 14.6 sin 75^o[/tex]

 [tex] = 9.30\times 10^{-3} N[/tex]

B) magnitude of force due upward current direction

[tex]F = BIL sin \theta[/tex]

  [tex]= 0.60 \times 10^{-4} \times 11 \times 14.6 sin (75^o + 90^o)[/tex]

[tex]= 2.49\times 10^{-3} N[/tex]

The magnetic force on a wire depends on the direction of the current relative to the Earth's magnetic field. In scenario (a), the force is calculated using an angle of 165° between the wire and magnetic field. For (b), the angle is 75°, and the force's magnitude is computed with the sin of that angle.

Magnetic Force on a Current-Carrying Wire

The magnetic force on a current-carrying wire is given by F = I x L x B x sin(θ), where I is the current, L is the length of the wire, B is the magnetic field strength, and θ is the angle between the wire and the direction of the magnetic field.

a) When the current is directed horizontally toward the east and the Earth's magnetic field is pointing 75° below the horizontal in a north-south plane, we can calculate the force using the cross product of the current direction and magnetic field. The angle between the wire and the magnetic field is 90° + 75° = 165°. Therefore, the magnitude of the force is F = 11 A * 14.6 m * 0.60 * 10^-4 T * sin(165°) and its direction is perpendicular to both the current and the magnetic field, following the right-hand rule.

b) If the current is directed vertically upward, the force is perpendicular to the wire and the magnetic field. The angle between the current direction and the magnetic field is now 75°, so the magnitude of the force is F = 11 A * 14.6 m * 0.60 * 10^-4 T * sin(75°), and its direction is again determined by the right-hand rule.

To solve this problem we will apply the concept related to the conservation of the Momentum. We will then start considering that the amount of initial momentum must be equal to the amount of final momentum. Considering that all the objects at the initial moment have the same initial velocity (Zero, since they start from rest) the final moment will be equivalent to the multiplication of the mass of each object by the velocity of each object, so

Initial Momentum = Final Momentum

[tex](m_B+m_1+m_2)v_i = m_1v_1+m_2v_2+m_Bv_B[/tex]

Here,

[tex]m_B[/tex] =  mass of Raft

[tex]m_1[/tex] = Mass of swimmers 1

[tex]m_2[/tex] = Mass of swimmers 2

[tex]v_i[/tex] = Initial velocity (of the three objects)

[tex]v_B[/tex] = Velocity of Raft

Replacing,

[tex](199+52+70)*0 = (52)(4)+(70)(-4)+199v_B[/tex]

Solving for [tex]v_B[/tex]

[tex]vB = \frac{72}{199}[/tex]

[tex]v_B = 0.3618m/s[/tex]

Therefore the velocity the rarft start to move is 0.3618m/s

To solve this problem we will apply the concepts related to the magnitude of the electric field. Defined as the proportion of the electric potential per unit of distance, that is,

[tex]E = \frac{V}{d}[/tex]

Where E is the magnitude of the electric field between the plates, V is the potential difference between the plates, and d is the separation of the plates

The breakdown field of air is [tex]3.0*10^6[/tex] N/C.

Replacing we have that the Potential is

[tex]V = Ed[/tex]

[tex]V = (3.0*10^6)(1.89*10^{-3})[/tex]

[tex]V = 5670V[/tex]

Therefore the potential difference between you and your friend just before the spark is 5670V

Answer:

[tex]\Delta E=58800\ J[/tex]

Explanation:

Given:

mass of lead chunk, [tex]m=200\ kg[/tex]

height of the fall, [tex]h= 30\ m[/tex]

So, increase in the internal energy of the system after the collision is :

[tex]\Delta E=m.g.h[/tex]

[tex]\Delta E=200\times 9.8\times 30[/tex]

[tex]\Delta E=58800\ J[/tex]

Answer: On the basis of speed they are all equivalent.

Yellow light = Fm radio wave = Green light = X-ray = AM radio wave = Infrared wave

Explanation:

Yellow light, Fm radio wave, Green light ,X-ray, AM radio wave and Infrared wave are all electromagnetic waves, and all electromagnetic waves move at the same vacuum speed which is the speed of light and is approximately 3.0x10^8 m/s.

They only differ in wavelength and frequency

c = λf

c (speed of light) = λ (wavelength) x f (frequency)

Therefore; on the basis of speed they are all equivalent.

Yellow light = Fm radio wave = Green light = X-ray = AM radio wave = Infrared wave

All electromagnetic waves, including yellow light, FM radio waves, green light, X-rays, AM radio waves, and infrared waves travel at the same speed in a vacuum, known as the speed of light, which is 3.00 × 10^8 m/s. Therefore, they are equivalent in speed.

The student has asked to rank various types of electromagnetic waves based on their speed in a vacuum. In a vacuum, all electromagnetic waves travel at the same speed, which is the speed of light, and is one of the fundamental constants of nature. The speed of light (c) in a vacuum is 3.00 × 108 meters per second (m/s).

Therefore, the ranking of the electromagnetic waves from fastest to slowest for yellow light, FM radio wave, green light, X-ray, AM radio wave, and infrared wave would be:

Since all of these waves travel at the same speed in a vacuum, they can be considered equivalent in terms of speed. However, they differ in wavelength and frequency.

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The weight of the measuring stick is balanced by the 1-kg rock suspended from it at the 0.75m mark according to the principle of torques. This results in the weight of the measuring stick being approximately 13.07 Newtons.

The problem you've asked falls right into the area of physics related to torques and forces. The 1-kg rock generates a force of approximately 9.8 N downwards (using the approximate gravity strength of 9.8 m/s^2). According to the principle of torques, for the stick to be balanced, the force at the 0.75m mark (1 - 0.25 = 0.75m), which is the weight of the stick, should balance the total torque generated by the weight of the rock. That means the weight of the stick (in Newtons) times 0.75 should equal the weight of the rock times 1. This gives us the equation: 0.75*W = 1*9.8, where W is the weight of the stick. Solving this gives us the weight of the stick as approximately 13.07 N.

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Answer:

q = square root (4KsL³/k)

The force of extension of the spring is equal to the force of repulsion between the two like charges. Two like charges(positive or negative) would always repel each other and two unlike charges would always attract each other. This electric force between the charges is what is responsible for the stretching of the spring. The electric force causes the spring to increase in length from L to 2L. Equating these forces, that is the electric force between the charges and the elastic force of the spring and rearranging the variables gives the expression to obtain q.

Explanation:

See the attachment below for full solution.

By equating the spring force with the electrostatic repulsion force at the new equilibrium length, we find that the charge on each end of the spring is [tex]q = 2\sqrt(ks ke l^3)[/tex]. This derivation uses Hooke's Law and Coulomb's Law for calculation.

To solve this problem, we need to understand the balance between the spring force and the electrostatic repulsion force. We start by looking at Hooke's Law for the spring force:

[tex]F_spr = -k_s (x - l)[/tex]

Where k_s is the spring constant, x is the stretched length (which is 2l in this case, since the equilibrium length doubles), and l is the original equilibrium length. The spring force is given by:

[tex]F_spr = -k_s (2l - l) \\= -k_s l[/tex]

The electrostatic repulsion force between two identical charges q separated by a distance 2l is given by Coulomb's Law:

[tex]F_e = \frac{k_e q^2}{(2l)^2} \\= \frac{k_e q^2}{4l^2}[/tex]

At equilibrium, the magnitudes of these two forces are equal:

[tex]k_s l = \frac{k_e q^2}{4l^2}[/tex]

Rearranging to solve for q:

[tex]q^2 = 4k_s k_e l^3[/tex]

[tex]q = \sqrt{4k_s k_e l^3} \\= 2\sqrt{k_s k_e l^3}[/tex]

So, each end of the spring must have a charge of[tex]q = 2\sqrt{k_s k_e l^3}.[/tex]

Answer:

(a). The weight of the system is 336.32 N.

(b).  The molar volume is 16.6 m³/k mol.

The mass based volume is 0.145 m³/kg.

Explanation:

Given that,

Weight of octane = 0.3 kmol

Volume = 5 m³

(a). Molecular mass of octane

[tex]M=114.28\ g/mol[/tex]

We need to calculate the mass of octane

Mass of 0.3 k mol of octane is

[tex]M=114.28\times0.3\times1000[/tex]

[tex]M=34.284\ kg[/tex]

We need to calculate the weight of the system

Using formula of weight

[tex]W=mg[/tex]

Put the value into the formula

[tex]W=34.284\times9.81[/tex]

[tex]W=336.32\ N[/tex]

(b). We need to calculate the molar volume

Using formula of molar volume

[tex]\text{molar volume}=\dfrac{volume}{volume of moles}[/tex]

Put the value into the formula

[tex]\text{molar volume}=\dfrac{5}{0.3}[/tex]

[tex]\text{molar volume}=16.6\ m^3/k mol[/tex]

We need to calculate the mass based volume

Using formula of mass based volume

[tex]\text{mass based volume}=\dfrac{volume}{mass}[/tex]

Put the value into the formula

[tex]\text{mass based volume}=\dfrac{5}{34.284}[/tex]

[tex]\text{mass based volume}=0.145\ m^3/kg[/tex]

Hence, (a). The weight of the system is 336.32 N.

(b).  The molar volume is 16.6 m³/k mol.

The mass based volume is 0.145 m³/kg.

Answer:

1.80 Meter

Explanation:

The bottom of a water (liquid) container appears to be raised up in comparison to its actual depth because of the refraction. How shallow will the bottom appear depends on the refractive index of the liquid.

The refractive index(n) and apparent and real depth are related as given below:

[tex]n = \frac{Real Depth}{Apparent Depth}[/tex]

[tex]Apparent Depth = \frac{Real Depth}{n}[/tex]

Given,

n = 1.333

Real depth = 2.40 m

[tex]Apparent Depth = \frac{2.40}{1.33}[/tex]

Thus, Apparent depth = 1.80 Meter

Answer:

[tex]3.61581\ Nm^2/C[/tex]

Explanation:

[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]

V = Volume of cube = [tex]0.04^3[/tex]

[tex]\rho[/tex] = Charge density = [tex]500\ nC/m^3[/tex]

Electric flux is given by

[tex]\phi=\dfrac{Q}{\epsilon_0}\\\Rightarrow \phi=\dfrac{\rho V}{\epsilon_0}\\\Rightarrow \phi=\dfrac{500\times 10^{-9}\times 0.04^3}{8.85\times 10^{-12}}\\\Rightarrow \phi=3.61581\ Nm^2/C[/tex]

The electric flux through the surface of the cube is [tex]3.61581\ Nm^2/C[/tex]

The electric flux through the surface of a cubical object inside a sphere can be found by using Gauss's Law. It is equal to the total charge enclosed by the cube, calculated by multiplying the charge density by the volume of the cube and then divided by the permittivity of free space. In this case, the electric flux is about 36.109 N.m^2/C.

The subject of this question is physics, specifically dealing with the concept of electric charges and electric fields. To find the electric flux through the cubical surface placed inside the sphere, we can use Gauss's Law.

According to Gauss's Law, the electric flux through a closed surface is equal to the total charge enclosed by the surface divided by permittivity constant (εo).

In this case, we first need to find the charge enclosed by the cube, which we can find by multiplying the charge density with the volume of the cube. Since the charge density is 500nC/m^3 and the volume is 4cm x 4cm x 4cm (converted to m^3), the enclosed charge is 32nC. Then, divide this by the permittivity of free space, resulting in about 36.109 N.m^2/C, which represents the electric flux through the surface of the cube.

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