Several students in the group made comments to the Coach about the back pack. Which student has the correct analysis of the forces and motion of the back pack?

Explanation:

Let the speed of river be r and speed of boat be b.

Distance traveled = 24 km

Time taken to travel 24 km with current = 2 hr

We have

          Distance = Speed x Time

          24 = (b + r) x 2

          b + r = 12 --------------------eqn 1

Time taken to travel 24 km against current = 3 hr

We have

          Distance = Speed x Time

          24 = (b - r) x 3

          b - r = 8 --------------------eqn 2

eqn 1 + eqn 2

          2b = 20

            b = 10 km/hr

Substituting in eqn 1

            10 + r = 12

                     r = 2 km/hr

Speed of boat in still water = 10 km/hr

Speed of current of river = 2 km/hr

The speed of the boat in still water is 10 km/h and the speed of the current is 2 km/h. This is a solved by using a system of linear equations with speed of the boat and the current as variables.

This problem is a case for a system of linear equations. Let's denote the speed of the boat in still water as b km/h and the speed of the current as c km/h. When the boat travels down the river, the boat and the current speeds are added, because they move in the same direction. When the boat travels up the river, the speeds are subtracted, because they move in opposite directions.

So we have the following equations:

The solution to this system of equations will give you the speed of the boat in still water and the speed of the current. Adding these two equations together, we get:

Therefore, b = 10 km/h, which is the speed of the boat in still water. Substitute b = 10 km/h into the first equation to find the speed of the current, c = 12 km/h - 10 km/h = 2 km/h.

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Answer:

The student needs to group variables into dimensionless quantities.  

Explanation:

Large experiments take a lot of time to perform because the significant variables need to be separated from the non-significant variables. However, for large quantities of variables, it is necessary to focus on the key variables.

One technique to do that is to use the Buckingham Pi Theorem. The theorem states that the physical variables can be expressed in terms or independent fundamental physical quantities. In other words:

P = n- k

n = total number of quantities

k = independent physical quantities.

A place to start with will be to find dimensionless quantities involving the mass, length, time, and at times temperature. These units are given as M, L, T, and Θ

The grouping helps because it eliminates unwanted and unnecessary experiments.

Answer:

I just took the test, It's hypothesis then design an experiment, the last one i got wrong but its not dependent. hope that helped a little.

Answer:

(a). The speed of transverse wave on the rope is 15.78 m/s.

(b). The wavelength is 0.122 m.

(c). The changed speed of transverse wave on the rope is 21.56 m/s.

The changed wavelength is 0.167 m.

Explanation:

Given that,

Frequency = 129 Hz

mass = 1.50 kg

Linear mass density of the rope = 0.0590 kg/m

(a). We need to calculate the speed of a transverse wave on the rope

Using formula of speed

[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]

Put the value into the formula

[tex]v=\sqrt{\dfrac{1.50\times9.8}{0.0590}}[/tex]

[tex]v=15.78\ m/s[/tex]

(b). We need to calculate the wavelength

Using formula of wavelength

[tex]\lambda =\dfrac{v}{f}[/tex]

Put the value into the formula

[tex]\lambda=\dfrac{15.78}{129}[/tex]

[tex]\lambda=0.122\ m[/tex]

(c). If the mass were increased to 2.80 kg.

We need to calculate the speed of a transverse wave on the rope

Using formula of speed

[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]

Put the value into the formula

[tex]v=\sqrt{\dfrac{2.80\times9.8}{0.0590}}[/tex]

[tex]v=21.56\ m/s[/tex]

We need to calculate the wavelength

Using formula of wavelength

[tex]\lambda =\dfrac{v}{f}[/tex]

Put the value into the formula

[tex]\lambda=\dfrac{21.56}{129}[/tex]

[tex]\lambda=0.167\ m[/tex]

Hence, (a). The speed of transverse wave on the rope is 15.78 m/s.

(b). The wavelength is 0.122 m.

(c). The changed speed of transverse wave on the rope is 21.56 m/s.

The changed wavelength is 0.167 m.

The speed of the transverse wave on the rope is 2.48 m/s and the wavelength is 0.019 m. If the mass were increased to 2.80 kg, the speed of the wave would stay the same but the wavelength would be different.

To find the speed of a transverse wave on the rope, we can use the equation:

v = √(T/μ)

where v is the speed of the wave, T is the tension in the rope, and μ is the linear mass density of the rope. Plugging in the values given in the question, we get:

v = √(1.50 kg * 9.8 m/s^2 / 0.0590 kg/m) = 2.48 m/s

To find the wavelength of the wave, we can use the equation:

λ = v/f

where λ is the wavelength, v is the speed of the wave, and f is the frequency of the tuning fork. Plugging in the values given in the question, we get:

λ = 2.48 m/s / 129 Hz = 0.019 m

If the mass were increased to 2.80 kg, the speed of the wave on the rope would remain the same. However, the wavelength would be different because it is determined by the frequency of the tuning fork and the speed of the wave, not the mass of the hanging weight.

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Answer:

the answer is at least 3 feet

Explanation:

In order to ensure that a cable is not affected by electromagnetic interference, it should be at least 3 feet  away from fluorescent lighting.

This is because, cables can be adversely affected by electromagnetic interference - which is a disturbance that affects an electrical circuit due to either electromagnetic induction or radiation emitted from an external source - and insulation alone cannot provide adequate protection for these cables.

Therefore, the cables should be kept a few feet away from flourescent lighting in order to prevent this interference.

To minimize the EMI on a cable from fluorescent lighting, it should be kept at a distance of at least 2 feet or 0.6 meters. The distance can vary depending on the cable type, the electromagnetic field size and the data's sensitivity.

To minimize the electromagnetic interference (EMI) effect on a cable from fluorescent lighting, it is recommended to maintain a distance of at least 2 feet or 0.6 meters. This is because the fluorescent light produces a magnetic field that can interact with the cable and cause electromagnetic interference. The above mentioned distance is considered a safe threshold, yet it can vary depending on the type of cable, the strength of the electromagnetic field produced by the light and the sensitivity of the data being transmitted.

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Answer: Fmin = (m₁ + m₂) μsg

Explanation:

To begin, we would first define the parameters given in the question.

Mass of the box = m₁

Mass of the board = m₂

We have a Frictionless surface given that Fr is acting as the frictional force between the box and the board.

from our definition of force, i.e. the the frictional force against friction experienced by the box, we have

Fr = m₁a ...................(1)

Also considering the force between the box and the board gives;

Fr = μsm₁g .................(2)

therefore equating both (1) and (2) we get

m₁a = μsm₁g

eliminating like terms we get

a = μsg

To solve for the minimum force Fmin  that must be applied to the board in order to pull the board out from under the box, we have

Fmin  = (m₁ + m₂) a  ...........(3)

where a = μsg, substituting gives

Fmin = (m₁ + m₂) μsg

cheers i hope this helps

The constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box is  [tex]\mu_{s} g({m_{1}+m_{2}})[/tex].

Given data:

The mass of small box is, [tex]m_{1}[/tex].

The mass of board is, [tex]m_{2}[/tex].

The length of board is, L.

The coefficient of static friction between the board and box is, [tex]\mu_{s}[/tex].

The linear force acting between the box and the board provides the necessary friction to box. Therefore,

[tex]F=F_{f}\\F=\mu_{s}m_{1}g\\m_{1} \times a = \mu_{s} \times m_{1}g\\a= \mu_{s} \times g[/tex]

a is the linear acceleration of board.

Then, the minimum force applied on the board is,

[tex]F_{min}=({m_{1}+m_{2}})a\\F_{min}=({m_{1}+m_{2}})(\mu_{s} \times g)\\F_{min}=\mu_{s} g({m_{1}+m_{2}})[/tex]

Thus, the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box is

[tex]\mu_{s} g({m_{1}+m_{2}})[/tex].

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The current will double

Explanation:

The relationship between voltage and current in an electric circuit is given by the following equation (Ohm's law):

[tex]V=IR[/tex]

where

V is the voltage

I is the current

R is the resistance

making R the subject,

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

Since in this experiment the resistance is kept constant, we can write:

[tex]\frac{V_1}{I_1}=\frac{V_2}{I_2}[/tex]

where

[tex]V_1=25 V[/tex] is the voltage in the 1st experiment

[tex]V_2=50 V[/tex] is the voltage in the 2nd experiment

[tex]I_1,I_2[/tex] are the currents in the 1st and 2nd experiment

We can re-arrange the equation as

[tex]\frac{I_2}{I_1}=\frac{V_2}{V_1}=\frac{50}{25}=2[/tex]

This means that the current will double in the 2nd experiment.

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Answer: Bodies of water

Explanation:

Large bodies of water, such as oceans, seas and large lakes, can affect the climate of an area

Answer:

The speed at the bottom of the driveway is3.67m/s.

Explanation:

Height,h= 5sin20°= 1.71m

Potential energy PE=mgh= 2000×9.8×1.71

PE= 33516J

KE= PE- Fk ×d

0.5mv^2= 33516 - (4000×5)

0.5×2000v^2= 33516 - 20000

1000v^2= 13516

v^2= 13516/1000

v =sqrt 13.516

v =3.67m/s

There is an omission of some sentences in the question which affects the answering of question B and C, so we will based the omission of the sentences on assumption in order to solve the question that falls under it.

NOTE: The omitted sentences are written in bold format

A tennis ball has a mass of 0.059 kg. A professional tennis player hits the ball hard enough to give it a speed of 41 m/s (about 92 miles per hour.) The ball moves toward the left, hits a wall and bounces straight back to the right with almost the same speed (41 m/s).

As indicated in the diagram below, high-speed photography shows that the ball is crushed about d = 2.0 cm at the instant when its speed is momentarily zero, before rebounding.

A) What is the average speed of the ball during the period from first contact with the wall to the moment the ball's speed is momentarily zero?

B) How much time elapses between first contact with the wall, and coming to a stop?

C) What is the magnitude of the average force exerted by the wall on the bal dring contact?

D) In contrast, what is the magnitude of the gravitational force of the Earth on the ball?

Answer:

a) [tex]V_{avg} = 20.5m/s[/tex]

b) 9.76 × 10⁻⁴s

c) 247.9 N

d) 5.8 N

Explanation:

Given  that;

Initial speed [tex](V_i)[/tex] = 0

Final speed [tex](V_f)[/tex] = 41 m/s

Distance (d) = 0.002

mass (m) = 0.059 kg

g = 9.8 m/s²

a)

The average speed of the ball can be calculated as;

[tex]V_{avg} = \frac{V_i+V_f}{2}[/tex]

[tex]V_{avg} = \frac{0+41}{2}[/tex]

[tex]V_{avg} = 20.5m/s[/tex]

b)

The time elapsed can be calculated by using the second equation of motion which is given as:

[tex]S=(\frac{V_i+V_f}{2})t[/tex]

If we make time (t) the subject of the formula; we have:

[tex](V_i+V_f)t=2S[/tex]

[tex]t= (\frac{2S}{V_I+V_f})[/tex]

[tex]=\frac{2(0.02)}{41+0}[/tex]

[tex]=\frac{0.04}{41}[/tex]

= 0.000976

=9.76 × 10⁻⁴s

c)

the magnitude of the average force (F) exerted by the wall on the bal dring contact can be determined using;

Force (F) = mass × acceleration

where acceleration [tex](a)= \frac{Vo}{t}[/tex]

[tex]\frac{41}{0.00976}[/tex]

acceleration (a) = 4200.82 m/s²

F = m × a

= 0.059 × 4200.82

= 247.85

≅ 247.9 N

d)

the magnitude of the gravitational force of the Earth on the ball

Force (F) = mass (m) × gravity (g)

= 0.059kg × 9.8 m/s²

= 5.782 N

≅ 5.8 N

The average speed of the tennis ball during contact with the wall is zero, and without the time of contact, we cannot determine the time elapsed or the average force exerted by the wall. However, the gravitational force on the ball is 0.5782 N.

The question relates to the change in momentum and the forces involved when a tennis ball bounces off a wall. Specifically, a tennis ball with a mass of 0.059 kg is hit at a speed of 41 m/s, bounces off a wall, and comes back at the same speed. To tackle the posed questions, it is essential to apply concepts from Newton's laws of motion and the conservation of momentum.

The average speed of the ball during contact is zero since the speed decreases uniformly from 41 m/s to zero.

Without the time of contact with the wall, this cannot be determined. Previous examples of collisions show time of contact can vary, so it must be provided to answer this part of the question.

To calculate the magnitude of the average force exerted by the wall on the ball, we would need the time of contact with the wall. Since it is not given, this cannot be calculated accurately.

The magnitude of the gravitational force of the Earth on the ball is calculated as the product of the mass of the ball and the acceleration due to gravity (9.8 m/s²), which is 0.059 kg * 9.8 m/s² = 0.5782 N.

Answer:

a constant downward net force.

Explanation:

The diver experiences a constant downward net force because while on his was up he decelerates until he gets to his maximum height when the acceleration is zero, during this phase force is not constant. While on his way down neglecting influences of air or atmosphere he falls with a constant downward net acceleration hence his net downward force will be constant.

Answer:

b.) 4h

Explanation:

Answer:

5541Hz

Explanation:

If the frequency of a wave is directly proportional to the velocity we have;

F = kV where;

F is the frequency

K is the constant of proportionality

V is the velocity

Since f = kV

K = f/v

K = F1/V1 = F2/V2

Given f1 = 412Hz v1 = 25.8m/s f2 = ? V2 = 347m/s

Substituting in the formula we have;

412/25.8=f2/347

Cross multiplying

25.8f2 = 412×347

F2 = 412×347/25.8

F2 = 5541Hz

The frequency heard will be 5541Hz

a) Final velocity after docking: +0.094 m/s

b) Kinetic energy loss: 31.6 J

c) Final velocity after docking: -0.056 m/s

d) Kinetic energy loss: 31.6 J

Explanation:

a)

Since the system of two satellites is an isolated system, the total momentum is conserved. So we can write:

[tex]p_i = p_f[/tex]

Or

[tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2)v[/tex]

where, in the reference frame in which the first satellite was originally at rest, we have:

[tex]m_1 = 4.50\cdot 10^3 kg[/tex] is the mass of the 1st satellite

[tex]m_2 = 7.50\cdot 10^3 kg[/tex] is the mass of the 2nd satellite

[tex]u_1 = 0[/tex] is the initial velocity of the 1st satellite

[tex]u_2 = +0.150 m/s[/tex] is the initial velocity of the 2nd satellite

v is their final velocity after docking

Solving for v,

[tex]v=\frac{m_2 u_2}{m_1 +m_2}=\frac{(7.50\cdot 10^3)(0.150)}{4.50\cdot 10^3 + 7.50\cdot 10^3}=0.0938 m/s[/tex]

b)

The initial kinetic energy of the system is just the kinetic energy of the 2nd satellite:

[tex]K_i = \frac{1}{2}m_2 u_2^2 = \frac{1}{2}(7.50\cdot 10^3)(0.150)^2=84.4 J[/tex]

The final kinetic energy of the two combined satellites is:

[tex]K_f = \frac{1}{2}(m_1 +m_2)v^2=\frac{1}{2}(4.50\cdot 10^3+7.50\cdot 10^3)(0.0938)^2=52.8 J[/tex]

Threfore, the loss in kinetic energy during the collision is:

[tex]\Delta K = K_f - K_i = 52.8 - 84.4=-31.6 J[/tex]

c)

In this case, we are in the reference  frame in which the second satellite is at rest. So, we have

[tex]u_2 = 0[/tex] (initial velocity of satellite 2 is zero)

[tex]u_1 = -0.150 m/s[/tex] (initial velocity of a satellite 1)

Therefore, by applying the equation of conservation of momentum,

[tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2)v[/tex]

And solving for v,

[tex]v=\frac{m_1 u_1}{m_1 +m_2}=\frac{(4.50\cdot 10^3)(-0.150)}{4.50\cdot 10^3 + 7.50\cdot 10^3}=-0.0563 m/s[/tex]

d)

The initial kinetic energy of the system is just the kinetic energy of satellite 1, since satellite 2 is at rest:

[tex]K_i = \frac{1}{2}m_1 u_1^2 = \frac{1}{2}(4.50\cdot 10^3)(-0.150)^2=50.6 J[/tex]

The final kinetic energy of the system is the kinetic energy of the two combined satellites after docking:

[tex]K_f = \frac{1}{2}(m_1 + m_2)v^2=\frac{1}{2}(4.50\cdot 10^3+ 7.50\cdot 10^3)(-0.0563)^2=19.0 J[/tex]

Therefore, the kinetic energy lost in the collision is

[tex]\Delta K = K_f - K_i = 19.0 -50.6 = -31.6 J[/tex]

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To find the speed of the tip of the woman's shadow, we use related rates and the similarities of triangles. The rate at which the shadow tip moves is found by setting up a proportion based on similar triangles and differentiating with respect to time. The result is that the tip of the shadow moves at 10.67 ft/sec when the woman is 50 ft from the light pole.

The question is about a real-world application of related rates, which is a concept in calculus where one rate is determined based on another rate. In this scenario, we have a woman walking away from a light pole, and we need to find out how fast the tip of her shadow is moving. To solve it, we use the similarities of triangles created by the woman and the light pole with their respective shadows.

Let's let the height of the light pole be P (16 ft), the height of the woman be W (6 ft), the distance of the woman from the pole be w (50 ft), and the distance of the tip of her shadow from the pole be s. We will use the fact that the ratios P/s and W/(s-w) are equal because the triangles are similar. Setting up the proportions, after some algebra, we find that ds/dt (the rate at which the tip of the shadow moves) is a function of dw/dt (the rate at which the woman walks).

By differentiating both sides of the proportion with respect to time t, applying the chain rule, and plugging in the known values, we can solve for ds/dt as follows:

ds/dt = P/W * dw/dt * (s/w) = (16/6) * 4 * (50/50) = 64/6 = 10.67 ft/sec

The tip of her shadow is moving along the ground at a rate of [tex]\frac{1600}{61}\) ft/sec[/tex] when she is 50 ft from the base of the pole

To solve this problem, we can use similar triangles to relate the woman's height to the height of the street light and their respective shadows. Let [tex]\(x\)[/tex] be the distance from the woman to the pole, [tex]\(s\)[/tex] be the length of her shadow, and [tex]\(h = 16\)[/tex] ft be the height of the street light. The woman's height is [tex]\(w = 6\) ft.[/tex] At any given moment, the triangles formed by the woman and her shadow and the street light and the woman's shadow are similar. Therefore, we have the proportion:

[tex]\[\frac{h}{w} = \frac{h + s}{x}\][/tex]

We can solve for [tex]\(s\):[/tex]

[tex]\[h \cdot x = w \cdot (h + s)\][/tex]

[tex]\[h \cdot x = w \cdot h + w \cdot s\][/tex]

[tex]\[h \cdot x - w \cdot h = w \cdot s\][/tex]

[tex]\[s = \frac{h \cdot x - w \cdot h}{w}\][/tex]

Now, we want to find the rate at which [tex]\(s\)[/tex] is changing with respect to time, denoted as [tex]\(\frac{ds}{dt}\).[/tex] To do this, we differentiate the expression for [tex]\(s\)[/tex] with respect to time [tex]\(t\):[/tex]

[tex]\[\frac{ds}{dt} = \frac{d}{dt}\left(\frac{h \cdot x - w \cdot h}{w}\right)\][/tex]

[tex]\[\frac{ds}{dt} = \frac{h}{w} \cdot \frac{dx}{dt}\][/tex]

Given that [tex]\(h = 16\) ft, \(w = 6\) ft, and \(\frac{dx}{dt} = 4\) ft/sec,[/tex] we can substitute these values into the equation:

[tex]\[\frac{ds}{dt} = \frac{16}{6} \cdot 4\][/tex]

[tex]\[\frac{ds}{dt} = \frac{64}{6}\][/tex]

[tex]\[\frac{ds}{dt} = \frac{160}{15}\][/tex]

[tex]\[\frac{ds}{dt} = \frac{1600}{150}\][/tex]

[tex]\[\frac{ds}{dt} = \frac{1600}{61}\][/tex]

Explanation:

Formula for centripetal acceleration of an object is as follows.

               a = [tex]\frac{v^{2}}{r}[/tex]

When an object is travelling in a circular path then it is difficult to measure its velocity.

Hence, for a circular object the formula for acceleration is as follows.

                a = [tex]\frac{4 \pi^{2} r}{T^{2}}[/tex]

     a = [tex]\frac{V^{2}}{r}[/tex],     and       V = [tex]\frac{d}{T} = \frac{2 \pi r}{T}[/tex]

     a = [tex]\frac{(\frac{[2\pi r]}{T})^{2}}{r}[/tex]

        = [tex]\frac{4 \pi^{2} r}{T^{2}}[/tex]

Thus, we can conclude that the magnitude of the acceleration of the toy is [tex]\frac{4 \pi^{2} r}{T^{2}}[/tex].

Answer:

115799.256V/m

Explanation:

Given V = 623v, d = 5.38mm = 0.00538m

Strength of electric field E = V/d

E = 623/0.00538

= 115799.256V/m

Answer:

1.

Explanation:

because 12,000 divided by 300 is 40 so its 40m

Answer:

Option (B)

Explanation:

In the stabilizing natural selection, the extreme traits from both the ends are eliminated by natural selection and natural selection favors the intermediate trait. So over time individuals having the intermediate traits are selected over the individuals having extreme traits.

So here the population of the bat which possesses moderate wing length is selected over the individual with extreme traits like individuals with short wings and long wings. As a result, the population of moderate length wing bats increased.

Therefore the correct answer is (B)- stabilizing natural selection.

Answer:

Net charge = 2.59nC

Explanation:

Gauss' Law states that the net electric flux is given by:

∮→E⋅→d/A = q enc/ϵ0

At this point, we have to solve the net electric flux through each side.

One corner is at (4.8, 3.9, 0).

The other corners are each 1.9m apart, so the other corners are at

(4.8, 5.8, 0), (6.7, 3.9, 0) and (6.7, 5.8, 0).

The other four corners are 1.9m away in the z-axis:

(4.8, 5.8, 2), (6.7, 3.9, 2) and (6.7, 5.8, 2).

Therefore, there are two planes (parallel to the x-z plane), one at

y = 3.9

and the other at

y = 5.8 which have a constant y-coordinate and are facing in the −^j and +^j respectively.

Area = 1.9m * 1.9m = 3.61m²

The flux through the first plane (area of 3.61m²) is given by:

E.A = (3.4i + 4.4 * 3.9²j + 3.0k) * (-3.61m²j) = -241.59564

The flux through the other plane is

E.A = (3.4i + 4.4 * 5.8²j + 3.0k)* (3.61m²j) = 534.33776

Now, for the other planes. There are no ^j components for the unit vectors for the area.

Therefore, even though they change in y-coordinate, those terms cancel out.

Therefore, for the planes with unit vector in the x-direction, we get:

E.A = (3.4i + 4.4y²j + 3.0k) * (1.9m * 1.9m) = ±12.274

And in the z-direction:

E.A = (3.4i + 4.4y²j + 3.0k) * (1.9m * 1.9m) = ±10.83

Now, when we sum all these fluxes together, the contribution from the x- and z-directions cancel out. Therefore, our net flux is:

-241.59564 + 534.33776 = 292.74212

Therefore, the enclosed charge is given by

q enc = ϵ0* (292.74212)

= 2.5856871136363E−9C

= 2.59E-9 nC--_- Approximated

= 2.59nC

Answer:

Q_enclosed = 1.576 nC

Explanation:

Given:

- The edge length of the cube L = 1.9 m

- One corner of the cube P_1 = ( 4.8 , 13.9 ) m

- The Electric Field vector is given by:

                             E = 3.4 i + 4.4*y^2 j + 3.0 k   N/C

Find:

What is the net charge contained by the cube?

Solution:

- The flux net Ф through faces parallel to y-z plane is:

                           net Ф_yz = E_x . A . cos (θ)

Where, E_x is the component of E with unit vector i.

            θ is the angle between normal vector dA and E.

Hence,

             net Ф_yz = 3.4 . 1.9^2 . cos (0) + 3.4 . 1.9^2 . cos (180)  

             net Ф_yz = 3.4 . 1.9^2  - 3.4 . 1.9^2 . cos (180)    

             net Ф_yz = 0.

- Similarly, The flux net Ф through faces parallel to x-y plane is:

                            net Ф_xy = E_z . A . cos (θ)

Where, E_z is the component of E with unit vector k.    

             net Ф_xy = 3 . 1.9^2 . cos (0) + 3 . 1.9^2 . cos (180)  

             net Ф_xy = 3 . 1.9^2  - 3 . 1.9^2 . cos (180)    

             net Ф_xy = 0                    

-The flux net Ф through faces parallel to x-z plane is:

                           net Ф_xz = E_y . A . cos (θ)

Where, E_y is the component of E with unit vector j.

             net Ф_xz = 4.4y_1^2 * 1.9^2 . cos (0) + 4.4y_2^2. 1.9^2 . cos (180)  

Where, The y coordinate for face 1 y_1 = 3.9 - 1.9 = 2, & face 2 y_2 = 3.9

             net Ф_xz = - 4.4*2^2*1.9^2 . cos (0) - 4.4*3.9^2. 1.9^2 . cos (180)

            net Ф_xz = -63.536 + 241.59564 = 178.0596 Nm^2/C      

- From gauss Law we have:

             Total net Ф_x,y,z = Q_enclosed / ∈_o

Where,

Q_enclosed is the charge contained in the cube

∈_o is the permittivity of free space = 8.85*10^-12

Hence,

             Total net Ф_x,y,z =  net Ф_xz +  net Ф_yz + net Ф_xy

             Total net Ф_x,y,z =  178.0596 + 0 + 0 = 178.0596 Nm^2/C  

We have,

             Q_enclosed = Total net Ф_x,y,z *  ∈_o

             Q_enclosed = 178.0596 *  8.85*10^-12

             Q_enclosed = 1.576 nC        

Answer:

The aanswers to the question are

(a) 5.33 m³

(b) 1.83 kg/m³

Explanation:

Volume of leak = 40 L, density of propane = 0.24g/cm³

Mass of leak = Volume × Density = 40000 cm³×0.24 g/cm³ = 9600 gram

Molar mass of propane = 44.1 g/mol Number of moles = 9600/44.1 = 217.69 moles

at 1 atmosphare and 298.15 K we have

PV = nRT therefore V = nRT/P = (217.69×8.3145×298.15)/‪101325‬ = 5.33 m³

The volume of the vapour = 5.33 m³

(b) Density = mass/volume

Recalculating the above for T = 293 K we have V = 5.33×293÷298.15 = 5.23 m³

Therefore density of propane vapor = 9600/5.23 = 1834.22 g/m³ or 1.83 kg/m³

Answer:

Explanation:

check the attached for the solution

Answer:

b.#NAME?

Explanation:

Remember, in Spreadsheet programs like Ms Excel several types of errors can occur such as value error.

However, since Seymour explains that there is a mistyped function name in the sheet it is more likely to display on the affected cell as #NAME?.

For example the function =SUM is wrongly spelled =SOM.

Therefore it is important to make sure the function name is spelled correctly.

Answer:

Magnitude the of the acceleration is 3.32[tex]m/s^2[/tex] and direction is south

Explanation:

[tex]v_{0} =70.9 m/s\\v=14 m/s\\S=727 m[/tex]

we know that

[tex]v^2 = v_{0}^2 +2aS[/tex]

by substituting the values we can get the required acceleration

[tex]v^2 = v_{0}^2 +2aS\\14^2 = 70.9^2 +2\times a\times 727\\a=3.32 m/s^2[/tex]

Magnitude the of the acceleration is 3.32[tex]m/s^2[/tex] and direction is south

Answer:

a. True

Explanation:

Equipotential lines are the imaginary lines in the space where actually the electric potential is same at each and every point.

Work is not required to move along such points of the equipotential line because the movement is always perpendicular to the electric field lines because these lines are always perpendicular to the electric field lines.

The electric potential for a point charge is given mathematically as:

[tex]V=\frac{1}{4\pi.\epsilon_0}\times \frac{Q}{r}[/tex]

where:

[tex]Q=[/tex] magnitude of the point charge

[tex]r=[/tex] radial distance form the charge

[tex]\epsilon_0=[/tex] permittivity of free space

Equipotential lines in physics are lines where the electric potential remains constant, perpendicular to electric field lines, and require no work to move a charge along them.

Equipotential lines are lines along which the electric potential remains constant. These lines are perpendicular to the electric field lines. It requires no work to move a charge along an equipotential line, but work is needed to move a charge from one equipotential line to another.

Answer:

A) Fx = 45.3N Fy = 21.1N

B) Fm = 45.3N R = 196N Ff = 35.4N

C) i) 0KJ ii) 11.5KJ iii) 11.5KJ iv) 0KJ

D) 23KJ

E) 0.047

Explanation:

a) Horizontal component of the force Fx = 50cos 25° = 45.3N

Vertical component = 50sin25° = 21.1N

b) Magnitude of the applied force = moving force Fm = Fx = 50cos25° =45.3N

Magnitude of the weight = mg = 20×9.8 = 196N

Magnitude of normal force which is the reaction is equal to the weight = mg = 20×9.8 = 196N

Frictional force = moving force = 50cos45° = 35.4N

c) since workdone = Force done × perpendicular distance in the direction of the force

- workdone on the moving force is 0Joules since it has no perpendicular distance

- workdone on weight is the weight × distance = (20×9.8)×12 = 11.5KJ

= work done on normal force = workdone by the weight = 11.5KJ

Workdone on the frictional force is is 0Joules since the force is along the horizontal (no perpendicular distance)

d) the total work done = work done by Applied Force +Weight + Normal Force + Frictional Force= 0+11.5+11.5 = 23KJ

e) the coefficient of friction = moving force / normal reaction = 45.3/196 = 0.047

Explanation:

A.

Horizontal component, Fy = F * cos(a)

= 50cos25

= 50 * 0.91

= 45.32 N

Vertical component,Fx = F * sin(a)

= 50sin25

= 50 * 0.42

= 21.13 N

B.

Applied force = 50 N

Weight,W = m * g

= 20 * 9.81

= 196.2 N

Normal force, N = W - Fx

= 200 - 21.13

= 179 N

Frictional force = Fy

= 45.32 N.

C.

Workdone by Normal and Weight forces are = 0, because they both act perpendicular to the movement.

Workdone by friction = Workdone by applied forces

= force * distance

= 12 * 45.32

= 543.84 J

D.

Total amount of work done on the crate = 0 (the movement with a constant speed).

E.

The coefficient of friction, u = Fy/(W - Fx)

= 45.5/(196.2 - 21.13)

= 0.26

Answer:

Explanation:

Given

Electric Field Strength [tex]E_0=2.07\times 10^5\ N/C[/tex]

Charge on electron [tex]q=1.6\times 10^{-19}\ C[/tex]

Force on any charged particle in an Electric field is given by

[tex]Force=charge\ on\ electron\times Electric\ Field\ strength [/tex]

[tex]F=1.6\times 10^{-19}\times 2.07\times 10^5[/tex]

[tex]F=3.312\times 10^{-14}\ N[/tex]

as the electric field is pointing downward therefore force will be acting downward on a positively charged particle but opposite for an electron i.e. in upward direction                                                  

The roofing tile will land around 3.1 meters away from the roof edge. This is found by using equations of motion and breaking the problem into vertical and horizontal components.

To solve this, we can break the problem down into two dimensions (horizontal and vertical) and use equations of motion. On the vertical, the roofing tile starts off 10 m above the ground, and falls under acceleration due to gravity. On the horizontal, the tile leaves the roof edge with a horizontal speed of 6 m/s * cos(30), and continues with this speed (since there's no horizontal acceleration).

The time it takes the tile to hit the ground can be found using the equation y = v(initial)*t - 0.5*g*t^2. Assuming upward is positive and taking the initial velocity on the vertical as 6 m/s * sin(30), y = -10 m, v(initial) = 3 m/s and g = 9.8 m/s^2, we can solve for t to get approximately 0.598 seconds.

The distance from the exit point on the roof can be found by multiplying the horizontal speed 5.2 m/s (6 m/s * cos(30)) by the time (0.598 sec) to get approximately 3.1 m.

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Answer:

hard line, soft line

Explanation:

Explanation:

Besides human error, other sources of error include friction on the cart (we assumed there is no friction, but in reality, that's never the case), and instrument errors (imprecision in the spring scale).

Besides human error, experiments can also encounter sampling errors, nonsampling errors, uncertainty factors, and spontaneous errors. These can be caused by an inaccurate sampling process, equipment malfunctions, variability in samples, and limitations in the measuring device.

Possible sources of error in an experiment, besides human error, could include sampling errors and non-sampling errors. Sampling errors occur when the sample utilized may not be large enough or not accurately representative of the larger population.

This can lead to inaccurate assumptions and conclusions. Non-sampling errors, on the other hand, are unrelated to the sampling process and could be a result of equipment or instrument malfunction (like a faulty counting device). An example for this would be an uncertainty in measurement induced by the fact that smallest division on a measurement ruler is 0.1 inches.

Further, variability in samples can also lead to errors called spontaneous errors. These can be due to natural phenomena such as exposure to ultraviolet or gamma radiation, or to intercalating agents. Lastly, uncertainty factors can contribute to errors as well. These factors include limitations of the measuring device, inaccuracies caused by the instrument itself (e.g., a paper cutting machine that causes one side of the paper to be longer than the other).

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Answer:

Time period of oscillation on moon will be equal to 3.347 sec

Explanation:

We have given mass which is attached to the spring m = 3.42 kg

Spring constant K = 12 N/m

We have to find the period of oscillation

Period of oscillation is equal to [tex]T=2\pi \sqrt{\frac{m}{K}}[/tex], here m is mass and K is spring constant

So period of oscillation [tex]T=2\times 3.14\times \sqrt{\frac{3.42}{12}}[/tex]

[tex]T=2\times 3.14\times 0.533=3.347sec[/tex]

So time period of oscillation will be equal to 3.347 sec

As it is a spring mass system and from the relation we can see that time period is independent of g

So time period will be same on earth and moon