How long would it take for 3.5 C of charge to pass through a cross-sectional area of a wire in which a current of 5 mA passes?

Answer:

With an  Environmental  Engineering  and a broadcasting minor

You can work as an On Air  personality that host  programs that provide your audience with  documentaries about the environments and project carried out by Environmental Engineer

and also you can work as a journalist that explore the world making research that will preserve the environment  and leveraging the media as a broadcaster to provide this research findings as a video for you audience  

Explanation:

In order to get a better understanding let define some terms

Environmental Engineer :

Environmental engineers resolve and help prevent environmental problems. They work in many areas, including air pollution control, industrial hygiene, toxic materials control, and land management. The duties of an environmental engineer range from planning and designing an effective waste treatment plant to studying the effects of acid rain on a particular area. An environmental engineer is sometimes required to work outdoors, though most of her work is done in a laboratory or office setting. Career opportunities for environmental engineers exist in consulting, research, corporate, and government positions.

Broadcasting:

Broadcasting is the distribution of audio or video content to a dispersed audience via any electronic mass communications medium, but typically one using the electromagnetic spectrum (radio waves), in a one-to-many model.

Answer:

14days, after registration

Explanation:

When you move into new jeey from another state you must have your vehicle inspected within ___.

answer: 14days after registration. To register a vehicle in New Jersey, one must have the following:

A valid driver license

A valid probationary license or a validated New Jersey Permit

Valid Insurance

Vehicle registration cards

Vehicles are inspected every two years,  and five years for a new vehicle. There are checklist used by the inspector to ascertain the conditions of the vehicle and ensure if it is road worthy. Inspection are carried out to reduce road accidents which can endanger the life of the driver and other road users.

Answer:

a. before

Explanation:

Did the displacement at this point reach its maximum of 2 mm before or after the interval of time when the displacement was a constant 1 mm?

from the graph given from a source. the vertical axis represents the displacement of the graph motion, whilst the horizontal side is representing the time variable of the motion .

displacement is distance in a specific direction.

before the displacement was maximum at 2mm was instant at time=0.04s.

But later was constant at 0.06s at a displacement point of 1mm

Answer:

a) Q = [tex]9*10^{-10}[/tex] C; E = [tex]2.03*10^5[/tex] V/m; V = 406.8 V

b) dE/dt = [tex]4.07*10^{11}[/tex] V/(m.s); No

c) [tex]J_d[/tex] = 3.6 [tex]A/m^2[/tex]; Equal

Explanation:

Given parameters are:

Area, A = 5 cm^2

Separation, d = 2 mm

Changing current, [tex]i_c[/tex] = 1.8 mA

At time t = 0 the charge [tex]Q_0[/tex] = 0

a) Here, we are asked to find charge, Q, electric field, E, and potential difference, V at time t = 0.5 [tex]\mu s[/tex]

[tex]Q = i_ct = 1.8*10^{-3}*5*10^{-7} = 9*10^{-10}[/tex] C

[tex]E = \sigma/\epsilon_0 = (Q/A)/\epsilon_0 = (9*10^{-10}/5*10^{-4})/(8.85*10^{-12}) = 2.03*10^5[/tex] [tex]\frac{V}{m}[/tex]

[tex]V = Ed = 2.03*10^{-5}*2*10^{-3} = 406.8[/tex] V

b) [tex]E = (Q/A)/\epsilon_0[/tex]

⇒ [tex]\frac{dE}{dt} = \frac{dQ}{dt} \frac{1}{\epsilon_0 A} = \frac{i_c}{\epsilon_0 A} = \frac{1.8*10^{-3}}{5*10^{-4}*8.85*10^{-12}} = 4.07*10^{11}[/tex] V/(m.s)

No, it is constant that does not vary in time because [tex]i_c[/tex] is constant.

c) the displacement current density, [tex]J_d = \epsilon_0\frac{dE}{dt} = \epsilon_0\frac{i_c}{\epsilon_0 A} = i_c/A[/tex]

⇒ [tex]J_d = 1.8*10^{-3}/(5*10^{-4}) = 3.6[/tex] [tex]A/m^2[/tex]

[tex]i_d =J_dA = 3.6*5*10^{-4} = 1.8*10^{-3}[/tex] A

So, [tex]i_c[/tex] and [tex]i_d[/tex] are equal.

Answer:

[tex]a)[/tex]The charge on the plates[tex]$Q=9 * 10^{-10} C ; E=2.03 * 10^{5} \mathrm{~V} / \mathrm{m} ; \mathrm{V}=406.8 \mathrm{~V}$[/tex]

[tex]b)[/tex]The time rate of change of the electric field between the plates[tex]$\mathrm{dE} / \mathrm{dt}=4.07 * 10^{11} \mathrm{~V} /(\mathrm{m} . \mathrm{s}) ; \mathrm{No}$[/tex]

[tex]c)[/tex]The displacement current density jD between the plates[tex]$J_{d}=3.6 \mathrm{~A} / \mathrm{m}^{2}$[/tex]Equals

Explanation:

Given parameters are:

Area, [tex]$A=5cm^{2}[/tex]

Separation, [tex]$\mathrm{d}=2 \mathrm{~mm}$[/tex]

Changing current, [tex]$i_{c}=1.8 \mathrm{~mA}$[/tex]

At time [tex]$\mathrm{t}=0$[/tex] the charge [tex]$Q_{0}=0$[/tex]

a) Here, we are asked to find charge, [tex]$Q$[/tex], electric field, [tex]$E$[/tex] and potential difference, [tex]$\mathrm{V}$[/tex] at time [tex]$\mathrm{t}=0.5 \mu \mathrm{s}$[/tex]

[tex]$Q=i_{c} t=1.8 \times10^{-3} \times 5 \times10^{-7}=9 \times10^{-10} \mathrm{C}$[/tex]

[tex]$E=\sigma / \epsilon_{0}=(Q / A) / \epsilon_{0}=\left(9 \times10^{-10} / 5 \times10^{-4}\right) /\left(8.85\times 10^{-12}\right)=2.03 \times10^{5} \frac{\mathrm{V}}{m}$[/tex]

[tex]$V=E d=2.03 \times10^{-5} \times2\times 10^{-3}=406.8 \mathrm{~V}$[/tex]

b) [tex]$E=(Q / A) / \epsilon_{0}$[/tex]

[tex]$\Rightarrow \frac{d E}{d t}=\frac{d Q}{d t} \frac{1}{\epsilon_{0} A}=\frac{i_{c}}{\epsilon_{0} A}=\frac{1.8 \times10^{-3}}{5\times 10^{-4} \times 8.85 \times 10^{-12}}=4.07 \times 10^{11} \mathrm{~V} /(\mathrm{m} . \mathrm{S})$[/tex]

No, it is constant that does not vary in time because is constant.

c) the displacement current density, [tex]$J_{d}=\epsilon_{0} \frac{d E}{d t}=\epsilon_{0} \frac{i_{c}}{\epsilon_{0} A}=i_{c} / A$[/tex]

[tex]$\Rightarrow J_{d}=1.8 \times 10^{-3} /\left(5 \times10^{-4}\right)=3.6 \mathrm{~A} / \mathrm{m}^{2}$[/tex]

[tex]$i_{d}=J_{d} A=3.6 \times5 \times10^{-4}=1.8\times 10^{-3} \mathrm{~A}$[/tex]

So, [tex]$i_{c}$[/tex] and [tex]$i_{d}$[/tex] are equal.

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Answer:

D. 43° 52`  

Explanation:

A bearing is an angle, measured clockwise from the north direction. When solving a bearing problem, it is good to represent the bearings in the given question with diagram.

The diagrammatically representation of the bearing of lines A and B, 16° 10` and 332° 18` respectively given in the question is shown in the figure attached.

At Point A, we will calculate angle ∠BAO.

Calculating the angle ∠BAO

∠BAO = 90° - 16° 10`

          = 73° 50`

At Point B, we will calculate angle ∠ABO.

Calculating the angle ∠ABO

∠ABO = 332° 18` - 270° 0`

          = 62° 18`

At Point O, we will calculate the include angle ∠BOA.

Calculating the angle ∠BOA

∠BAO + ∠ABO + ∠BOA = 180°  (sum of angles in a triangle)

73° 50` + 62° 18` + ∠BOA = 180°

136° 8` + ∠BOA = 180°

∠BOA = 180° - 136° 8`

∠BOA = 43° 52`

The value of the included angle BOA is 43° 52

Answer:

α = 2  rad/s²

Explanation:

Newton's second law for rotation:

τ = I * α   Formula  (1)

where:

τ : It is the torque applied to the body.  (N*m)

I :  it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Data

I =  8.0 kg * m²   :moment of inertia of the disk

R =  1.6 m : radius of the disk

F = 10.0 N : tangential force applied to the disk

Torque applied to the disk

The torque is defined as follows:

τ = F*R

τ = 10.0 N* 1.6 m

τ = 16 N*m

Angular acceleration of the disk ( α  )

We replace data in the formula (1):

τ = I * α

16 = 8 *α

α = 16 / 8

α = 2  rad/s²

Answer:

4.5m/s

Explanation:

Linear speed (v) = 42.5m/s

Distance(x) = 16.5m

θ= 49.0 rad

radius (r) = 3.67 cm

= 0.0367m

The time taken to travel = t

Recall that speed = distance / time

Time = distance / speed

t = x/v

t = 16.5/42.5

t = 0.4 secs

tangential velocity is proportional to the radius and angular velocity ω

Vt = rω

Angular velocity (ω) = θ/t

ω = 49/0.4

ω = 122.5 rad/s

Vt = rω

Vt = 0.0367 * 122.5

Vt =4.5 m/s

The tangential speed of a point on the 'equator' of the baseball is approximately 1.7983 m/s, based on the provided radius of the ball and its angular speed.

The question is essentially asking for the tangential speed of the baseball, which describes the speed of a point on its outer edge (or 'equator') as the baseball spins or rotates. The tangential speed can be calculated using the formula v = rω, where v is the tangential speed, r is the radius, and ω is the angular speed. The angular speed is essentially the rate at which an object is rotating or spinning, measured in radians per second.

Given from the prompt, we know that ω = 49.0 rad and r = 3.67 cm = 0.0367 m (because we need the radius in meters). Inserting these values into the formula we get: v = (0.0367 m)(49.0 rad/s) = 1.7983 m/s. So, a point on the 'equator' of the baseball is moving at a tangential speed of approximately 1.7983 m/s.

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Answer:

D

Explanation:

Gene flow is the transfer of genetic variation from one population to another

Answer:

The compression in the spring is 0.012 meters.

Explanation:

It is given that,

Mass of the child, m = 25 kg

Spring constant of the spring, [tex]k=2\times 10^4\ N/m[/tex]

When the child makes a nice big bounce, she finds that at the bottom of the bounce she is accelerating upward at [tex]9.8\ m/s^2[/tex]. Let x is the compression in the spring. The force of gravity is balanced by the force of the spring as :

[tex]mg=kx[/tex]

[tex]x=\dfrac{mg}{k}[/tex]

[tex]x=\dfrac{25\ kg\times 9.8\ m/s^2}{2\times 10^4\ N/m}[/tex]

x = 0.012 meters

So, the compression in the spring is 0.012 meters. Hence, this is the required solution.

Final answer:

The spring is compressed by 2.45 cm when the child is accelerating upward at 9.8 m/s² at the bottom of the bounce, as calculated using the forces acting on the child and Hooke's Law.

Explanation:

To determine how much the spring is compressed, we must consider the forces acting on the child at the bottom of the bounce. The upward force the spring exerts must be equal to the sum of the gravitational force on the child and the force required to accelerate the child upward at 9.8 m/s².

The gravitational force acting on the child is Fg = m × g, where m is the mass of the child (25 kg) and g is the acceleration due to gravity (9.8 m/s2). Thus, Fg = 25 kg × 9.8 m/s² = 245 N.

The additional force needed to accelerate the child upward at 9.8 m/s² is also Fa = m × a, yielding Fa = 245 N. The total force exerted by the spring then is Fs = Fg + Fa = 490 N.

To find the compression of the spring, we use Hooke's Law, Fs = k × x, where k is the spring constant (2.0 × 104 N/m) and x is the compression.

Solving for x, we get x = Fs / k = 490 N / (2.0 × 104 N/m) = 0.0245 m or 2.45 cm.

Final answer:

The normal force on a 25.2 kg block is 246.96 N on a level surface, 212.66 N on a 30.8° incline, and 315.36 N in an elevator accelerating upward at 2.78 m/s².

Explanation:

To calculate the magnitude of the normal force on a 25.2 kg block under various circumstances, we use different physics principles for each scenario:

For the block on a horizontal surface, N = (25.2 kg)(9.8 m/s2) = 246.96 N.

For the block on a 30.8° incline, N = (25.2 kg)(9.8 m/s2)(cos(30.8°)) = 212.66 N.

For the block in an accelerating elevator, N = (25.2 kg)(9.8 m/s2 + 2.78 m/s2) = 315.36 N.

Explanation:

If an object is thrown downward with an initial velocity of v₀​, then the distance it travels is given by s = 4.9 t²+v₀t

Now an object is thrown downward from a cliff 400 m high and it travels 138.3 m in 3 sec. We need to find initial velocity of the​ object.

           s = 4.9 t²+v₀t

           138.3 = 4.9  x 3²+ v₀ x 3

             3v₀ = 94.2

               v₀ = 31.4 m/s

Initial velocity of the​ object = 31.4 m/s

Answer:

The car is 3.4 s in the air.

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

The vertical position of the car can be obtained by the following equation:

y = y0 + v0 · t · sin α + 1/2 · g · t²

Where:

y = vertical position of car at time t.

y0 = initial vertical position.

v0 = initial velocity.

t = time.

α = launching angle.

g = acceleration of gravity.

The vertical component of the position vector when the car reaches the ground is -28 m (considering the edge of the cliff as the origin of the system of reference) and the initial vertical position is therefore 0 m. The launching angle is 22° below the horizontal (see figure). Then, we only have to find the initial velocity to solve the equation of vertical position for the time of flight.

To find the initial velocity, we have to use two equations: the equation of velocity of the car at the time it reaches the edge of the cliff and the equation of position of the car to find that time:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

If we place the origin of the frame of reference at the point where the car starts rolling, then the initial position is zero. Since the car starts from rest, the initial velocity, v0, is zero. Then, we can find the time it takes the car to travel the 54 m down the incline:

x = x0 + v0 · t + 1/2 · a · t²    (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

54 m = 1/2 · 4.4 m/s² · t²

2 · 54 m / 4.4 m/s² = t²

t = 5.0 s

With this time, we can find the velocity of the car when it reaches the edge of the cliff:

v = v0 + a · t   (v0 = 0)

v = a · t

v = 4.4 m/s² · 5.0 s

v = 22 m/s

Then, the initial velocity of the falling car is 22 m/s. Using the equation of vertical position:

y = y0 + v0 · t · sin α + 1/2 · g · t²   (y0 = 0)

y = v0 · t · sin α + 1/2 · g · t²

-28 m = 22 m/s · t · sin 22° - 1/2 · 9.81 m/s² · t²

0 = 28 m + 22 m/s · t · sin 22° - 4.91 m/s² · t²

Solving the quadratic equation for t using the quadratic formula:

t =3.4 s  (the other values is negative and, thus, discarded).

The car is 3.4 s in the air.

Final answer:

By solving the equation h = 1/2 g t^2 for the time it takes an object to fall 28 m, we find that the car is in the air for approximately 2.39 seconds before it hits the ocean.

Explanation:

To calculate how long the car is in the air, we need to analyze the car's vertical motion separately from its horizontal motion. The car falls 28 m, which we can use with the acceleration of gravity to find the time it takes to hit the water.

We use the equation h = ½ g t^2 where h is the height the car falls (28 m), and g is the acceleration due to gravity (9.81 m/s2). We're looking for t, the time in seconds.

28 m = ½ (9.81 m/s2) t^2

Rearranging for t: t = √(2 * 28 m / 9.81 m/s2)

t = √(5.70 s2)

t = 2.39 s

The car is in the air for approximately 2.39 seconds before hitting the water.

The 2-kg block of solid iron has twice the mass of the 1-kg block.

The correct option is C) mass.

Mass is a measure of the amount of matter in an object, and it is not affected by its shape or size. Since both blocks are made of solid iron, they have the same density. Therefore, the 2-kg block of solid iron has twice the mass of the 1-kg block.

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Answer:

I=0.0361 kg.m^2

Explanation:

Torque is the rotational equivalent of a force

Torque= perpendicular distance r X Force F

Torque T = I(moment of inertia) X α (angular acceleration)

T= Iα

r= 0.0285m

F= 1.9 x 10^3

T=0.0285 x 1.9 x 10^3

T= 54.15Nm

I=T/α

I=54.15/150

I=0.361 kg.m^2

Answer:

vise grip

Explanation:

Manual in-line stabilization (MILS) of the cervical spine is a type of airway management when dealing with  patients in traumatic condition ..it is a means that is performed by grasping the mastoid process of the patient, so as to prevent the movement of the cervical column during intubation of the trachea

MLS provides a means of stability to the cervical column for a patient in trauma. During this technique, a patient is restricted from moving his or her cervical collar. The vise grip can be used for a patient with neck injury. The technique is used to roll a patient to face up to prevent further injuries.

Answer: the technique which is best for manual in-line stabilization of a person floating faceup on the surface is vice grip.

Explanation:

Vice grip is a rescue technique used to prevent further injury to a guest who is suspected of having suffered a spinal injury as the typical posture is floating faceup on the surface.

Operation of a heat engine is best described as using input heat to perform work and rejects excess thermal energy to a lower temperature reservoir.

This is defined as the form of energy that is transferred between two substances at different temperatures.

Heat engine uses heat to perform work which results in the rejection of excess thermal energy to a lower temperature reservoir which was why option C was chosen as the most appropriate choice.

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The correct description of a heat engine's operation is that it uses heat to perform work, while also rejecting a portion of the heat to a cooler reservoir; it is a cyclical process where energy is partially converted from thermal to mechanical form according to the Second Law of Thermodynamics.

The statement that best describes the operation of a heat engine is (c) A heat engine uses input heat to perform work and rejects excess thermal energy to a lower temperature reservoir. In thermodynamics, a heat engine is a system that converts heat or thermal energy to mechanical work. This process involves transferring energy from a high-temperature source, the hot reservoir (Qh), and partly converting this energy into work (W), while the rest is rejected as waste heat into a lower temperature sink, the cold reservoir (Qc).

Based on the Second Law of Thermodynamics, no heat engine can convert 100% of the input thermal energy into work because there is always some waste heat that must be expelled to a colder environment. The efficiency of a heat engine is measured as the ratio of work done to the heat input from the hot reservoir. The theoretical upper limit of this efficiency depends on the temperatures of the hot and cold reservoirs and is given by the Carnot efficiency, which is always less than 100%.

The force exerted on the bumper is 1397 N

Explanation:

We can solve this problem by using the equilibrium of the torques: in fact, the torque exerted on one side of the jack must be equal to the torque exerted on the other side of the jack.

Therefore, we can write:

[tex]F_h d_h = F_b d_b[/tex]

where

[tex]F_h = 163 N[/tex] is the force applied to the end of the jack handle

[tex]d_h = 15 cm[/tex] is the distance between the force applied on the handle and the pivot

[tex]F_b[/tex] is the force exerted by the jack on the car bumper

[tex]d_b = 1.75 cm[/tex] is the distance between this force and the car bumper

And solving for [tex]F_b[/tex], we find:

[tex]F_b = \frac{F_h d_h}{d_b}=\frac{(163)(15)}{1.75}=1397 N[/tex]

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Answer:

False.  the system does not complete the circle movement

Explanation:

For this exercise, we must find the rope tension at the highest point of the path,

          -T - W = m a

The acceleration is centripetal

           a = v² / R

           T = ma - mg

           T = m (v² / R - g)

The minimum tension that the rope can have is zero (T = 0)

          v² / R - g = 0

          v = √ g R

Let's find out what this minimum speed is

          v = √ 9.8 1

          v = 3.13 m / s

We see that the speed of the body is less than this, so the system does not complete the movement.

Answer:

1100 Hz is a discriminitive stimulus signal (SD), while 1300 Hz is a discriminitive stimuli for extinction (S∆ )

Explanation:

This is used to describe Discrimination Training  in learning and behavior. The 1100 Hz is a discriminitive stimulus signal (SD), while 1300 Hz is a discriminitive stimuli for extinction (S∆ )

The 1100 Hz tone acts as a conditioned stimulus (CS), predicting food, while the 1300 Hz tone serves as an unconditioned stimulus (UCS), with no associated response. Understanding pitch and loudness is key in differentiating stimuli in conditioning.

The question relates to classical conditioning concepts in behavioral psychology, where the 1100 Hz tone is conditionally associated with the provision of food. In this context, the 1100 Hz tone functions as a conditioned stimulus (CS), since it predicts the arrival of food, prompting the rat to press the lever. On the other hand, the 1300 Hz tone does not forecast the coming of food and, thus, operates as an unconditioned stimulus (UCS), having no effect on the rat's lever-pressing behavior. It's also important to note the perception of frequency, which we refer to as pitch, and the perception of intensity, which is understood as loudness. The ability to discern differences in pitch allows for different responses to the 1100 Hz and 1300 Hz tones.

Explanation:

It is given that,

Mass of the car 1, [tex]m_1=900\ kg[/tex]

Initial speed of car 1, [tex]u_1=15i\ m/s[/tex] (east)

Mass of the car 2, [tex]m_2=750\ kg[/tex]

Initial speed of car 2, [tex]u_1=20j\ m/s[/tex] (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]

[tex]900\times 15i +750\times 20j=(900+750)V[/tex]

[tex]13500i+15000j=1650V[/tex]

[tex]V=(8.18i+9.09j)\ m/s[/tex]

The magnitude of speed,

[tex]|V|=\sqrt{8.18^2+9.09^2}[/tex]

V = 12.22 m/s

(b) Let [tex]\theta[/tex] is the direction the wreckage move just after the collision. It is given by :

[tex]tan\theta=\dfrac{v_y}{v_x}[/tex]

[tex]tan\theta=\dfrac{9.09}{8.18}[/tex]

[tex]\theta=48.01^{\circ}[/tex]

Hence, this is the required solution.

To solve this problem, we can use the principle of conservation of momentum. We calculate the momentum of each car before the collision, then find the total momentum before the collision. Since the cars stick together after the collision, we can calculate the velocity of the wreckage and determine its direction using trigonometry.

To solve this problem, we can use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.  

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Explanation:

Minimum of aerobic activity 150 minutes, or a mix of moderates and intensive exercise 75 minutes of vigorous aerobic activity a week. We can  extend this practice throughout the week with the instructions. Mild aerobic workouts might include practices like quick strolling or swimming, while activity like running can include strong aerobic activity.

Answer:

Option B

0.3 m/s2 South

Explanation:

Acceleration, [tex]a=\frac {v-u}{t}[/tex] where v and u are final and initial velocities respectively, t is the time taken

Substituting 14.1 m/s for v, 17.7 m/s for u and 12 s for t then

[tex]a=\frac {14.1 m/s- 17.7 m/s}{12}=-0.3 m/s^{2}[/tex]

Since this is negative acceleration, it's direction is opposite hence 0.3 m/s2 South

Answer:

[tex]n=101.7\times 10^6[/tex]

Explanation:

It is given that,

Frequency of the radio wave, [tex]f=101.7\ MHz=101.7\times 10^6\ Hz[/tex]

We know that the number of vibrations per second is called frequency of an object. We need to find the number of vibrations per second. Clearly, the number of vibrations per second represented in a radio wave is [tex]101.7\times 10^6[/tex]. Hence, this is the required solution.

Answer:

[tex]E = 86.4 \times 10^6 J[/tex]

Explanation:

Given data:

light energy = 20 MJ

Electric consumption is 1.0 kW

Duration of energy consumption is 24 hr

Energy consumption is given as

[tex]E = Power \times time[/tex]

[tex]E = 1 \times 10^3 \times 24 \times 3600[/tex]

[tex]E = 8.64 \times 10^6 J = 86.4 MJ [/tex]

[tex]E = 86.4 \times 10^6 J[/tex]

Answer

The answer for the first one I think is false.

The second one would be true i think. I hope i got it right and have a wonderful day

Answer:

True

False

Explanation:

From 0 to E, the train moves a distance of 55 m.

From F to J, the train moves a distance of 59 m.

The total distance is 55 + 59 = 114 m.

The displacement is the difference between the final position and initial position.  Here, the distance between J and 0 is -4 m.

radiation

convection

conduction

trust me dawg

The method of thermal energy transfer in this three cases:

A: Radiation

B: Convection

C: Conduction

Heat transfer is defined by thermodynamic systems as "The transfer of heat over the system boundary caused by a temperature difference between the system and its surroundings."

There are several ways for heat to go from one place to another. Conduction, convection, and radiation are some of the several ways that heat is transferred.

Conduction is the process of energy being transferred from one medium particle to another while the particles are in close proximity to one another.

The flow of fluid molecules from higher temperature regions to lower temperature regions is referred to as convection.

Radiant heat is the name for thermal radiations. Emission of electromagnetic waves results in the production of thermal radiation. These waves remove the energy from the body that is releasing them.

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Answer:

28.6 m/s

Explanation:

the verified expert clearly isnt an expert no shade tho

Answer:

3.07 m/s

Explanation:

We know that from kinematics equation

[tex]v^{2}=u^{2}+2as[/tex] and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity

Making u the subject then

[tex]u=\sqrt {v^{2}-2gs}[/tex]

Substituting v for 6.79 m/s, s for 1.87 m and g as 9.81 m/s2 then

[tex]u=\sqrt {6.79^{2}-(2\times 9.81\times 1.87)}=3.068338313 m/s\approx 3.07 m/s[/tex]

Final answer:

To find the ball's initial speed, use the kinematic equation [tex]v^2[/tex] = [tex]u^2[/tex] + 2*g*d, where u is the initial velocity and v is the final velocity. The initial speed calculated is approximately 3.07 m/s.

Explanation:

To find the ball's initial speed when thrown straight up into the air, we can use the principles of kinematics under the influence of gravity. The known variables are the final velocity (v = 6.79 m/s), the distance fallen (d = 1.87 m), and the acceleration due to gravity (g = 9.81 [tex]m/s^2[/tex]).

We need to remember that the ball is moving upward against gravity, so we will consider g to be negative in our calculations. We will use the following kinematic equation which relates velocity, acceleration, and displacement:

[tex]v^2[/tex] = [tex]u^2[/tex] + 2*g*d

Where u is the initial velocity, v is the final velocity, g is the acceleration due to gravity, and d is the displacement.

Let's solve for u:

[tex]v^2 = u^2[/tex] + 2 * (-g) * d

[tex]u^2 = v^2[/tex] - 2 * g * d

[tex]u^2[/tex] = [tex](6.79 m/s)^2[/tex] - 2 * (9.81 [tex]m/s^2[/tex]) * (1.87 m)

[tex]u^2[/tex] = 46.1041 [tex]m^2/s^2[/tex] - 36.6894 [tex]m^2/s^2[/tex]

[tex]u^2[/tex] = 9.4147 [tex]m^2/s^2[/tex]

u = sqrt(9.4147 [tex]m^2/s^2[/tex])

u ≈ 3.07 m/s

Therefore, the initial speed of the ball was approximately 3.07 m/s.

Answer:

2.71207 Hz

Explanation:

v = Speed of sound in air = 343 m/s

[tex]v_r[/tex] = Relative speed between the speakers and the student = 1.02 m/s

f' = Actual frequency of sound = 456 Hz

Frequency of sound heard as the student moves away from one speaker

[tex]f_1=f'\dfrac{v-v_r}{v}\\\Rightarrow f_1=456\dfrac{343-1.02}{343}\\\Rightarrow f_1=454.64396\ Hz[/tex]

Frequency of sound heard as the student moves closer to the other speaker

[tex]f_2=f'\dfrac{v+v_r}{v}\\\Rightarrow f_2=456\dfrac{343+1.02}{343}\\\Rightarrow f_2=457.35603\ Hz[/tex]

The difference in the frequencies is

[tex]f_2-f_1=457.35603-454.64396=2.71207\ Hz[/tex]

The student hears 2.71207 Hz

[tex]1 \times 10^{-10.1} \mathrm{Wm}^{-2}[/tex] is the intensity of the sound.

Answer: Option B

Explanation:

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about [tex]1 \times 10^{-12} \mathrm{Wm}^{-2}[/tex]). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

                     [tex]\text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)[/tex]

Where,

I = Intensity of the sound produced

[tex]I_{0}[/tex] = Standard Intensity of sound of 60 decibels = [tex]1 \times 10^{-12} \mathrm{Wm}^{-2}[/tex]

So for 19 decibels, determine I as follows,

                   [tex]19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)[/tex]

                  [tex]\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}[/tex]

                  [tex]\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9[/tex]

When log goes to other side, express in 10 to the power of that side value,

                  [tex]\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}[/tex]

                  [tex]I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}[/tex]