Which of the following types of light cannot be studied with telescopes on the ground?a. Visible light b. X-rays c. Radio waves d. All of the above

Answer:

1.457881265 kgm²

0.02842 Nm/rad

0.13962 rad/s

Explanation:

M = Mass = 3.97 kg

R = Radius = 85.7 cm

[tex]\tau[/tex] = Torque = 0.0688 Nm

[tex]\theta[/tex] = Angle of rotation = 2.42 rad

Moment of inertia about the center of the disk is given by

[tex]I=\dfrac{1}{2}MR^2\\\Rightarrow I=\dfrac{1}{2}\times 3.97\times 0.857^2\\\Rightarrow I=1.457881265\ kgm^2[/tex]

The rotational inertia of the disk about the wire is 1.457881265 kgm²

Torque is given by

[tex]\tau=\kappa \theta\\\Rightarrow \kappa=\dfrac{\tau}{\theta}\\\Rightarrow \kappa=\dfrac{0.0688}{2.42}\\\Rightarrow \kappa=0.02842\ Nm/rad[/tex]

The torsion constant is 0.02842 Nm/rad

Time period is given by

[tex]T=2\pi\sqrt{\dfrac{I}{\kappa}}[/tex]

Angular frequency is given by

[tex]\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\sqrt{\dfrac{\kappa}{I}}\\\Rightarrow \omega=\sqrt{\dfrac{0.02842}{1.457881265}}\\\Rightarrow \omega=0.13962\ rad/s[/tex]

The angular frequency of this torsion pendulum when it is set oscillating is 0.13962 rad/s

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.

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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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 mass of the mountain is 3.002 × 10^16 kg and its fraction of Earth's mass is approximately 4.92 × 10^-8. These results are unreasonable due to the large mass of the mountain compared to Earth and the assumptions made in the question.

To calculate the mass of the mountain, we can use Newton's Law of Universal Gravitation. The gravitational force exerted by the mountain is equal to 2.00% of the person's weight. Since weight is equal to mass multiplied by acceleration due to gravity, we can set up the equation:

0.02mg = GMm / d^2

where G is the gravitational constant, M is the mass of the mountain, m is the mass of the person, and d is the distance between them. Since the person's weight is equal to mg, we can rewrite the equation as:

0.02mg = (GMm / d^2)

Dividing both sides by mg gives us:

0.02 = (GM / d^2)

Now we can solve for the mass of the mountain (M):

M = (0.02d^2 / G)

Substituting the given values (d = 10.0 km, G = 6.673 × 10^-11 Nm²/kg²),

M = (0.02 × 10000^2) / (6.673 × 10^-11)

M = 3.002 × 10^16 kg

The mass of the mountain is 3.002 × 10^16 kg.

Comparing the mass of the mountain with that of Earth, we can use the equation:

Mountain's Mass / Earth's Mass = 3.002 × 10^16 / (6 × 10^24)

Mountain's Mass / Earth's Mass = 4.92 × 10^-8

The mass of the mountain is approximately 4.92 × 10^-8 of Earth's mass.

These results are unreasonable because the mass of the mountain and its fraction of Earth's mass are too large. It is unlikely for a mountain to have such a massive mass compared to the entire Earth.

The premises that are unreasonable or inconsistent in this scenario are the assumption that the gravitational force exerted by the mountain is 2.00% of the person's weight and the assumption that the distance between the mountain and the person is 10.0 km.

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The calculated mass of the mountain is 2.937 × 10¹⁷ kg, which is 4.91 × 10⁻⁸ of Earth's mass. This result is unreasonable because the mass and fraction are too large. The assumption about the gravitational force exerted by the mountain is inconsistent.

Let's start by calculating the mass of the mountain, given the gravitational force it exerts.

(a)

Given data:

The weight of the person is W = mg, where m is the mass of the person and g is the gravitational acceleration (9.8 m/s²).

The gravitational force F is given by Newton's law of gravitation: F = G * (m_p * M) / r², where:

Given that F = 0.02 * mg, we can equate:

0.02 * mg = G * (m * M) / (10,000 m)²

Cancel out m and solve for M:

M = (0.02 * g * (10,000 m)²) / G

M = 2.937 × 10¹⁷ kg

(b)

Mass of Earth ME = 5.972 × 10²⁴ kg

Fraction of Earth's mass = M / ME

Fraction = 2.937 × 10¹⁷ kg / 5.972 × 10²⁴ kg = 4.91 × 10⁻⁸

(c)

The mass of the mountain and its fraction of the Earth's mass are excessively large. Such a massive mountain would be geologically and physically improbable.

(d)

The gravitational force assumed to be exerted by the mountain is too large. In reality, a mountain would exert a much smaller percentage of gravitational force on a person.

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.

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:

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.

[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]

Answer:2.13 m/s

Explanation:

Given

mass of ball [tex]m=24 gm[/tex]

Length of cord [tex]L=0.463 m[/tex]

acceleration due to gravity [tex]g=9.8 m/s^2[/tex]

minimum velocity after which cord will slack

[tex]\frac{mv^2}{r}=mg[/tex]

[tex]v=\sqrt{rg}[/tex]

[tex]v=\sqrt{0.463\times 9.8}[/tex]

[tex]v=2.13 m/s[/tex]

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:

Planet Z will have the following properties;

Active Volcanoes

Active Tectonics

An Atmosphere produced by outgassing

Explanation:

The little terrestrial worlds have heat shorter than the much bigger terrestrial worlds, so the bigger worlds tend to have active volcanism and tectonics. These active volcanism and tectonics are likely to erase ancient craters. The active volcanism and tectonics would create an atmosphere by producing gases.

It is know that the Terrestrial worlds that are not far from the star have higher surface temperature.

Fast rate of rotation can cause winds and strong storms but here it is slower compared to earth. Also, a tilt of axis causes seasons.

The properties the star have are active volcanoes, active tectonics and an atmosphere produced by outgassing.

Final answer:

Planet Z's characteristics such as geological activity, seasons, and atmosphere can be inferred from its size, orbital distance, and rotation rate. A planetary mass similar to Earth's suggests active geology and an atmosphere from outgassing, while a proper distance from the sun allows for liquid water and polar ice caps. The planet's rotation influences the presence of seasons and potential for strong winds and violent storms.

Explanation:

Based on Planet Z's size, orbital distance, and rotation rate, it is possible to infer several characteristics that this planet might have. The level of geological activity on a planet is often proportional to its mass, suggesting that planets similar in size to Earth and Venus are more likely to exhibit geological activity such as active volcanoes or tectonics. Similarly, a planet's distance from its sun can influence the presence of liquid water, with those at optimum distances having the potential for erosion due to liquid water and possibly polar ice caps. A slower rotation might lead to more extreme temperature differences between day and night, which could impact atmospheric conditions and lead to strong winds and violent storms due to the larger temperature gradient.

Planetary rotation also contributes significantly to the development of seasons; hence, how Planet Z rotates will affect whether it experiences seasons. A planet that has active geology and volcanism will likely have an atmosphere that is at least partially produced by outgassing, as seen on Earth, and could also support active tectonics. Lastly, if the planet is not geologically active, it may have a surface crowded with impact craters, similar to the Moon and Mercury, which have less geological activity to renew their surfaces.

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:

[tex]0.54^{\circ}[/tex]

3.99322032 mm

Explanation:

n = Lines per mm = 125

Seperation between slits is given by

[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{125}\\\Rightarrow d=0.008\ mm[/tex]

[tex]\lambda_1[/tex] = 498 nm

[tex]\lambda_2[/tex] = 569 nm

We have the expression

[tex]dsin\theta_1=m\lambda_1[/tex]

For first maximum m = 1

[tex]\theta_1=sin^{-1}\dfrac{m\lambda_1}{d}\\\Rightarrow \theta_1=sin^{-1}\dfrac{1\times 498\times 10^{-9}}{0.008\times 10^{-3}}\\\Rightarrow \theta_1=3.57^{\circ}[/tex]

[tex]\theta_2=sin^{-1}\dfrac{m\lambda_2}{d}\\\Rightarrow \theta_2=sin^{-1}\dfrac{1\times 569\times 10^{-9}}{0.008\times 10^{-3}}\\\Rightarrow \theta_2=4.08^{\circ}[/tex]

Angular separation is given by

[tex]\Delta \theta=\theta_2-\theta_1\\\Rightarrow \Delta \theta=4.08-3.57\\\Rightarrow \Delta \theta=0.54^{\circ}[/tex]

Angular separation is [tex]0.54^{\circ}[/tex]

Now

[tex]\lambda_1[/tex] = 589 nm

[tex]\lambda_2[/tex] = 589.59 nm

[tex]\Delta \lambda=\lambda_2-\lambda_1\\\Rightarrow \Delta \lambda=589.59-589\\\Rightarrow \Delta \lambda=0.59]\ nm[/tex]

We have the relation

[tex]\dfrac{\lambda}{\Delta \lambda}=mN\\\Rightarrow N=\dfrac{\lambda}{m\Delta \lambda}\\\Rightarrow N=\dfrac{589}{2\times 0.59}\\\Rightarrow N=499.15254[/tex]

Width is given by

[tex]w=\dfrac{N}{n}\\\Rightarrow w=\dfrac{499.15254}{125}\\\Rightarrow w=3.99322032\ mm[/tex]

The width is 3.99322032 mm

The angular separation  \theta of the first maxima of these spectral lines generated by this diffraction grating is 0.54°

The width which this grating needs to be to allow you to resolve the two lines 589.00 and 589.59 nanometers is 3.99322032 mm

n = Lines per mm

= 125

The Separation between slits is given by:

d= 1/n

d= 1/125

= 0.008mm.

Where

line 1 = 498nm

line 2 = 569nm

The first maximum m= 1 will be:

θ1=  3.57°

θ2= 4.08°

The angular separation would be:

θ2- θ1= 0.54°.

Now, to find the width is:

w= N/n

= 499.15254/125

= 3.99322032 mm.

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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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Answer:

675 Joules

Explanation:

Considering that work can be calculated with the following formula:

W = Fx D

Where:

W = work

F = force

D = distance

We can directly use this formula in case the applied force remains constant

In case the force does not remain constant, we can calculate the work as a change, this is :

ΔW = ΔFxΔD

For this scenario, we have:

W₁ = F₁xD₁

F₁ = 210 N, D = 0 m

W₁ = 210Nx0m = 0 Joules

W₂ = F₂xD₂

F₂ = 45 N , D₂ = 15 m

W₂ = 45Nx15m = 675 Joules

Finally: ΔW = Total work performed when moving the car 15 m

ΔW = W₂ - W₁ = 675 Joules - 0 Joules = 675 Joules

Final answer:

The amount of work done on the car while pushing it is 1575 Joules.

Explanation:

Work is defined as the product of force and displacement. In this case, the force required to keep the car moving decreases as the car gets going. Work can be calculated using the formula:

Work = Force × Distance

Given that the force decreased from 210 N to 45 N over a distance of 15 m, we can calculate the work done as follows:

Work = (210 N + 45 N) / 2 × 15 m = 1575 J

Therefore, the amount of work done on the car while pushing it is 1575 Joules.

Answer:

C. all above is true.

Explanation:

Energy releasing reactions are exothermic. Total energy of products ( [tex]  N_2 H_2[/tex]  ) is less than the total energy of reactants ( [tex]  H_2 + N_2 [/tex] ) gives negative enthalpy change.

hint: prefix exo- means "outside, external".

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:

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:

(1) passed through the foil

Explanation:

Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.

Answer:

Option 1

Explanation:

The correct answer is option 1

The gold foil experiment was conducted by Rutherford. This experiment was conducted to study the Atom.

In the experiment Alpha rays from the emitter are passed through gold foil and there was a receiver that was present there to intercept the alpha rays.

The outcome of the result was that most of the alpha particle pass through foil undeflected and very few rays revert back on the original path from the heavy mass present at the center.

Later this heavy mass was known Nucleus.

Hence, most alpha particles passed through the foil.

Explanation:

a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.

1300 m east = 1300 i

2400 m north = 2400 j

Drops the penny from a cliff 640 m high = -640 k

Displacement of penny = 1300 i + 2400 j - 640 k

b) Displacement of man for return trip = -1300 i - 2400 j

    [tex]\texttt{Magnitude = }\sqrt{(-1300)^2+(-2400)^2}=2729.47m[/tex]

    Magnitude of his displacement = 2729.47 m

Answer:

Explanation:

d1 = 1300 m east

d2 = 2400 m north

d3 = 640 m downward

(a)

The displacement of penny is given by

[tex]\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}[/tex]

[tex]\overrightarrow{d}=1300\widehat{i}+2400 \widehat{j}-640\widehat{k}[/tex]

(b) For the return journey of man, the displacement is given by

[tex]\overrightarrow{d}=-\overrightarrow{d_{1}}-\overrightarrow{d_{2}}[/tex]

[tex]\overrightarrow{d}=-1300\widehat{i}-2400 \widehat{j}[/tex]

The magnitude of the displacement is given by

[tex]d=\sqrt{1300^{2}+2400^{2}}=2729.47 m[/tex]

Answer:

a)

6.33 x 10⁸ N/C

Direction : Towards negative charge.

b)

1.11125 x 10²⁰ m/s²

Direction : Towards positive charge.

Explanation:

a)

[tex]Q_{1}[/tex] = magnitude of negative charge = 25 x 10⁻⁶ C

[tex]Q_{2}[/tex] = magnitude of positive charge = 50 x 10⁻⁶ C

[tex]r_{1}[/tex] = distance of negative charge from point P = 0.02 m

[tex]r_{2}[/tex] = distance of positive charge from point P = 0.08 m

Magnitude of electric field at P due to negative charge is given as

[tex]E_{1} = \frac{kQ_{1}}{r_{1}^{2} } = \frac{(9\times10^{9})(25\times10^{-6})}{0.02^{2} } = 5.625\times10^{8} N/C[/tex]

Magnitude of electric field at P due to positive charge is given as

[tex]E_{2} = \frac{kQ_{2}}{r_{2}^{2} } = \frac{(9\times10^{9})(50\times10^{-6})}{0.08^{2} } = 0.703125\times10^{8} N/C[/tex]

Net electric field at P is given as

[tex]E = E_{1} + E_{2}\\E = 5.625\times10^{8} + 0.703125\times10^{8} \\E = 6.33\times10^{8} N/C[/tex]

Direction:

Towards the negative charge.

b)

[tex]m[/tex] = mass of the electron placed at P = 9.31 x 10⁻³¹ C

[tex]Q_{1}[/tex] = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

Acceleration of the electron due to the electric field at P is given as

[tex]a = \frac{qE}{m}\\ a = \frac{(1.6\times10^{-19})(6.33\times10^{8})}{(9.11\times10^{-31})}\\a = 1.11125\times10^{20} ms^{-2}[/tex]

Direction: Towards the positive charge Since a negative charge experience electric force in opposite direction of the electric field.

Answer:

Explanation:

qA = - 25 x 10^-6 C

qB = 50 x 10^-6 C

AP = 2 cm

BP = 8 cm

(a)

Electric field at P due to the charge at A

[tex]E_{A}=\frac{kq_{A}}{AP^{2}}[/tex]

[tex]E_{A}=\frac{9\times 10^{9}\times 25\times 10^{-6}}{0.02^{2}}[/tex]

EA = 5.625 x 10^8 N/C

Electric field at P due to the charge at B

[tex]E_{B}=\frac{kq_{B}}{BP^{2}}[/tex]

[tex]E_{A}=\frac{9\times 10^{9}\times 50\times 10^{-6}}{0.08^{2}}[/tex]

EB = 0.70 x 10^8 N/C

The resultant electric field at P due to both the charges is

E = EA+ EB = (5.625 + 0.7) x 10^8

E = 6.325 x 10^8 N/C towards left

(b)  mass of electron, m = 9.1 x 10^-31 kg

Let a be the acceleration of electron

Force on electron, F = charge of electron x electric field

F = q x E

[tex]a = \frac{qE}{m}[/tex]

[tex]a = \frac{1.6\times 10^{-19}\times 6.325\times 10^{8}}{9.1\times 10^{-31}}[/tex]

a = 1.11 x 10^20 m/s^2

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]

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:

In computing the volume of a cube,

Maximum possible error = +/-1350cm³

Relative error = 0.05

Percentage error = 5%

In computing the surface area of a cube,

Maximum possible error = +/-180cm²

Relative error = 0.0333

Percentage error = 3.33%

Explanation:

A cube is a three dimensional solid object with six (6) faces, twelve (12) edges and eight(8) vertices.

The volume of a cube = x³

Where x= length of the edge of a cube

X = 30cm +/- 0.5cm

Differentiate V with respect to x (V = Volume of a cube)

dV/dx = 3 x²

dV = 3 x² . dx

dV= 3 × 30² × (+/-0.5)

= 2700(+/-0.5)

= +/-1350cm³

Maximum possible error =

+/- 1350cm³

Relative error = Maximum error /surface area

= ΔV/V

Recall that V = x³

V= (30)³

A = 27000cm³

Substitute the values for and V into the formula for Relative error

Relative error = 1350 / 270000

Relative error = 0.05

% error = Relative error × 100

= 0.05× 100

= 5%

Surface Area of a cube = 6x²

A = 6x²

Differentiate A with respect to x

dA/dx= 12x

dA = 12x . dx

dA= 12 × 30 (0.5)

= +/- 180cm²

Maximum possible error =

+/- 180cm²

Relative error = Maximum error / total area

= dA/dx

Recall that A = 6x²

A = 6(30)²

A = 5400cm²

Substitute the values for and A into the formula for Relative error

Relative error = 180/ 5400

Relative error = 0.0333(4 decimal place)

% error = Relative error × 100

= 0.0333 × 100

= 3.33%

Final answer:

Use of differentials to estimate maximum and relative errors in volume and surface area calculations for a cube.

Explanation:

Differentials for Cube:

Surface Area:

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

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.

Answer:

D

Explanation:

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

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.

Answer:

Small sports car.

Explanation:

Lets take

mass of the small car = m

mass of the truck = M

As we know that when car collide with the massive truck then due to change in the moment of the car both car as well as truck will feel force.We also know that from Third law of Newton's ,it states that every action have it reaction with same magnitude but in the opposite direction.

Therefore

F = m a

a=Acceleration of the car

[tex]a=\dfrac{F}{m}[/tex]

F= M a'

a'=Acceleration of the massive truck

[tex]a'=\dfrac{F}{M}[/tex]

Here given that M > m that is why a > a'

Therefore car will experiences more acceleration.