A circular loop of wire is in the plane of the paper. The north pole of a bar magnet is being moved toward the center of the loop from a position in front of the paper. The direction of the induced current in the loop:______.
a) is north.b) is south.c) is counterclockwise.d) is clockwise.d) cannot be determined.

Answer:

157.9 kg

Explanation:

Density: This can be defined as the ratio of the mass of a body and it's volume.

The S.I unit of density is kg/m³.

From the question,

Density = Mass/volume

D = m/v............................ Equation 1

Where D = Density of gold, m = mass of gold, v = volume of gold.

make m the subject of the equation

m = Dv.................... Equation 2

Since the gold is a cube,

v = l³................... Equation 3

Where l = length of the gold cube.

Substitute equation 3 into equation 2

m = Dl³............... Equation 4

Given: D = 19300 kg/m³, l = 0.2015 m

Substitute into equation 4

m = 19300(0.2015)³

m = 157.9 kg.

Explanation:

It is given that,

Wavelength of a wave, [tex]\lambda=35\ cm=0.35\ m[/tex]

Amplitude, [tex]A=8.4\ cm[/tex]

Period of the wavelength, t = 1.2 s

The wave is traveling in a string in the positive x direction. We need to write the wave equation for this conditions.

The general equation of the wave when it is traveling in +x direction is given by :

[tex]y=A\sin(kx-\omega t)[/tex]

A is amplitude

k is propagation constant

[tex]k=\dfrac{2\pi}{\lambda}\\\\k=\dfrac{2\pi}{0.35}\\\\k=17.95\ m^{-1}\approx 18\ m^{-1}[/tex]

[tex]\omega[/tex] is angular frequency

[tex]\omega=\dfrac{2\pi}{T}\\\\\omega=\dfrac{2\pi}{1.2}\\\\\omega=5.23\ s^{-1}[/tex]

So, the wave equation is given by :

[tex]y(x,t)=(0.084) \sin (18 x - 5.2 t)[/tex]

Hence, this is the required solution.

Answer:

20,000 kg m/s

Explanation:

Given:

v₀ = 50 m/s

a = -5 m/s²

t = 2 s

Find: v

v = at + v₀

v = (-5 m/s²) (2 s) + (50 m/s)

v = 40 m/s

p = mv

p = (500 kg) (40 m/s)

p = 20,000 kg m/s

The correct answer is B. thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom

Explanation:

Marvin Zuckerman was an important American Psychologists mainly known for his research about personality and the creation of a model to study this aspect of human psychology. This model purposes five factors define personality, these are the thrill and adventure-seeking that involves seeking for adventures and danger; experience seeking that implies a strong interest in participating in new activities; disinhibition that implies being open and extrovert; and susceptibility to boredom that implies avoiding boredom or repetition. Thus, option B correctly describes the characteristics used in Zuckerman's test.

Answer:

v = 0.477m/s

Explanation:

You have the following wave function:

[tex]y(x,t)=35sin(2\pix-3t+1.5)[/tex]       (1)

where y is the vertical displacement of the wave for the position x.

The general form of a wave function can be written as follow:

[tex]y(x,t)=Asin(kx-\omega t+\phi)[/tex]         (2)

by comparing the equation (2) and (1) you have:

A: amplitude of the wave = 35

k: wave number = 2π

w: angular frequency of the wave = 3

φ: phase of the wave = 1.5

The speed of the wave is given by the following formula:

[tex]v=\frac{\omega}{k}[/tex]        

you replace the values of the parameters in the previous formula:

[tex]v=\frac{3}{2\pi}=0.477\frac{m}{s}[/tex]

The speed of the wave is 0.477m/s

Answer:

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

Explanation:

Let suppose that lion and Thomson's gazelle are running at constant speed before and after collision and that collision is entirely inelastic. Given the absence of external force, the Principle of Momentum Conservation is applied such that:

[tex]\vec p_{L} + \vec p_{G} = \vec p_{F}[/tex]

Where:

[tex]\vec p_{L}[/tex] - Linear momentum of the lion, measured in kilograms-meters per second.

[tex]\vec p_{G}[/tex] - Linear momentum of the Thomson's gazelle, measured in kilograms-meters per second.

[tex]\vec p_{F}[/tex] - Linear momentum of the lion-Thomson's gazelle, measured in kilograms-meters per second.

After using the definition of momentum, the system is expanded:

[tex]m_{L}\cdot \vec v_{L} + m_{G}\cdot \vec v_{G} = (m_{L} + m_{G})\cdot \vec v_{F}[/tex]

Vectorially speaking, the final velocity of the lion-gazelle system is:

[tex]\vec v_{F} = \frac{m_{L}}{m_{L}+m_{G}}\cdot \vec v_{L} + \frac{m_{G}}{m_{L}+m_{G}}\cdot \vec v_{G}[/tex]

Where:

[tex]m_{L}[/tex], [tex]m_{G}[/tex] - Masses of the lion and the Thomson's gazelle, respectively. Measured in kilograms.

[tex]\vec v_{L}[/tex], [tex]\vec v_{G}[/tex], [tex]\vec v_{F}[/tex] - Velocities of the lion, Thomson's gazelle and the lion-gazelle system. respectively. Measured in meters per second.

If [tex]m_{L} = 177\,kg[/tex], [tex]m_{G} = 32\,kg[/tex], [tex]\vec v_{L} = 81.8\cdot j\,\left[\frac{km}{h} \right][/tex] and [tex]\vec v_{G} = 59.0\cdot i\,\left[\frac{km}{h} \right][/tex], the final velocity of the lion-gazelle system is:

[tex]\vec v_{F} = \frac{177\,kg}{177\,kg+32\,kg}\cdot \left(81.8\cdot j\right)\,\left[\frac{km}{h} \right] + \frac{32\,kg}{177\,kg+32\,kg}\cdot \left(59.0\cdot i\right)\,\left[\frac{km}{h} \right][/tex]

[tex]\vec v_{F} = 9.033\cdot i + 69.276\cdot j\,\left[\frac{km}{h} \right][/tex]

The speed of the system is the magnitude of the velocity vector, which can be found by means of the Pythagorean theorem:

[tex]\|\vec v_{F}\| = \sqrt{\left(9.033\frac{km}{h} \right)^{2}+\left(69.276\frac{km}{h} \right)^{2}}[/tex]

[tex]\|\vec v_{F}\| \approx 69.862\,\frac{km}{h}[/tex]

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

Answer:

this sound wave get to the other by either undergoing either reflection,absorption or transmission

Explanation:

When a sound wave meets an obstacle, some of the sound is reflected back from the front surface and some of the sound passes into the obstacle material, where it is absorbed or transmitted through the material like the building walls

Thus, Reflection and absorption are dependent on the wavelength of the sound. Hence percentage of the sound transmitted through an obstacle depends on how much sound is reflected and how much is absorbed.

Answer:

x = 1.1*10^-7 m

Explanation:

You can consider that the electron travels in the horizontal positive x direction. Furthermore, you can consider that the electric field between the plates is totally vertical and points upward (which means that the negative plates is above the line of motion of the proton).

In order to calculate the vertical distance traveled by the electron when it has passed the region between the plates, you take into account the acceleration experienced by the proton, due to the electric field between the plates.

You first calculate the electric field between the plates, by using the following formula:

[tex]E=\frac{\sigma}{\epsilon_o}[/tex]        (1)

σ: surface charge density of each plate = ±1.0*10^6 C/m^2

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

You replace the values of the parameters in the equation (1):

[tex]E=\frac{1.0*10^6C/m^2}{8.85*10^{-12}C^2/Nm}=1.12*10^{17}\frac{N}{C}[/tex]

Next, you calculate the acceleration of the electron produced by the electric force. You use the second Newton law for the electric force:

[tex]F_e=qE=ma[/tex]       (2)

q: charge of the electron = 1.6*10^-19C

m: mass of the proton = 1.67*10^-27 kg

You solve the equation (2) for a:

[tex]a=\frac{qE}{m}=\frac{(1.6*10^{-19})(1.12*10^{17}N/C)}{1.67*10^{-27}kg}\\\\a=1.07*10^{25}\frac{m}{s^2}[/tex]

Next, you calculate the time that proton takes to reach the negative capacitor. You use the following formula:

[tex]y=v_{oy}t+\frac{1}{2}at^2[/tex]     (3)

y: vertical distance from the center of the capacitor to the negative plate = 3.5cm/2 = 1.75cm = 0.0175m

voy: vertical initial speed of the proton = 0m/s (the proton enter to the region of electric field horizontally)

You solve the equation (3) for t:

[tex]y=\frac{1}{2}at^2\\\\t=\sqrt{\frac{2y}{a}}=\sqrt{\frac{2(0.0175m)}{1.05*10^{25}m/s^2}}=5.77*10^{-14}s[/tex]

Next, you calculate the horizontal distance traveled by the proton in the time calculated. You use the following formula:

[tex]x=v_ot[/tex]         (4)

x: length of the capacitor = 3.5cm = 0.035m

vo: initial speed of the proton = 1.9*10^6 m/s

[tex]x=(1.9*10^6m/s)(5.77*10^{-14}s)=1.1*10^{-7}m[/tex]

The horizontal distance traveld by the electron when it impacts the negative plate of teh capacitor is 1.1*10^-7m

Answer:

0.79 cm

Explanation:

The computation is shown below:-

Particle acceleration is

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

We will take d which indicates distance as from the negative plate, so the travel by proton is 0.800 cm - d at the same time

[tex]d = \frac{1}{2} a_et^2\\\\0.800 cm - d = \frac{1}{2} a_pt^2\\\\\frac{d}{0.800 cm - d} = \frac{a_e}{a_p} \\\\\frac{d}{0.800 cm - d} = \frac{m_p}{m_e} \\\\\frac{d}{0.800 cm - d} = \frac{1836m_e}{m_e}[/tex]

After solving the equation we will get 0.79 cm from the negative plate.

Therefore it is 0.79 cm far from the negative pate i.e the point at which the electron and proton pass each other

The point at which the electron and proton pass each other will be 0.79 cm.

When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.

The given data in the problem is;

d' is the distance between the two parallel plates= 0.800 cm

The acceleration is given as;

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

The distance from Newton's law is found as;

[tex]d = ut+\frac{1}{2} at^2 \\\\ u=0 \\\\ d= \frac{1}{2} at^2 \\\\ d-d' = \frac{1}{2} a_pt^2 \\\\ 0.800-d= \frac{1}{2} a_pt^2 \\\\\ \frac{d}{0.800-d} =\frac{a}{a_p} \\\\ \frac{d}{0.800-d} =\frac{m_p}{m} \\\\ \frac{d}{0.800-d} =\frac{1836m_e}{m_e} \\\\ d=0.79 \ cm[/tex]

Hence the point at which the electron and proton pass each other will be 0.79 cm.

To learn more about the charge refer to the link;

https://brainly.com/question/24391667

Answer: In a pulley, the ideal mechanical advantage is equal to the number of rope segments pulling up on the object. The more rope segments that are supporting to do the lifting work, the less pressure that is needed for the job.

Explanation:

Complete Question

The complete question is  shown on the first uploaded image  

Answer:

The distance which the car skid is  [tex]l = \frac{v_i^2 }{2 * \mu_k * g }[/tex]

Explanation:

From the question we are told that  

     The  initial velocity of the car is  [tex]v_i[/tex]

     The  coefficient of kinetic friction is  [tex]\mu_k[/tex]

According to the law of energy conservation

    The initial Mechanical Energy =  The final  Mechanical Energy  

                           [tex]M_i = M_f[/tex]  

The initial mechanical energy is  mathematically represented as  

              [tex]M_i = KE _o + PE_e[/tex]

where KE is the initial kinetic energy which is  mathematically represented as

        [tex]KE = \frac{1}{2} m v_i^2[/tex]

And  PE  is  the initial potential energy which is  zero given that the car is  on the ground

     Now  

           [tex]M_f = W_{\mu}[/tex]

Where  [tex]W_{\mu}[/tex] is  the work which friction exerted on the car  which is  mathematically represented as

       [tex]W_{\mu} = m* \mu_k * g * l[/tex]

Where  [tex]l[/tex] is the distance covered by the car before it slowed down

        [tex]\frac{1}{2} m v_i^2 = m* \mu_k * g * l[/tex]

=>     [tex]l = \frac{v_i^2 }{2 * \mu_k * g }[/tex]

Answer:

Q = 20.22 x 10³ W = 20.22 KW

Explanation:

First we need to find the volume of water dropped.

Volume = V = πr²h

where,

r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m

h = height drop = 1.45 cm = 0.0145 m

Therefore,

V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

m = ρV

where,

m = mass = ?

ρ = density of water = 1000 kg/m³

Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

m = 1.038 kg

Now, we calculate the heat required to vaporize this amount of water.

q = mH

where,

H = Heat of vaporization of water = 22.6 x 10⁵ J/kg

Therefore,

q = (1.038 kg)(22.6 x 10⁵ J/kg)

q = 23.46 x 10⁵ J

Now, for the rate of heat transfer:

Rate of Heat Transfer = Q = q/t

where,

t = time = (18.6 min)(60 s/1 min) = 1116 s

Therefore,

Q = (23.46 x 10⁵ J)/1116 s

Q = 20.22 x 10³ W = 20.22 KW

Answer:

Part A : 336 N/m

Part B: 19.75 cm

Explanation:

Part A: Using hooks law,

F = ke................... Equation 1

Where F = Force applied to the spring, k = spring constant, e = extension.

Note: From the question, the force applied to the spring is the weight of the mass hung on it.

Hence,

F = mg............... Equation 2

Substitute equation 2 into equation 1.

mg = ke

make k the subject of the equation

k = mg/e.............. Equation 3

Given: m = 2.4 kg, e = 18-11 = 7 cm = 0.07 m

Constant: g = 9.8 m/s²

Substitute these value into equation 3

k = 2.4(9.8)/(0.07)

k = 336 N/m.

Part B,

Similarly,

mg = ke.

make e the subject of the equation

e = mg/k................ Equation 4

Given: m = 3.0 kg, k = 336 N/m, g = 9.8 m/s²

Substitute into equation 4

e = 3.0(9.8)/336

e = 0.0875 m

e = 8.75 cm

Length of the spring = original length+ extension

Length of the spring = 11+8.75

Length of the spring = 19.75 cm.

Answer:

Option B (remain vertically under the plane) is the correct option.

Explanation:

Some other alternatives are given really aren't linked to the specified scenario. So choice B is the perfect solution to that.

Answer:

The two small charged spheres are now 4.382 cm apart

Explanation:

Given;

distance between the two small charged sphere, r = 7.59 cm

The force on each of the charged sphere can be calculated by applying Coulomb's law;

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

where;

F is the force on each sphere

q₁ and q₂ are the charges of the spheres

r is the distance between the spheres

[tex]F = \frac{kq_1q_2}{r^2} \\\\kq_1q_2 = Fr^2 \ \ (keep \ kq_1q_2 \ constant)\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\r_2 = \sqrt{\frac{F_1r_1^2}{F_2}} \\\\r_2 = r_1\sqrt{\frac{F_1}{F_2}}\\\\(r_1 = 7.59 \ cm, \ F_2 = 3F_1)\\\\r_2 = 7.59cm\sqrt{\frac{F_1}{3F_1}}\\\\r_2 = 7.59cm\sqrt{\frac{1}{3}}\\\\r_2 = 7.59cm *0.5773\\\\r_2 = 4.382 \ cm[/tex]

Therefore, the two small charged spheres are now 4.382 cm apart.

Answer:

There are two components for a two-dimensional coordinate system/vector.

Explanation:

For two-dimensional vectors, such as velocity, acceleraton, etc, there are two components, the x- and y-components.

These components could be rotated or translated, depending on the coordinate system.

Instead of rectangular cartesian system, the components could also be in the form of polar coordinates, such as radius and theta (angle).

For three-dimensional vectors, such as velocity in space, there are three components, in various coordinate systems.

Answer:

D is determined by distance between the telescopes.

Explanation:

Answer:

8 ft/s

Explanation:

This is a straight forward question without much ado.

It is given from the question that she walks with a speed of 8 ft/s

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1

Answer:

Velocity = 131 m/s

Speed = 131 m/s

Explanation:

Equation of motion, s = f(t) = 12t² + 35 t + 1

To get velocity of the particle, let us find the first derivative of s

v (t) = ds/dt = 24t + 35

At t = 4

v(4) = 24(4) + 35

v(4) = 131 m/s

Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s

Answer:

The person is running at a speed of 2.564 m/s

Explanation:

Given;

mechanical energy dissipated per kilogram per step, E/kg/S = 0.6 J/kg/S

mass of the person, m = 52 kg

power developed by the person, P = 80 W

mechanical energy of the person per step, E = 0.6 J/kg x 52 kg

                                                        [tex]E_{step}[/tex] = 31.2 J

mechanical energy for the total step, [tex]E_{total}[/tex] = 31.2 J x S

P = E / t

[tex]P_{avg} = \frac{E_{total}}{t} \\\\P_{avg} = \frac{E_{step}*S}{t}\\\\\frac{P_{avg}}{E_{step}} = \frac{S}{t} \\\\\frac{S}{t} = \frac{80}{31.2} \\\\\frac{S}{t} = 2.564 \ m/s[/tex]

Therefore, the person is running at a speed of 2.564 m/s

Answer:

 a = 4.88 m / s²

Explanation:

We can solve this exercise using the expressions of kinematics in one dimension

         v² = v₀² - 2a x

where v is the velocity, v₀ is the initial velocity, at acceleration and ax is the distance, the negative sign is because the velocity decreases because

          a = (v₀² - v²) / 2x

let's calculate

         a = (8.6² - 5.3²) / (2   4.7)

         a = 4.88 m / s²

Answer:

5000N

Explanation:

According to Newton's second law of motion, the net force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) of the body caused by the force. i.e

∑F = m x a             -------------(i)

From the question, the net force is the combined effect of the thrust (F) and the friction force (Fₓ). i.e

∑F = F + Fₓ             -------------(ii)

Where;

Fₓ = -200N       [negative sign because the friction force opposes motion]

Combine equations(i) and (ii) together to get;

F + Fₓ = m x a

F = ma - Fₓ         -------------(iii)

Where;

m = mass of car = 1200kg

a = acceleration of the car = 4m/s²

Now substitute the values of m, a and Fₓ into equation (iii) as follows;

F = (1200 x 4) - (-200)

F = 4800 + 200

F = 5000N

Therefore, the force the thrust is providing is 5000N

Answer:

2000 N

Explanation:

20 km/h = 5.56 m/s

100 km/h = 27.78 m/s

F = ma

F = m Δv/Δt

F = (200 kg + 80 kg) (27.78 m/s − 5.56 m/s) / (3 s)

F = 2074 N

Rounded to one significant figure, the force is 2000 N.

Answer:

  I = 2.18 10⁻⁴ W / m²

Explanation:

The two-slit interference pattern is described by the expression for constructive interference.

             d sin θ = m λ

If we also want to know the distribution of intensities we must perform the su of the electric field of the two waves, and find the intensity as the square of the velvet field, obtaining the expression

              I = I_max cos² ((π d /λ L) y)

where d is the separation of the slits, λ  the wavelength, L the distance to the screen e and the separation of the interference line with respect to the central maximum

let's reduce the magnitudes to the SI system

λ  = 583 nm = 583 10⁻⁹ m

L = 75.0 cm = 75.0 10⁻² m

d = 0.640 mm = 0.640 10⁻³ m

y = 0.900 mm = 0.900 10⁻³ m

let's calculate the intensity of this line

        I = 5 10⁻⁴ cos² ((π 0.640 10⁻³ /583 10⁻⁹ 0.75 10⁻²) 0.900 10⁻³)

        I = 5 10⁻⁴ cos2 (413.84)

         I = 5 10⁻⁴ 0.435

        I = 2.18 10⁻⁴ W / m²

Answer is given below

Explanation:

Answer:

its a magnesium because it only metal in option and metals are capable to conduct electricity.

Answer:

Magnesium (Mg)

Explanation:

According to the periodic table, Metals have a high electrical conductivity than non-metals. Magnesium is a metal so it has a high electrical conductivity. Rest of them (Cl, Ne, and O ) are non-metals.

Answer:

Explanation:

When a body is launched in air and allowed to fall freely under the influence of gravity, the motion experienced by the body is known as a projectile motion. The body is launched at a particular velocity and at an angle theta to the horizontal. The velocity of the body ca be resolved towards the horizontal component and the vertical component.

Along the horizontal Ux = Ucos(theta)

Along the vertical Uy = Ucos(theta)

Ux and Uy are the velocities of the body along the horizontal and vertical components respectively.

This means that the only factor connecting horizontal and vertical components of projectile motion is its velocity since we are able to calculate the velocity of the body along both components irrespective of its initial velocity.

Answer:

Explanation:

Rotational kinetic energy of flywheel

= 1/2 Iω² where I is moment of inertia , ω is angular velocity

for wheel I = 1/2 m R² where m is mass and R is radius of flywheel

Putting the values

I = 1/2 x 370 x .5²

= 46.25 kg m²

ω = 2πn where n is frequency of revolution per second

ω = 2 x 3.14 x 540 = 3391.2

Rotational kinetic energy = .5 x 46.25 x 3391.2²

= 265.94 x 10⁶ J

If this energy is transferred to a car of mass 1600kg , velocity acquired by it be v , then

kinetic energy of car = rotational kinetic energy of flywheel

= 1/2 m v ² =  265.94 x 10⁶

.5 x 1600 v² = 265.94 x 10⁶

v² = 33.24 x 10⁴

v = 5.76 x 10²

= 576 m /s

Answer:

the result is the quantization of __Energy__ of the particle

Explanation:

Answer:

1 yr = 24 * 3600 * 365 = 3.2 * 10E7 sec

C = 6 R = 1.5 * 10E8 * 6 = 9 * 10E8 km     circumference of orbit

v = C / t = 9 * 10E8 km / 3 * 10E7  sec = 30 km / sec = 18 mi/sec