A circular loop in the plane of a paper lies in a 0.75 T magnetic field pointing into the paper. The loop's diameter changes from 18.0 cm to 6.8 cm in 0.46 s.
A) Determine the direction of the induced current and justify your answer.
B) Determine the magnitude of the average induced emf.
C) If the coil resistance is 2.5 Ω, what is the average induced current?

Answers

Answer 1

Answer:

Explanation:

A.the direction of induced current will be clockwise

B: Changing 18cm and 6.8cm into 0.18m and 0.68

2.5

Divide them both by 2 to find the radius . Now we have 0.09 and .034m.

Now use Φ=(π*0.09^2)(.75 T)cos0 and the 0.019wb

(π*0.034^2)(.75 T)cos0 and the 0.00272wb

ow use ε=-N(ΔΦ/Δt)

For ΔΦ, 0.091-0.0027=0.0883

C.

To find the current, use I=ε/R

0.0883/2.5= 0.035A


Related Questions

A circular loop of wire has radius of 9.50 cm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0295 W/m^2, and the wavelength of the wave is 6.40 m.

Required:
What is the maximum emf induced in the loop?

Answers

Answer:

The maximum emf induced in the loop is 0.132 Volts

Explanation:

Given;

radius of the circular loop, r = 9.5 cm

intensity of the wave, I = 0.0295 W/m²

wavelength, λ = 6.40 m

The intensity of the wave is given as;

[tex]I = \frac{B_o^2*c }{2\mu_o}[/tex]

where;

B₀ is the amplitude of the field

c is the speed of light = 3 x 10⁸ m/s

μ₀ is permeability of free space = 4π x 10⁻⁵ m/A

[tex]I = \frac{B_o^2*c }{2\mu_o}\\\\B_o^2 = \frac{I*2\mu_o}{c} \\\\B_o^2 = \frac{0.0295*2*4\pi*10^{-7}}{3*10^8} \\\\B_o^2 = 2.472 *10^{-16}\\\\B_o = \sqrt{2.472 *10^{-16}}\\\\ B_o = 1.572*10^{-8} \ T[/tex]

Area of the circular loop;

A = πr²

A = π(0.095)²

A = 0.0284 m²

Frequency of the wave;

f = c / λ

f = (3 x 10⁸) / (6.4)

f = 46875000 Hz

Angular velocity of the wave;

ω = 2πf

ω = 2π(46875000)

ω = 294562500 rad/s

The maximum induced emf is calculated as;

emf = B₀Aω

       = (1.572 x 10⁻⁸)(0.0284)(294562500)

       = 0.132 Volts

Therefore, the maximum emf induced in the loop is 0.132 Volts

A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just on the verge of skidding to the outside of the curve. A front view of a car driving on a banked curve. The cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car. Which forces are directly responsible for producing the car’s centripetal acceleration? Coriolis force centripetal force frictional force normal force gravitational force

Answers

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

If the magnetic field steadily decreases from BBB to zero during a time interval ttt, what is the magnitude III of the induced current

Answers

Answer:

Using ohms law

The current is found from Ohm's Law.

I = V /R = E /R = Bxy /Rt.

Consider the following spectrum where two colorful lines (A and B) are positioned on a dark background. The violet end of the spectrum is on the left and the red end of the spectrum is on the right. A B 5. (1 point) What is the name for this type of spectrum? 6. (1 point) Transition A is associated with an electron moving between the n= 1 and n= 3 levels. Transition B is associated with an electron moving between the n= 2 and n= 5 levels. Which transition is associated with a photon of longer wavelength?

Answers

Answer:

Explanation:

a )

This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .

b ) The wavelength of a photon  is inversely proportional to its energy .  Photon  due to transition between n = 1 and n = 3 will have higher energy than

that due to transition between n = 2 and n = 5 . So the later photon ( B)  will have greater wavelength or photon  due to transition between n = 2 and n = 5 will have greater wavelength .

Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg. Its maximum angular velocity is 1200 rpm.
How long does it take the flywheel to reach top angular speed of 1200 rpm?

Answers

Answer:

t = 2.95 min

Explanation:

Given that,

The diameter of flywheeel, d = 1.5 m

Mass of flywheel, m = 250 kg

Initial angular velocity is 0

Final angular velocity, [tex]\omega_f=1200\ rpm = 126\ rad/s[/tex]

We need to find the time taken by the flywheel to each a speed of 1200 rpm if it starts from rest.

Firstly, we will find the angular acceleration of the flywheel.

The moment of inertia of the flywheel,

[tex]I=\dfrac{1}{2}mr^2\\\\I=\dfrac{1}{2}\times 250\times (0.75)^2\\\\I=70.31\ kg-m^2[/tex]

Now,

Let the torque is 50 N-m. So,

[tex]\alpha =\dfrac{\tau}{I}\\\\\alpha =\dfrac{50}{70.31}\\\\\alpha =0.711\ rad/s^2[/tex]

So,

[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }\\\\t=\dfrac{126-0}{0.711}\\\\t=177.21\ s[/tex]

or

t = 2.95 min

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength is 4.0 mV/m at a point 1.5 m away from the center of the circle. At what rate is the magnetic field changing?

Answers

Answer:

The rate at which the magnetic field changes is  [tex]\frac{\Delta B }{\Delta t } = - 5.33*10^{-3} \ T/ s[/tex]

Explanation:

From the question we are told that

   The  electric field strength is [tex]E = 4.0 mV/m = 4.0 *10^{-3} V/m[/tex]

   The  radius of the  circular region where the electric field is induced is

   [tex]d = 1.5 \ m[/tex]

Generally the induced electric field is mathematically represented as

     [tex]E = - \frac{r}{2} * \frac{\Delta B }{\Delta t }[/tex]

The  negative sign show that the induced electric field is acting in opposite direction to the change in magnetic field

Where  [tex]\frac{\Delta B }{\Delta t }[/tex] is the change in magnetic field

So  

       [tex]\frac{\Delta B }{\Delta t } = - \frac{2 * E }{r}[/tex]

substituting values

       [tex]\frac{\Delta B }{\Delta t } = - \frac{2 * 4.0 *10^{-3}}{ 1.5 }[/tex]

       [tex]\frac{\Delta B }{\Delta t } = - 5.33*10^{-3} \ T/ s[/tex]

The _____________ is the thermonuclear fusion of hydrogen to form helium operating in the cores of massive stars on the main sequence

Answers

I believe it’s the CNO Cycle

If an astronomer wants to find and identify as many stars as possible in a star cluster that has recently formed near the surface of a giant molecular cloud (such as the Trapezium cluster in the Orion Nebula), what instrument would be best for her to use

Answers

Answer:

Infrared telescope and camera

Explanation:

An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.

Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, Infrared images is better used, since they are able to penetrate the surrounding clouds of dust, and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.

Determine the position in the oscillation where an object in simple harmonic motion: (Be very specific, and give some reasoning to your answer.) has the greatest speed has the greatest acceleration experiences the greatest restoring force experiences zero restoring force g

Answers

Answer:

Explanation:

The greatest speed is attained at middle point or equilibrium point or where displacement from equilibrium point is zero .

When the object remains at one of the extreme point it experiences greatest acceleration but at that point velocity is zero . Due to acceleration , its velocity goes on increasing till it come to equilibrium point . At this point acceleration becomes zero . After that its velocity starts decreasing because of negative acceleration . Hence at middle point velocity is maximum .

The greatest acceleration is attained at maximum displacement or at one of the two extreme end .

Greatest restoring force too will be at position where acceleration is maximum because acceleration is produced by restoring force .

Restoring force is proportional to displacement or extension against restoring force . So it will be maximum when displacement is maximum .

Zero restoring force exists at equilibrium position or middle point or at point where displacement is zero . It is so because acceleration at that point is zero .

You set two parallel slits 0.1 mm apart at a distance of 2 m from a screen and illuminate them with light of wavelength 450 nm. The distance between a bright spot in the interference pattern and the dark spot adjacent to it is Group of answer choices

Answers

Answer:

Δx = 9 x 10⁻³ m = 9 mm

Explanation:

The formula for fringe spacing in Young's Double Slit Experiment is given as follows:

Δx = λL/d

where,

Δx = fringe spacing = ?

λ = wavelength of light = 450 nm = 450 x 10⁻⁹ m

L = Distance between slits and screen = 2 m

d = distance between slits = 0.1 mm = 0.1 x 10⁻³ m

Therefore,

Δx = (450 x 10⁻⁹ m)(2 m)/(0.1 x 10⁻³ m)

Δx = 9 x 10⁻³ m = 9 mm

A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. (a) At the moment contact is made with the battery the voltage across the capacitor is

Answers

Answer:

(a) D.  Zero.

(b) C.  Equal to the battery's terminal voltage.

Explanation:

The question is incomplete, see the complete question for your reference and information.

A resistor and a capacitor are connected in series across an ideal battery having a constant voltage

across its terminals. At the moment contact is made with the battery

(a) the voltage across the capacitor is

A) equal to the battery's terminal voltage.

B) less than the battery's terminal voltage, but greater than zero.

C) equal to the battery's terminal voltage.

D) zero.

(b) the voltage across the resistor is

A) equal to the battery's terminal voltage.

B) less than the battery's terminal voltage, but greater than zero.

C) equal to the battery's terminal voltage.

D) zero

A RC circuit is a circuit that is composed of both resistors and capacitors connect to a source of current or voltage.

basically when a voltage source is applied to an RC circuit, the capacitor, C charges up through the resistance, R

A proton moving at 4.80 106 m/s through a magnetic field of magnitude 1.74 T experiences a magnetic force of magnitude 7.00 10-13 N. What is the angle between the proton's velocity and the field? (Enter both possible answers from smallest to largest. Enter only positive values between 0 and 360.)

Answers

Answer:

31.55° and 148.45°

Explanation:

Formula for calculating the force experiences by the proton placed in a magnetic field is as expressed below;

F = qvBsinθ where;

F is the magnetic force experienced by the proton

q is the charge on the proton

v is the velocity of the proton

B is the magnetic field

θ is the angle between the proton's velocity and the field (Required)

Given parameters

F =  7.00 * 10⁻¹³N

q = 1.602*10⁻¹⁹C

v = 4.80 * 10⁶ m/s

B = 1.74 T

θ  =?

From the formula F = qvBsinθ;

sinθ = F/qvB

sinθ = 7.00 * 10⁻¹³/1.602*10⁻¹⁹* 4.80 * 10⁶*1.74

sinθ =  7.00 * 10⁻¹³/13.38*10⁻¹³

sinθ = 0.5231689 * 10⁰

sinθ = 0.5231689

θ = sin⁻¹0.5231689

θ = 31.55°

The following are the positive values of the angle between 0° and 360°

Sin is positive in the first and second quadrant. In the second quadrant the angle is equal to 180°-31.55° = 148.45°.

Hence the possible values of the angle from smallest to largest are 31.55° and 148.45°

We repeat the experiment from the video, but this time we connect the wires in parallel rather than in series. Which wire will now dissipate the most heat?
Both wires will dissipate the same amount of heat.
A. The Nichrome wire (resistance 2.7)
B. The copper wire (resistance 0.1)

Answers

Answer: B. The copper wire (resistance 0.1)

Explanation: When resistance is in parallel, voltage (V) is the same but current is different for every resistance. Current (i) is related to voltage and resistance (R) by Ohm's Law

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

So, since both wires are in parallel, they have the same voltage but because the copper wire resistance is smaller than Nichrome wire, the first's current will be bigger.

Every resistor in a circuit dissipates electrical power (P) that is converted into heat energy. The dissipation can be found by:

P = [tex]i^{2}*R[/tex]

As current for copper wire is bigger than nichrome, power will be bigger and it will dissipate more heat.

In conclusion, the copper wire will dissipate more heat when connected in parallel.

An 1300-turn coil of wire that is 2.2 cmcm in diameter is in a magnetic field that drops from 0.14 TT to 0 TT in 9.0 msms . The axis of the coil is parallel to the field.
What is the emf of the coil? (in V)

Answers

Answer:

The induced  emf is  [tex]\epsilon =7.68 \ V[/tex]

Explanation:

From the question we are told that

     The number of turns is  [tex]N = 1300 \ turns[/tex]

    The diameter is  [tex]d = 2.2 \ cm = 2.2*10^{-2}[/tex]

     The initial magnetic field is  [tex]B_i = 0.14 \ T[/tex]

      The final magnetic field is  [tex]B_f = 0 \ T[/tex]

      The  time taken is  [tex]dt = 9.0ms = 9.0*10^{-3} \ s[/tex]

 

The radius is mathematically evaluated as

      [tex]r = \frac{d}{2 }[/tex]

substituting values

     [tex]r = \frac{2.2 *10^{-2}}{2 }[/tex]

     [tex]r = 1.1*10^{-2} \ m[/tex]

The induced emf is mathematically represented as

    [tex]\epsilon =- N * \frac{d\phi }{dt }[/tex]

Where  [tex]d\phi[/tex] is the change in magnetic field which is mathematically represented as

        [tex]d\phi = dB * A * cos\theta[/tex]

=>   [tex]d\phi = [B_f - B_i ] * A * cos\theta[/tex]

Here  [tex]\theta = 0[/tex] given that the axis of the coil is parallel to the field

Also A is the cross-sectional area which is mathematically represented as

       [tex]A = \pi r^2[/tex]

substituting values

      [tex]A = 3.142 * [1.1*10^{-2}]^2[/tex]

       [tex]A = 3.8 *10^{-4] \ m^2[/tex]

So

    [tex]d\phi = [0 - 0.14 ] * 3.8*10^{-4}[/tex]

    [tex]d\phi = -5.32*10^{-5} \ weber[/tex]

So  

     [tex]\epsilon =- 1300 * \frac{-5.32*10^{-5} }{ 9.0*10^{-3} }[/tex]

    [tex]\epsilon =7.68 \ V[/tex]

Two long parallel wires separated by 4.0 mm each carry a current of 24 A. These two currents are in the same direction. What is the magnitude of the magnetic field at a point that is between the two wires and 1.0 mm from one of the two wires

Answers

Answer:

Explanation:

Magnetic field at a a point R distance away

B = μ₀ / 4π X 2I / R where I is current

Magnetic field due to one current

=  10⁻⁷ x 2 x 24 / 1 x 10⁻³

48 x 10⁻⁴ T

Magnetic field due to other current

=  10⁻⁷ x 2 x 24 / 3x 10⁻³

16 x 10⁻⁴ T

Total magnetic field , as they act in opposite direction, is

= (48 - 16 ) x 10⁻⁴

32 x 10⁻⁴ T .

An electromagnetic standing wave in air of frequency 750 MHz is set up between two conducting planes 80.0 cm apart. At how many positions between the planes could a point charge be placed at rest so that it would remain at rest?

Answers

Answer:

The positions the point charge will be placed at rest and still remain at rest are 20 cm, 40 cm, and 60 cm between the ends.

Explanation:

Given;

distance between the conducting planes, d = 80 cm

frequency of the electromagnetic wave, f = 750 MHz

speed of light, c = 3 x 10⁸ m/s²

Determine the wavelength

λ = C/f

where;

λ is the wavelength

C is the speed of light

f is the frequency

λ = C/f

λ = (3 x 10⁸) / (750 x 10⁶)

λ = 0.4 m = 40 cm

One complete cycle = one wavelength = 40 cm

half of the wavelength ( λ / 2) = 20 cm

one wavelength + half wavelength (3λ / 2) = 60 cm

The positions of the wave at zero amplitude (between 0 and 80 cm) = 20 cm, 40 cm, 60 cm

Thus, the positions the point charge will be placed at rest and still remain at rest are 20 cm, 40 cm, and 60 cm between the ends.

A person standing 180m from the foot of a high building claps hi
hand and hears the echo 0.03minutes later. What is the speed
sound in air at that temperature?
A) 331m/s
B) 240m/s C) 200m/s D) 300m/s

Answers

Answer:

C) 200 m/s

Explanation:

The sound travels a total distance of 360 m in 0.03 minutes.

v = (360 m) / (0.03 min × 60 s/min)

v = 200 m/s

a 5 charge is locataed 1.25 m to the left of a -3 charge. What is the magnitude and direction of the electrostatic force on the postive charge

Answers

Answer:

The force is 86.5×10^9 N towards the negative charge (to the right)

Explanation:

The electrostatic force on the charges is given by Coulomb's law;

F= Kq1q2/r^2

This an inverse square law.

F= electrostatic force on the charges

K= constant of Coulomb's law

q1 and q2= magnitude of the charges

Since K= 9.0×10^9Nm^2C^2

F= 9.0×10^9 × 5 × 3/(1.25)^2 = 135×10^9/1.56

F= 86.5×10^9 N

The force is 86.5×10^9 N towards the negative charge.

A system is a group of objects that’s analyzed as one unit. Consider a car moving along a road that has a flat section and a hill. The energy of the car at any given time is equal to the energy that its engine provides minus the energy that the car. When the car moves along the flat section, all of its energy is , which is calculated from its velocity and . When the car moves uphill, some of its energy is transformed to , which is calculated from its gravity, height, and .

Answers

Answer:

a) Em= K +U,  b) Em= K

Explanation:

The system in this case is formed by the mobilizes and the hill.

Let's write the expressions correctly and completely.

a) When the car moves in the path, the mechanical energy is the siua of the kinetic energy of the car and the potential energy of the car when going up the hill.

              Em = K + U

be) when the car moves in the flat part all the mechanical energy is formed by its kinetic energy that is calculated with the mass and speed of the car

             Em = K

c) When the car goes up the hill the energy the mechanical energy is conserved, but part of the kinetic energy is transformed into potential energy.

Answer:

leaves

kinetic energy

mass

potential energy

mass

Explanation:

An object on a rope is lowered steadily decreasing speed. Which is true?
A) The tope tensions is greater than the objects weight
B) the rope tension equals the objects weight
C)the rope tension is less than the objects weight
D) the rope tension can’t be compared to the objects weight

Answers

Answer:

C) the rope tension is less than the objects weight

Explanation:

According to Newton's Second Law, when an unbalanced or net force is applied to a body, it produces an acceleration in the body in the direction of the net force itself.

In this scenario, we have two forces acting on the object. First is the weight of object acting downward. Second is the tension in the rope acting upwards.

Since, the object is being lowered in the direction of weight. Therefore, weight of the object must be greater than the tension in the rope. So, the net force has the downward direction and the object is lowered. Hence, the correct option is:

C) the rope tension is less than the objects weight

A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an angle 25° in respect to normal to surface of coil. The magnetic field entering coil varies 0.02 (T) in every 2 seconds. The two ends of coil are connected to a resistor of 20 (Ω).
A) Calculate Emf induced in coil
B) Calculate the current in resistor
C) Calculate the power delivered to resistor by Emf

Answers

Answer:

a) 2.278 x 10^-5 volts

b) 1.139 x 10^-6 Ampere

c) 2.59 x 10^-11 W

Explanation:

The radius of the wire r = 2 mm = 0.002 m

the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = [tex]\pi r^{2}[/tex]

==> A = 3.142 x [tex]0.002^{2}[/tex] = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = 2.278 x 10^-5 volts

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

[tex]I[/tex] = E/R

where R is the resistor

[tex]I[/tex] = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere

c) power delivered to the resistor is given as

P = [tex]I[/tex]E

P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W

In each of the following situations, a wave passes through an opening in an absorbing wall. Rank the situations in order from the one in which the wave is best described by the ray approximation to the one in which the wave coming through the opening spreads out most nearly equally in all directions in the hemisphere beyond the wall.
(a) The sound of a low whistle at 1 kHz passes through a doorway 1 m wide.
(b) Red light passes through the pupil of your eye.
(c) Blue light passes through the pupil of your eye.
(d) The wave broadcast by an AM radio station passes through a doorway 1 m wide. (e) An x-ray passes through the space between bones in your elbow joint.

Answers

Answer:

A) geometric optics, B) geometric optics , c) geometric optics ,

e) geometric optics, f)  geometric optics

Explanation:

For this exercise we must use the condition for interference and diffraction so that these phenomena are relevant the wavelength must be comparable to the gap spacing

                λ> = a

Lam when the spacing is much greater than the wavelength, the description of geometric optics is more and more exact

let's analyze each situation

a) let's find the wavelength

                v = λ f

                λ= v / f

                λ= 343/1000

                λ = 0.343 m

                 0.343 << 1m

therefore the description of the geometric optics of

b) red light passes through the pupil of the eye

red light has a wavelength of 700 num or more, the lojo pupil has a maximum of 8 me

         λ = 700 10⁻⁹ m = 7 10⁻⁷ m

         a = 8 mm 10⁻³

longitudinal is much less therefore the geometric optics is correct

c) luz azul lam = 450 nm = 450 10⁻⁹ m = 4.5 10⁻⁷ m

again the wavelength is much less than the diameter of the pupil, for which the description with the optics is generally sufficient

d) a radio A transmits up to a maximum of f = 1400 Khz = 1,400 10⁶ Hz

let's find the wavelength

          c = λf

          λ = c / f

           λ= 3 108 / 1,400 106

          λ= 2.14 102 m

in this case the wavelength is greater than the width of the gate, so the description of diffraction should be used to explain the phenomenon

e) X-rays have wavelength lam = 10-10 m

the separation of the elbow bones is of the order of a few millimeters, for local the wavelength is much less than the separation, therefore with the relations of geometric optics it is sufficient

A valuable statuette from a Greek shipwreck lies at the bottom of the Mediterranean Sea. The statuette has a mass of 10,566 g and a volume of 4,064 cm3. The density of seawater is 1.03 g/mL.

a. What is the weight of the statuette?
b. What is the mass of displaced water?
c. What is the weight of displaced water?
d. What is the buoyant force on the statuette?
e. What is the net force on the statuette?
f. How much force would be required to lift the statuette?

Answers

Answer:

A) W = 103.55 N

B) mass of displaced water = 4186 g

C) W_displaced water = 41.06 N

D) Buoyant force = 41.06 N.

E) ZERO

F) 62.54 N

Explanation:

We are given;

mass of statuette;m = 10,566 g = 10.566 kg

volume = 4,064 cm³

Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³

A) The dry weight of the statuette can be calculated as;

W = mg

So;

W = 10.556 × 9.81

W = 103.55 N

B) Mass of displaced water is calculated from;

Density = mass/volume

So, mass = Density × Volume

m = 1.03 × 4,064 = 4186 g

C) Weight of displaced water is given by;

W_displaced water = (m_displaced water) × g

W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N

D) The buoyant force is the same as the weight of the displaced water.

Thus, Buoyant force = 41.06 N.

E) The apparent weight of the statuette is calculated from;

Apparent weight = Dry weight - Weight of displaced water

Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.

F. From E above, The Force required to lift the statuette = 62.54 N

Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the right as positive and forces pointing to the left as negative. Rank positive forces as larger than negative forces.
1. q1=+1nC
q2=-1nC
q3 =-1nC
2. q1= -1nC
+ q2 = + 1nC
q3= +1nC
3. q1= +1nC
q2= +1nC
q3= +1nC
4. q1= +1nC
q2= + 1nC
q3= -1nC
5. q1= -1nC
q2= - 1nC
q3= -1nC
6. q1=+1nC
q2=-1nC
q3 =+1nC

Answers

Answer:

Plss see attached file

Explanation:

Which type of reaction is shown in this energy diagram?

Answers

Answer:

Option C

Explanation:

The graph shows endothermic reaction because the reactants are lower in energy and the products are higher is energy. Endothermic reactions absorb energy having products with higher energy.

Answer:

C

Explanation:

In an endothermic reaction, the energy-time graph shows reactants are at a lower energy level than the products.

A container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37, what is the index of refraction of the second fluid

Answers

Answer:

The refractive index of fluid 2 is 1.78

Explanation:

Refractive index , n = real depth/apparent depth

For the first fluid, n = 1.37 and apparent depth = 9.00 cm.

The real depth of the container is thus

real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm

To find the refractive index of fluid index of fluid 2, we use the relation  

Refractive index , n = real depth/apparent depth.

Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.

So, n = 12.33 cm/6.86 cm = 1.78

So the refractive index of fluid 2 is 1.78

Since the same container is used, real depth of fluid 1 is equal to the real depth of fluid 2. The index of refraction of the second fluid is 1.8

Given that a container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37,

Then,

Index of refraction = [tex]\frac{Real depth}{Apparent depth}[/tex]

Real depth = Index of refraction x apparent depth

Since the same container is used, we can make an assumption that;

real depth of fluid 1 = real depth of fluid 2

That is,

1.37 x 9 = n x 6.86

Where n = Index of refraction for the second fluid.

make n the subject of formula

n = 12.33 / 6.86

n = 1.79

Therefore, the index of refraction of the second fluid is 1.8 approximately.

Learn more about refraction here: https://brainly.com/question/10729741

A 50 g ball of clay traveling at speed v0 hits and sticks to a 1.0 kg brick sitting at rest on a frictionless surface.

Required:
a What is the speed of the block after the collision?
b. What percentage of the ball's initial energy is "lost"?

Answers

Answer:

(a) The speed of the block after the collision is 0.0476v0.

(b) The percentage of the ball's initial energy lost, is 0 % (energy is conserved)

Explanation:

Given;

mass of ball of clay, m₁ = 50 g = 0.05 kg

mass of brick, m₂ = 1 kg

initial velocity of the ball of clay, u₁ = v0

initial velocity of the brick, u₂ = 0

Since the clay ball sticks with the brick after collision, it is inelastic collision.

Therefore, let their final velocity = v

(a) What is the speed of the block after the collision?

Apply the principle of conservation linear momentum

m₁u₁ + m₂u₂ = v (m₁ + m₂)

0.05v₀ + 1(0) = v( 0.05 + 1)

0.05v₀ = 1.05v

v = 0.05v₀ / 1.05

v = 0.0476v₀

Thus, the speed of the block after the collision is 0.0476 of its initial velocity.

(b). What percentage of the ball's initial energy is "lost"?

Initial kinetic energy of the ball = ¹/₂mv₀²

                                                    = ¹/₂ x 0.05 x v₀²

                                                    = 0.025v₀²

Final kinetic energy of the ball, = ¹/₂(m₁ + m₂)v²

                                                    = ¹/₂ x 1.05 x 0.0476v₀²

                                                    = 0.025v₀²

Change in kinetic energy = 0.025v₀² - 0.025v₀²

                                           = 0

percentage change in the initial kinetic energy of the ball;

= (0 / 0.025v₀²) x 100%

= 0 x 100%

= 0 %

Therefore, the percentage of the ball's initial energy lost, is 0 % (energy is conserved)

A solenoid of length 2.40 m and radius 1.70 cm carries a current of 0.190 A. Determine the magnitude of the magnetic field inside if the solenoid cons

Answers

Complete question:

A solenoid of length 2.40 m and radius 1.70 cm carries a current of 0.190 A. Determine the magnitude of the magnetic field inside if the solenoid consists of 2100 turns of wire.

Answer:

The magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻ T.

Explanation:

Given;

length of solenoid, L = 2.4 m

radius of solenoid, R = 1.7 cm = 0.017 m

current in the solenoid, I = 0.19 A

number of turns of the solenoid, N = 2100 turns

The magnitude of the magnetic field inside the solenoid is given by;

B = μnI

Where;

μ is permeability of free space = 4π x 10⁻⁷ m/A

n is number of turns per length = N/L

I is current in the solenoid

B = μnI = μ(N/L)I

B = 4π x 10⁻⁷(2100 / 2.4)0.19

B = 4π x 10⁻⁷ (875) 0.19

B = 2.089 x 10⁻⁴ T

Therefore, the magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.

Two factors that regulate (control) glandular secretion.

Answers

Answer:

The factors include age and puberty

Explanation:

Glandular secretion release chemicals such as hormones in response to the body’s metabolic needs.

As an individual ages , the metabolic rate of the body also reduces . This is due to the stress and ageing of the cells of the body. This explains why glandular secretion is optimal with young people and Lower in older people. It also explains why the immune system of a young person is mostly stronger than older people.

Puberty is another factor which affects glandular secretion as during puberty there is usually a high amount of hormonal changes due to high levels of secretions of some hormones. These hormones could however inhibit the other glandular secretions.

greater than: The electric potential energy of a proton at point A is _____ the electric potential energy of an proton at point B.

Answers

Answer:

[similar to]

Explanation:

it is the missing word

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