A customs inspector was suspecting that some of the 12 plastic spheres, which were shipped out of the country, had something in them. Each sphere weighted the same and had hard walls everywhere. Inspector thought that it was possible to hide something inside each sphere. He was correct, and was able to use a simple experiment in determining which sphere had diamonds inside. How did he do it?

Answers

Answer 1

Answer:

use a hammer to hit it

Explanation:

if u hit it u will be able to hear the shattered noise


Related Questions

In your own words, discuss how energy conservation applies to a pendulum. Where is the potential energy the most? Where is the potential energy the least? Where is kinetic energy the most? Where is kinetic energy the least?

Answers

Answer:

Explanation:

Energy conservation applies to the swinging of pendulum . When the bob is at one extreme , it is at some height from its lowest point . So it has some gravitational potential energy . At that time since it remains at rest its kinetic energy is zero or the least . As it goes down while swinging , its potential energy decreases and kinetic energy increases following conservation of mechanical energy . At the At the lowest point , its potential energy is least  and kinetic energy is maximum .

In this way , there is conservation of mechanical energy .

An astronomy student, for her PhD, really needs to estimate the age of a cluster of stars. Which of the following would be part of the process she would follow?
A. plot an H-R diagram for the stars in the cluster
B. count the number of M type stars in the cluster
C. measure the Doppler shift of a number of the stars in the cluster
D. search for planets like Jupiter around the stars in the center of the cluster
E. search for x-rays coming from the center of the cluster

Answers

Answer:

A. plot an H-R diagram for the stars in the cluster.

Explanation:

A star cluster can be defined as a constellation of stars, due to gravitational force, which has the same origin.

The astronomy student would have to plot an H-R diagram for the stars in the cluster and determine the age of the cluster by observing the turn-off point. The turn-off is majorly as a result of gradual depletion of the source of energy of the star. Thus, it projects off the constellation.

You need to design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2

Answers

Answer:

Hello your question has some missing parts and the required diagram attached below is the missing part and the diagram

Digital circuits require actions to take place at precise times, so they are controlled by a clock that generates a steady sequence of rectangular voltage pulses. One of the most widely

used integrated circuits for creating clock pulses is called a 555 timer.  shows how the timer’s output pulses, oscillating between 0 V and 5 V, are controlled with two resistors and a capacitor. The circuit manufacturer tells users that TH, the time the clock output spends in the high (5V) state, is TH =(R1 + R2)*C*ln(2). Similarly, the time spent in the low (0 V) state is TL = R2*C*ln(2). Design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2?

ANSWER : R1 = 144.3Ω,   R2 =  72.2 Ω

Explanation:

Frequency = 10 MHz

Time period = 1 / F =  0.1 u s

Duty cycle = 75% = 0.75

Duty cycle can be represented as :   Ton / T

Also: Ton = Th = 0.75 * 0.1 u s  = 75 n s

TL = T - Th = 100 ns - 75 n s = 25 n s

To find the value of R2 we use the equation for  time spent in the low (0 V) state

TL = R2*C*ln(2)

hence R2 = TL / ( C * In 2 )

c = 500 pF

Hence R2 = 25 / ( 500 pF * 0.693 )  = 72.2 Ω

To find the value of R1 we use the equation for the time the clock output spends in the high (5V) state,

Th = (R1 + R2)*C*ln(2)

  from the equation make R1 the subject of the formula

R1 =  (Th - ( R2 * C * In2 )) / (C * In 2)

R1 = ( 75 ns - ( 72.2 * 500 pF * 0.693)) / ( 500 pF * 0.693 )

R1 = ( 75 ns  - ( 25 ns ) / 500 pf * 0.693

     = 144.3Ω

Two stars of masses M and 6M are separated by a distance D. Determine the distance (measured from M) to a point at which the net gravitational force on a third mass would be zero.

Answers

Answer:

0.29D

Explanation:

Given that

F = G M m / r2

F = GM(6m) / (D-r)2

G Mm/r2 = GM(6m) / (D-r)2

1/r2 = 6 / (D-r)2

r = D / (Ö6 + 1)

r = 0.29 D

See diagram in attached file

The first step to merging is entering the ramp and _____.
A. honking to indicate your location
B. matching your speed
C. signaling your intent
D. telling your passengers where you're going

Answers

Answer:

  B. matching your speed

Explanation:

To merge safely, you must identify a gap in traffic and match your speed to the speed of the gap. Before you make your move to fill the gap, you should signal your intent.*

_____

* At least one resource says "The first step ... is to make sure you're traveling at the same speed ..." Then it goes on to say "Use your indicator. Do it early ...." The accompanying animation shows blinkers being activated on the ramp before the merge lane is entered. Apparently, "the first step" is not necessarily the first thing you do.

Answer:

It's C "signaling your intent"

Explanation:

The key thing to look at is they are asking the rest of the first step and that;s C

For the cellar of a new house, a hole is dug in the ground, with vertical sides going down 2.10 m. A concrete foundation wall is built all the way across the 8.90 m width of the excavation. This foundation wall is 0.189 m away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force that the water causes on the foundation wall. For comparison, the weight of the water is given by 2.10 m ✕ 8.90 m ✕ 0.189 m ✕ 1000 kg/m3 ✕ 9.80 m/s2 = 34.6 kN.

Answers

Answer:

The  force on the foundation wall is   [tex]F_f = 191394 \ N[/tex]

Explanation:

From the question we are told that

     The  depth of the hole's vertical side is  [tex]d = 2.10 \ m[/tex]

      The  width of the hole is  [tex]b = 8.90 \ m[/tex]

      The distance of the concrete wall from the front of the cellar is  [tex]c = 0.189 \ m[/tex]

Generally the area which the water from the drainage covers is mathematically represented as

        [tex]A = d * b[/tex]

substituting values

        [tex]A = 2.10 * 8.90[/tex]

       [tex]A = 18.69 \ m^2[/tex]

Now the gauge pressure exerted on the foundation wall is mathematically evaluated as

          [tex]P_g = \rho * d_{avg} * g[/tex]  

Here  is the average height foundation wall where the pressure of the water is felt and it is evaluated as

         [tex]d_{avg} = \frac{h_1 + h_2 }{2}[/tex]

where [tex]h_1[/tex] at the height at bottom of the hole which is equal to  [tex]h_1 = 0[/tex]

and  [tex]h_2[/tex] is the height at the top of the hole [tex]h_2 = d = 2.10[/tex]

        [tex]d_{avg} = \frac{0 + 2.10 }{2}[/tex]

       [tex]d_{avg} = 1.05[/tex]

Where  [tex]\rho[/tex] is the density of water with constant value [tex]\rho = 1000 \ kg/m^3[/tex]

substituting values

          [tex]P_g = 1000 * 1.05 * 9.8[/tex]

         [tex]P_g = 10290 \ Pa[/tex]

Then the force exerted by the water on the foundation wall mathematically represented as

      [tex]F_f = P_g * A[/tex]

substituting values

      [tex]F_f = 10290 * 18.69[/tex]

       [tex]F_f = 191394 \ N[/tex]

Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface temperature 1000K and emissivity 0.6:

Answers

Answer:

6.0 × [tex]10^{11}[/tex] W/[tex]m^{2}[/tex]

Explanation:

From Wien's displacement formula;

Q = e A[tex]T^{4}[/tex]

Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = [tex]\frac{Q}{A}[/tex] = e[tex]T^{4}[/tex]

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × [tex](1000)^{4}[/tex]

                             = 0.6 × 1.0 × [tex]10^{12}[/tex]

                             = 6.0 × [tex]10^{11}[/tex] [tex]\frac{W}{m^{2} }[/tex]

Therefore, the emissive intensity coming out of the surface is 6.0 × [tex]10^{11}[/tex] W/[tex]m^{2}[/tex].

You are in the frozen food section of the grocery store and you notice that your hand gets cold when you place it on the glass windows of the display cases. Your friend says this is because coolness is transferred from the display case to your hand. What do you think?

Answers

Answer:

I think my friend got it all wrong, as coolness can not be transferred but heat was actually transferred between my hand and the glass windows

Explanation:

In thermodynamics, coolness can not be transferred, only heat can be transferred

Here is how the mechanism of why i felt cold works, my body gave out heat, hence there was heat transfer from a region of high to a low heat region, equilibrium was reached and I started feeling the coolness in my hands.

Now, let's see what happens when the cannon is high above the ground. Click on the cannon, and drag it upward as far as it goes (15 m above the ground). Set the initial velocity to 14 m/s, and fire several pumpkins while varying the angle. For what angle is the range the greatest?


choices:


A. 45∘

B. 20∘

C. 30∘

D. 40∘

E. 50∘

Answers

Answer:

B. 20°

Explanation:

Range in projectile is defined as the distance covered in the horizontal direction. It is expressed as R = U²sin2Ф/g

U is the initial velocity of the body (in m/s)

Ф is the angle of projection

g is the acceleration due to gravity.

Given U = 14m/s, g = 9.8m/s and range R = 15 m

we will substitute this value into the formula to get the projection angle Ф as shown;

15 = 15²sin2Ф/9.8

15*9.8 = 15²sin2Ф

147 = 225sin2Ф

sin2Ф = 147/225

sin2Ф = 0.6533

2Ф = sin⁻¹0.6533

2Ф = 40.79°

Ф = 40.79°/2

Ф = 20.39° ≈ 20°

Hence, the range is greatest at angle 20°

A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is:
A. 0
B. 50 km/hr
C. 100 km/hr
D. 200 km/hr
E. cannot be calculated without knowing the acceleration

Answers

Answer:

The average velocity for this trip is 0 km/hr

Explanation:

We know that average velocity = total displacement/total time.

Now, its displacement is d = final position - initial position.

Since the  car starts and ends at its initial position at Hither, if we assume its initial position is 0 km, then its final position is also 0 km.

So, its displacement is d = 0 km - 0 km = 0 km.

Since the total time for the round trip is 2 hours, the average velocity is

total displacement/ total time = 0 km/2 hr = 0 km/hr.

So the average velocity for this trip is 0 km/hr  

At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is increasing at a rate of 3.0 J/s, how fast is the current changing

Answers

Answer:

The current is changing at the rate of 0.20 A/s

Explanation:

Given;

inductance of the inductor, L = 5.0-H

current in the inductor, I = 3.0 A

Energy stored in the inductor at the given instant, E = 3.0 J/s

The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

E = 22.5 J/s

This energy is increased by 3.0 J/s

E = 22.5 J/s + 3.0 J/s = 25.5 J/s

Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

I² = 10.2

I = √10.2

I = 3.194 A/s

The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

≅0.20 A/s

Therefore, the current is changing at the rate of 0.20 A/s.

The rate at which the current is changing is;

di/dt = 0.2 A/s

We are given;

Inductance; L = 5 H

Current; I = 3 A

Rate of Increase of energy; dE/dt = 3 J/s

Now, the formula for energy stored in inductor is given as;

E = ¹/₂LI²

Since we are looking for rate at which current is changing, then we differentiate both sides of the energy equation to get;

dE/dt = LI (di/dt)

Plugging in the relevant values gives;

3 = (5 × 3)(di/dt)

di/dt = 3/(5 × 3)

di/dt = 0.2 A/s

Read more at; https://brainly.com/question/13112120

You are using a hydrogen discharge tube and high quality red and blue light filters as the light source for a Michelson interferometer. The hydrogen discharge tube provides light of several different wavelengths (colors) in the visible range. The red light in the hydrogen spectrum has a wavelength of 656.3 nm and the blue light has a wavelength of 434.0 nm. When using the discharge tube and the red filter as the light source, you view a bright red spot in the viewing area of the interferometer. You now move the movable mirror away from the beam splitter and observe 158 bright spots. You replace the red filter with the blue filter and observe a bright blue spot in the interferometer. You now move the movable mirror towards the beam splitter and observe 114 bright spots. Determine the final displacement (include sign) of the moveable mirror. (Assume the positive direction is away from the beam splitter.)

Answers

Answer:

final displacement = +24484.5 nm

Explanation:

The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;

Δr = 2d2 - 2d1 = 150λ1

So, 2d2 - 2d1 = 150λ1

Dividing both sides by 2 to get;

d2 - d1 = 75λ1 - - - - eq1

Where;

d1 = distance between the fixed mirror and the beam splitter

d2 = position of moveable mirror from splitter when 158 bright spots are observed

Now, the path difference between the two waves when 114 bright spots were observed is;

Δr = 2d'2 - 2d1 = 114λ1

2d'2 - 2d1 = 114λ1

Divide both sides by 2 to get;

d'2 - d1 = 57λ1

Where;

d'2 is the new position of the movable mirror from the splitter

Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.

(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2

d2 - d1 - d'2 + d1 = 75λ1 - 57λ2

d2 - d'2 = 75λ1 - 57λ2

We are given;

(λ1 = 656.3 nm) and λ2 = 434.0 nm.

Thus;

d2 - d'2 = 75(656.3) - 57(434)

d2 - d'2 = +24484.5 nm

A 25 kg box sliding to the left across a horizontal surface is brought to a halt in a distance of 15 cm by a horizontal rope pulling to the right with 15 N tension.

Required:
a. How much work is done by the tension?
b. How much work is done by gravity?

Answers

B)is pills everything to the surface of the earth not sure about A

The work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.

The given parameters;

mass of the box, m = 25 kgdistance traveled by the box, d = 15 cm = 0.15 mtension on the rope, T = 15 N

The work done by the tension is calculated as follows;

W = Fd

W = 15 x 0.15

W = 2.25 J

The work done by gravity is calculated as;

W = (25 x 9.8) x 0.15

W = 36.75 J

Thus, the work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.

Learn more here: https://brainly.com/question/19498865

Why would physics be used to study light emitted by a star?
O A. Stars form interesting shapes in the sky.
B. Light is very pretty.
O C. The positions of stars control our lives.
O D. Light is a form of energy.

Answers

Answer:

O D.

Explanation:

Physics has an aspect that deals with the study of energy

Answer:

D. Light is a form of energy

Explanation:

In your words, describe how momentum is related to energy.

Answers

Answer:

you need momentum in order to release energy. For example, if you need to push something heavy and you get a running head start, then it will be easier.

Explanation:

A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 26 m/s when a 60 kg skydiver drops out by releasing his grip on the glider.
What is the glider's speed just after the skydiver lets go?

Answers

Answer:

The glider’s speed after the skydiver lets go is 26 m/s

Explanation:

To calculate the glider’s speed just after the skydiver lets go, we will need to use the conservation of momentum

Mathematically;

mv = mv + mv

so 680 * 26 = (680-60)v + 60 * 26

17680 = 620v + 1560

17680-1560 = 620v

16120 = 620v

v = 16120/620

v = 26 m/s

You throw a stone vertically upward with a speed of 26.0 m/s. (a) How fast is it moving when it reaches a height of 15.0 m? (b) How much time is required to reach this height when it's falling down? a. 19.5 m/s , b. 4.51 s a. 17.9 m/s , b. 0.620 s a. 19.5 m/s , b. 0.800 s a. 17.9 m/s , b. 4.28 s a. 380 m/s , b. 8 s

Answers

Answer:

ok well

Explanation:

teghe

Answer:

v = 19.5 m/s

t = 4.51 s

Explanation:

a)

given:

height is 15m from the ground

initial velocity Vi = 26 m/s

acceleration a or g = 9.81 m/s²

formula: Vf² = Vi² + 2aΔy

26² = Vi² + 2 (9.81) 15

Vi = 19.5 m/s

now you can calculate the time by using the equations below:

Δy = 1/2 (Vi + Vf) t

Vf = Vi + a t

Δy = Vi t + 1/2 a t

time must be 4.51 s

If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to:

Answers

Answer:

5.19 x 10³Hz

Explanation:

The capacitive reactance, [tex]X_{C}[/tex], which is the opposition given to the flow of current through the capacitor is given by;

[tex]X_C = \frac{1}{2\pi fC }[/tex]

Where;

f = frequency of the signal through the capacitor

C = capacitance of the capacitor.

Also, from Ohm's law, the voltage(V) across the capacitor is given by the product of current(I) and the capacitive reactance. i.e;

V = I x [tex]X_{C}[/tex]             [Substitute the value of

=> V = I x [tex]\frac{1}{2\pi fC}[/tex]      [Make f the subject of the formula]

=> f = [tex]\frac{I}{2\pi VC}[/tex]                    ---------------------(i)

From the question;

I = 3.33mA = 0.00333A

C = 8.50nF = 8.50 x 10⁻⁹F

V = 12.0V

Substitute these values into equation (i) as follows;

f = [tex]\frac{0.00333}{2 * 3.142 * 12.0 * 8.50 * 10^{-9}}[/tex]            [Taking [tex]\pi[/tex] = 3.142]

f = 5.19 x 10³Hz

Therefore, the frequency is closest to f = 5.19 x 10³Hz

Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If of the light passes through this combination, what is the angle between the transmission axes of the two filters

Answers

Answer:

The angle between the transmission axes of the filters is 65°

Explanation:

The complete question is

Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 18% of the light passes through this combination, what is the angle between the transmission axes of the two filters.

From Malus law,

[tex]I = I_{0} cos^{2} \beta[/tex]    ....1

where [tex]I[/tex] is the intensity of the polarized light,

[tex]I_{o}[/tex] is the intensity of the incident light

β the angle between the transmission axes of the two filters

Since the intensity is reduced to 18% or 0.18 of its initial value, this means that

[tex]cos^{2} \beta[/tex] = 0.18

substituting into the equation above, we have

[tex]I = 0.18I_{0}[/tex]    ....2

equating the two equations, we have

[tex]I_{0}cos^{2} \beta[/tex] = [tex]0.18I_{0}[/tex]

[tex]cos^{2}\beta[/tex] = [tex]\frac{0.81I_{0} }{I_{0} }[/tex] = 0.18

[tex]cos \beta[/tex] = [tex]\sqrt{0.18}[/tex] = 0.424

[tex]\beta[/tex] = [tex]cos^{-1} 0.424[/tex] = 64.9 ≅ 65°


Science activity
Imagine that some settlers have left Earth and gone to the Moon, taking
their recipe books with them. The first cake they baked was a disaster. It had
far too little moisture and was about six times the size they had expected.
the cake recipe was:

1.25 N butter
1.50 N sugar

4 eggs
1.50 N flour

20 ml milk
ANALYTICAL THINKING
Q. Why was the cake so big? Why was it se
dry?​

Answers

Answer:

Answer in explanation

Explanation:

The reason for the big size and less moisture of the cake is due to difference in weight of the ingredients on the surface of moon. So, the same has the lesser weight on the surface of the moon than it has on the surface of earth. Or in other words, The same weight of the ingredients will have greater mass and thus the greater quantity on the surface of earth than the surface of earth. For example, on earth 1.25 N butter will have a mass:

m = W/g = 1.25 N/(9.8 m/s²) = 0.13 kg

But, on moon:

m = W/g = 1.25 N/(1.625 m/s²) = 0.77 kg

Hence, it is clear that the mass of the same weight of the substance becomes 6 times greater on the surface of moon. This explains why the cake was so big.

Now, coming to the second part about the dryness of the cake. The main and only source of moisture in recipe is the eggs bu the eggs are taken in a quantity of numbers. So they are exactly the same on moon as well. While all the other ingredients are increased, the same amount of eggs are not sufficient to provide them with enough moisture. Hence the cake was dry.

Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

Answers

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

a football is kicked toward a goal keeper with an initial speed of 20m/s at an angle of 45 degrees with the horizontal .at the moment the ball is kicked the goal keeper is 50m from the player .at what speed and in what direction must the goalkeeper run in order to catch the ball at the same height at which it was kicked​

Answers

Answer:

3.18 m/s

Explanation:

Given that

Initial speed of the ball, u = 20 m/s

Angle of inclination, θ = 45°

Distance from the ball, h = 50 m

Using equations of projectile to solve this, we have

We start by finding the time of flight, T

T = 2Usinθ/g

T = (2 * 20 * sin45)/9.8

T = (40 * 0.7071) / 9.8

T = 28.284/9.8

T = 2.89 s

Next we find the Range, R

R = u²sin2θ/g

R = (20² * sin 90) / 9.8

R = (400 * 1) / 9.8

R = 400/9.8 = 40.82 m

Distance the gk must cover

40.82 - 50 m

-9.18 m or 9.18 m in the opposite direction.

Speed of the GK = d/t

9.18 / 2.89 = 3.18 m/s

Recent technological developments like high-resolution satellite imagery and diagnostic positron emission tomography (PET scans) have refined and extended the camera’s capacity to provide information. Which passage assertion does this information support most strongly?

Answers

Answer:

D) Photography can be used to both control and benefit society.

Explanation:

High-resolution satellite imagery and diagnostic positron emission tomography (PET scans) have been used to both control and benefits the society in the sense that it has helped to take records of information of crime, traffic offenders such drunk drivers and over speeding drivers, e.t.c. it helps control by given their information and automatically penalizing them or ensuring the agency penalized them and also benefit the society by preventing people from committing crime thereby, protecting them from offenders.

A cylinder is closed by a piston connected to a spring of constant 2.20 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. The piston has a cross sectional area of 0.0100 m^2 and negligible mass. What is the pressure of the gas at 250 °C?

Answers

Answer:

1.3515x10^5pa

Explanation:

Plss see attached file

1. Water flows through a hole in the bottom of a large, open tank with a speed of 8 m/s. Determine the depth of water in the tank. Viscous effects are negligible.

Answers

Answer:

3.26m

Explanation:

See attached file

What is dark energy?

Answers

Explanation:

Dark Energy. Dark Energy is a hypothetical form of energy that exerts a negative, repulsive pressure, behaving like the opposite of gravity. It has been hypothesised to account for the observational properties of distant type Ia supernovae, which show the universe going through an accelerated period of expansion

The planets how and block are near each other in the Dorgon system. the Dorgons have very advanced technology, and a Dorgon scientist wants to increase the pull of gravity between the two planets. Which proposals would the scientist make to accomplish this goal? check all that apply.

Answers

Answer:

Decreasing the distance between Hox and Blox, increasing the mass of Hox, or increasing the mass of Hox and Blox.

Explanation:

The gravity force is directly proportional to the mass of the bodies and inversely proportional to the square of the distance that separates them.

Or

If we decrease the distance between both planets (Hox and Blox), the gravitational pull between them will increase.  

On the other hand, if we keep the distance between Hox and Blox, but we increase the mass of one of them, or increase the mass of both, the gravitational pull between them will also increase.

Two long parallel wires are separated by 11 cm. One of the wires carries a current of 54 A and the other carries a current of 45 A. Determine the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current.

Answers

Explanation:

It is given that,

The separation between two parallel wires, r = 11 cm = 0.11 m

Current in wire 1, [tex]q_1=54\ A[/tex]

Current in wire 2, [tex]q_2=45\ A[/tex]

Length of wires, l = 4.3 m

We need to find the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current. The magnetic force per unit length is given by :

[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 54\times 45\times 4.3}{2\pi \times 0.11}\\\\F=0.0189\ N[/tex]

So, the magnetic force on a 4.3 m length of the wire  on both of currents is F=0.0189 N.

Since you analyzed the charging of a capacitor for a DC charging voltage, how is it possible that you

Answers

Answer:

 I = E/R   e^{-t/RC}

Explanation:

In a capacitor charging circuit you must have a DC power source, the capacitor, a resistor, and a switch. When closing the circuit,

                  E -q / c-IR = 0

we replace the current by its expression and divide by the resistance

                   I = dq / dt

                 

                   dq / dt = E / R  -q / RC

                   dq / dt = (CE -q) / RC

we solve the equation

                   dq / (Ce-q) = -dt / RC

we integrate and evaluate for the charge between 0 and q and for the time 0 and t

                   ln (q-CE / -CE) = -1 /RC   (t -0)

eliminate the logarithm

              q - CE = CE [tex]e^{-t/RC}[/tex]

               q = CE (1 + 1/RC  e^{-t/RC} )

In general the teams measure the current therefore we take the derivative to find the current

               i = CE (e^{-t/RC} / RC)

               I = E/R   e^{-t/RC}

This expression is the one that describes the charge of a condensate in a DC circuit

An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal butopposite charge on its plates. All the geometric parameters of the capacitor (plate diameter andplate separation) are now DOUBLED. If the original energy stored in the capacitor was U0, howmuch energy does it now store?

Answers

Answer:

U_f = (U_o)/2)

Explanation:

The capacitance of a given capacitor is given by the formula;

C = ε_o•A/d

While energy stored in plates capacitor is given as; U_o = Q²/2C

Now,we are told that that all the dimensions of the capacitor plate is doubled. Thus, we now have;

C' = ε_o•4A/2d

Hence, C' = 2C

so capacitance is now doubled

Thus, the final energy stored between the plates of capacitor is given as;

U_f = Q²/2C'

From earlier, we saw that C' = 2C.

Thus;

U_f = Q²/2(2C)

U_f = Q²/4C

Rearranging, we have;

U_f = (1/2)(Q²/2C)

From earlier, U_o = Q²/2C

Hence,

U_f = (1/2)(U_o)

Or

U_f = (U_o/2)

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