Two forces A and B act at a point. If their resultant is [given by] (B - A) in the direction of B, then
A. A and B are equal
B. A is greater than B
C. the angle between A and B is 0°
D. the angle between A and B is 90°
E. the angle between A and B is 180°

(A)

Explanation:

We can see that the resistors are connected in parallel so all of them have the same voltage of 100 V. We also know that

[tex]P = VI[/tex]

Since resistor Y dissipates 100 W of power, we can solve for the current as

[tex]I = \dfrac{P}{V} = \dfrac{100\:\text{W}}{100\:\text{V}} = 1.0\:\text{A}[/tex]

The current in resistor Y is

Answer:

The percentage of people who wanted both the swimming pool and the soccer complex is 0.6 + 0.6 – 0.95 = 0.25. This can also be seen in the Venn diagram.

Explanation:

Edmentum

After being released, the restoring force exerted by the spring performs

1/2 (5200 N/m) (0.090 m)² = 12.06 J

of work on the block. At the same time, the block's weight performs

- (0.260 kg) g (0.090 m) ≈ -0.229 J

of work. Then the total work done on the block is about

W ≈ 11.83 J

The block accelerates to a speed v such that, by the work-energy theorem,

W = ∆K   ==>   11.83 J = 1/2 (0.260 kg) v ²   ==>   v ≈ 9.54 m/s

Past the equilibrium point, the spring no longer exerts a force on the block, and the only force acting on it is due to its weight, hence it has a downward acceleration of magnitude g. At its highest point, the block has zero velocity, so that

0² - v ² = -2gy

where y is the maximum height. Solving for y gives

y = v ²/(2g) ≈ 4.64 m

Answer:

c) 2Q

Explanation:

From the given information:

The pressure inside a pipe can be expressed by using the formula:

[tex]\Delta P = \dfrac{128 \mu L Q}{\pi D^4}[/tex]

Since the diameter in both pipes is the same, we can say:

[tex]D = D_A = D_B[/tex]

where;

length of the first pipe A [tex]L_A = L[/tex] and the length of the second pipe B [tex]L_B = 2L[/tex]

Since the difference in pressure is equivalent in both pipes:

Then:

[tex]\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}[/tex]

[tex]\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}[/tex]

[tex]\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}[/tex]

[tex]\mathbf{Q_1 = 2Q}[/tex]

The flow rate in pipe B is 2Q of the flow rate of the pipe A

The flow rate is defined as the flow of the fluid across the cross section in per unit time.

From the given information:

The pressure inside a pipe can be expressed by using the formula:

[tex]\Delta p=\dfrac{128\mu LQ}{\pi D^4}[/tex]

Since the diameter in both pipes is the same, we can say:

[tex]D=D_A=D_B[/tex]

where;

length of the first pipe A  [tex]L_A=L[/tex] and the length of the second pipe B  

[tex]L_B=2L[/tex]

Since the difference in pressure is equivalent in both pipes:

Then:

[tex]\dfrac{128\mu L_1Q_1}{\pi D_1^4}=\dfrac{128\mu L_2Q_2}{\pi D_2^4}[/tex]

[tex]\dfrac{L_1Q_1}{D_1^4}=\dfrac{L_2Q_2}{D_2^4}[/tex]

[tex]\dfrac{LQ_1}{D_1^4}=\dfrac{2LQ}{D_2^4}[/tex]

[tex]Q_1=2Q[/tex]

Hence the flow rate in pipe B is 2Q of the flow rate of the pipe A

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Answer:

  E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]

Explanation:

In this exercise you are asked to calculate the electric field between two plates, the electric field is a vector

         E_ {total} = E₁ + E₂

         E_ {total} = 2 E

where E₁ and E₂ are the fields of each plate, we have used that for the positively charged plate the field is outgoing and for the negatively charged plate the field is incoming, therefore in the space between the plates for a test charge the two fields point in the same direction

to calculate the field created by a plate let's use Gauss's law

          Ф = ∫ E . dA = q_{int} /ε₀

As a Gaussian surface we use a cylinder with the base parallel to the plate, therefore the direction of the electric field and the normal to the surface are parallel, therefore the scalar product is reduced to the algebraic product.

           E 2A = q_{int} / ε₀

where the 2 is due to the surface has two faces

indicate that the surface has a uniform charge for which we can define a surface density

           σ = q_{int} / A

           q_{int} = σ A

we substitute

           E 2A = σ A /ε₀

           E = σ / 2ε₀  

therefore the total field is

           E_ {total} = σ /ε₀

let's substitute the density for the charge of the whole plate

           σ= Q / L²

            E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]

Answer:

c) charges exert forces on other charges.

Explanation:

When two different materials are rubbed together, there is a transfer of electrons from one material to the other material so this causes one object to become positively charged and the other object is negatively charged so they will attract each other not repel each other. Charges exert forces on other charges i.e. opposite charges attract each other whereas similar charges repel each other so in both cases force are exerted on one another.

Answer:

i think bro is becuse they want to get money also paying to this app  if i put paying right im latin but that is my opinion and also becuse they want we learn somethings

Explanation:

Answer:

433.15K

Explanation:

(320°F − 32) × 5/9 + 273.15 = 433.15K

Answer:

Their main disadvantage is that they are fairly easy to break and if they do, it results in small splinters of glass and the release of mercury which is quite toxic if absorbed into the body.

Answer:

B

Explanation:

no reason for this answer

Following the definition of the center of mass, "In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero."

(see explanation below)

Answer:

Part A - 4.084 mJ

Part B - 0.908 mJ

Part C - 8.168 mJ

Explanation:

Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?

Since capacitors C₂ and C₃ are in series, their equivalent capacitance is C',

1/C' = 1/C₂ + 1/C₃      (Since C₁ = C₂ = C₃ = C)

1/C' = 1/C + 1/C

1/C' = 2/C

C' = C/2

Since C' is in parallel with C₁, the equivalent capacitance for the circuit is C" = C₁ + C' = C + C/2 = 3C/2

C" = 3C/2

The energy stored in the circuit, W = 1/2C"V² where C" = equivalent capacitance = 3C/2 and V = voltage = 15.0 V

W = 1/2C"V²

W = 1/2(3C/2)V²

W = 3CV²/4

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = 3CV²/4

W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/4

W = 3 × 24.2 × 10⁻⁶ F × 225 V²/4

W = 16335/4 × 10⁻⁶ FV²

W = 4083.75 × 10⁻⁶ J

W = 4.08375 × 10⁻³ J

W = 4.08375 mJ

W ≅ 4.084 mJ

Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?

If the capacitors are connected in series, their equivalent resistance is C'

and 1/C' = 1/C₁ + 1/C₂ + 1/C₃

Since C₁ = C₂ = C₃ = C

1/C' = 1/C + 1/C + 1/C

1/C' = 3/C

C' = C/3

The energy stored in the circuit, W = 1/2C'V² where C' = equivalent capacitance = C/3 and V = voltage = 15.0 V

W = 1/2C'V²

W = 1/2(C/3)V²

W = CV²/6

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = CV²/6

W = 24.2 × 10⁻⁶ F (15.0 V)²/6

W = 24.2 × 10⁻⁶ F × 225 V²/6

W = 5445/6 × 10⁻⁶ FV²

W = 907.5 × 10⁻⁶ J

W = 0.9075 × 10⁻³ J

W = 0.9075 mJ

W ≅ 0.908 mJ

Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?

If the capacitors are connected in parallel, their equivalent resistance is C'

and C' = C₁ + C₂ + C₃

Since C₁ = C₂ = C₃ = C

C' = C + C + C

C' = 3C

The energy stored in the capacitor network, W = 1/2C'V² where C' = equivalent capacitance = 3C and V = voltage = 15.0 V

W = 1/2C'V²

W = 1/2(3C)V²

W = 3CV²/2

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = 3CV²/2

W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/2

W = 3 × 24.2 × 10⁻⁶ F × 225 V²/2

W = 16335/2 × 10⁻⁶ FV²

W = 8167.5 × 10⁻⁶ J

W = 8.1675 × 10⁻³ J

W = 8.1675 mJ

W ≅ 8.168 mJ

Answer:

Work done = 2289.084 Joules

Explanation:

Given the following data;

Mass = 102 Kg

Height = 2.29

Time = 1.32 seconds

We know that acceleration due to gravity, g = 9.8 m/s²

a. To find the work done by the person;

Here, work would be done in the form of gravitational potential energy.

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

G.P.E = mgh

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the formula, we have;

Work done = 102 * 2.29 * 9.8

Work done = 2289.084 Joules

Answer:

d = 30kg/cm³

Explanation:

d = m/v

d = 1080kg/(3cm*4cm*3cm)

d = 30kg/cm³

Answer:

The correct option is B. 7.5 * 10¹⁴ Hz

Explanation:

Frequency = (speed) / (wavelength)

= (3 x 10⁸ m/s) / (4 x 10⁻⁷ m)

= (3/4 x 10¹⁵) ( m / m - s )

= (0.75 x 10¹⁵) /sec

= 7.5 x 10¹⁴ Hz

= 750,000 GHz

Answer:

Mark Brainliest please

answer is

Explanation:

For any wave,              

                          Frequency = (speed) / (wavelength)

                                             = (3 x 10⁸ m/s) / (4 x 10⁻⁷ m)

                                             = (3/4 x 10¹⁵)  ( m / m - s )

                                             = (0.75 x 10¹⁵)  /sec

                                             =  7.5 x 10¹⁴  Hz

                                             =   750,000 GH

Answer:

to the right

Explanation:

if the car turns to the lift, the body forces energy to the left side, so according to the first law of Newton, the body will move to the right side to resist the sudden motion.

Answer:

a) t = 3.6 s

b) d = 23 m

c) v = 13 m/s

Explanation:

Let t be the time the accelerating rider rides

the distance she travels is

d = ½3.6t²

the distance for the other cyclist is

d =3.5(t + 3)

½3.6t² = 3.5(t + 3)

1.8t² - 3.5t - 10.5 = 0

quadratic formula, positive answer

t = (3.5 + √(3.5² - 4(1.8)(-10.5))) / (2(1.8))

t = 3.575786...

d = ½(3.6)(3.575786²) = 23.015...

v = 3.6(3.575786) = 12.8728...

Answer:Lesson Objectives

Describe how the action of waves produces different shoreline features.

Discuss how areas of quiet water produce deposits of sand and sediment.

Discuss some of the structures humans build to help defend against wave erosion.

Vocabulary

arch

barrier island

beach

breakwater

groin

refraction

sea stack

sea wall

spit

wave-cut cliff

wave-cut platform

Introduction

Waves are important for building up and breaking down shorelines. Waves transport sand onto and off of beaches. They transport sand along beaches. Waves carve structures at the shore.

Wave Action and Erosion

All waves are energy traveling through some type of material, such as water (Figure below). Ocean waves form from wind blowing over the water.

Ocean waves are energy traveling through water.

The largest waves form when the wind is very strong, blows steadily for a long time, and blows over a long distance.

The wind could be strong, but if it gusts for just a short time, large waves won’t form. Wave energy does the work of erosion at the shore. Waves approach the shore at some angle so the inshore part of the wave reaches shallow water sooner than the part that is further out. The shallow part of the wave ‘feels’ the bottom first. This slows down the inshore part of the wave and makes the wave ‘bend.’ This bending is called refraction.

Wave refraction either concentrates wave energy or disperses it. In quiet water areas, such as bays, wave energy is dispersed, so sand is deposited. Areas that stick out into the water are eroded by the strong wave energy that concentrates its power on the wave-cut cliff (Figure below).

The wave erodes the bottom of the cliff, eventually causing the cliff to collapse.

Other features of wave erosion are pictured and named in Figure below. A wave-cut platform is the level area formed by wave erosion as the waves undercut a cliff. An arch is produced when waves erode through a cliff. When a sea arch collapses, the isolated towers of rocks that remain are known as sea stacks.

(a) The high ground is a large wave-cut platform formed from years of wave erosion. (b) A cliff eroded from two sides produces an arch. (c) The top of an arch erodes away, leaving behind a tall sea stack.

Wave Deposition

Rivers carry sediments from the land to the sea. If wave action is high, a delta will not form. Waves will spread the sediments along the coastline to create a beach (Figure below). Waves also erode sediments from cliffs and shorelines and transport them onto beaches.

Sand deposits in quiet areas along a shoreline to form a beach.

Beaches can be made of mineral grains, like quartz, rock fragments, and also pieces of shell or coral (Figure below).

Quartz, rock fragments, and shell make up the sand along a beach.

Waves continually move sand along the shore. Waves also move sand from the beaches on shore to bars of sand offshore as the seasons change. In the summer, waves have lower energy so they bring sand up onto the beach. In the winter, higher energy waves bring the sand back offshore.

Some of the features formed by wave-deposited sand are in Figure below. These features include barrier islands and spits. A spit is sand connected to land and extending into the water. A spit may hook to form a tombolo.

Examples of features formed by wave-deposited sand.

Shores that are relatively flat and gently sloping may be lined with long narrow barrier islands (Figure below). Most barrier islands are a few kilometers wide and tens of kilometers long.

(a) Barrier islands off of Alabama. A lagoon lies on the inland side. (b) Barrier islands, such as Padre Island off the coast of Texas, are made entirely of sand. (c) Barrier islands are some of the most urbanized areas of our coastlines, such as Miami Beach.

In its natural state, a barrier island acts as the first line of defense against storms such as hurricanes. When barrier islands are urbanized (Figure above), hurricanes damage houses and businesses rather than vegetated sandy areas in which sand can move. A large hurricane brings massive problems to the urbanized area.

Protecting Shorelines

Intact shore areas protect inland areas from storms that come off the ocean (Figure below).

Dunes and mangroves along Baja California protect the villages that are found inland.

Explanation:

Answer: Below

Explanation: Correct on Edmentum

=F×s×cosa=2×g×0,8×cos90°= 0

The work done by gravity on a ball of 2 kg which is moving with a constant speed of 6 meter per second is zero. Thus, the correct option is A.

Work is the energy transfer to or from an object through the application of force along with the displacement. For a constant force aligned with the direction of motion, the work done is equal to the product of the force strength which is applied and the distance traveled by the object.

Work = Force × Displacement

Force = Mass × Acceleration

Acceleration of the ball is zero as it is moving with a constant speed. Therefore, the work done by the gravity is zero.

Therefore, the correct option is A.

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Explanation:

The equivalent capacitance of capacitors in parallel can be determined as

[tex]C_{eq} = C_1 + C_2 + C_3[/tex]

[tex]\:\:\:\:\:= 40\:\text{F} + 10\:\text{F} + 50\:\text{F} = 100\:\text{F}[/tex]

The frequency is 4,6E14 Hz.

What is the frequency?

Frequency is the fee at which modern changes direction in step with 2nd. it's far measured in hertz (Hz), a worldwide unit of degree wherein 1 hertz is identical to 1 cycle in line with 2d. Hertz (Hz) = One hertz is the same as 1 cycle in step with the second. Cycle = One entire wave of alternating present-day voltage.

Frequency describes the number of waves that pass a hard and fast place in a given quantity of time. So if the time it takes for a wave to skip is half of 2d, the frequency is 2 per 2nd. If it takes 1/one hundred of an hour, the frequency is a hundred in step with hour.

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Answer:

B

Explanation:

P1/P1 = 40/20

=2

Answer:

θ₄ = 37.2º

Explanation:

For this exercise it must be solved in two parts, the first part we look for the critical angle, for this we use the law of refraction with the angle in the middle of transmission of tea = 90º

        n₁ sin θ₁= n₂ sin  90

        θ₁ = sin⁻¹ [tex]\frac{n_2}{n_1}[/tex]

        θ₁ = sin⁻¹ (1.33 / 1.43)

        θ₁ = 68.4º

They indicate that the angle of incidence is half of the critical angle

        θ₃ = 68.4 / 2 = 34.2º

Let's use the law of refraction again

         n₁ sin θ₃ = n₂ sin θ₄

         sin θ₄ = [tex]\frac{n_1}{n_2}[/tex]   sin θ₃

         sin θ₄ = [tex]\frac{1.43}{1.33}[/tex]  sin 34.2

         θ₄ = sin⁻¹ 0.604345

        θ₄ = 37.2º

Answer:

Answer:

vdhdbdnnsnsbdhhshzbhshsbbsbd is not ask you to be able and r in the exam qq and

Answer:

ok

Explanation:

Answer:

L = 169.5 m

Explanation:

Using Ohm's Law:

V = IR

where,

V = Voltage = 1.5 V

I = Current = 10 mA = 0.01 A

R = Resistance = ?

Therefore,

1.5 V = (0.01 A)R

R = 150 Ω

But the resistance of a wire is given by the following formula:

[tex]R = \frac{\rho L}{A}[/tex]

where,

ρ = resistivity = 1 x 10⁻⁶ Ω.m

L = length of wire = ?

A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²

A = 1.13 x 10⁻⁶ m²

Therefore,

[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]

L = 169.5 m

(B) 1.00 m

Explanation:

Since the meter stick is traveling with Jill, it will have the same speed as she does so relative to Jill, the meter stick is stationary so its length remains 1.00 m as measured by her.

When astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c and measure the meter stick to be 1 meter. Hence, option B is correct.

Length contraction is defined as the phenomenon of the moving object being shorter than its appropriate length, measured in the object's rest frame.

When the object travels with the speed of light, the length of the object gets more contracted than its original length, relative to the observer. It is also known as the Lorentz-Fitgerald contraction.

Length contraction, L = L₀√(1-v²/c²), where L is the original length, L₀ is the contracted length. c² is known as the velocity of light. v² is the velocity of the speed of the object.

From the given,

speed of the spaceship = 0.280c      (c is the speed of the light)

Length contraction, L = L₀ √(1-v²/c²)

The stick also travels in the spaceship. Hence, the length of the meter stick does not change. It remains at its original length of one meter.  Thus, the ideal solution is option B.

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Answer:

No. As we gain mass the force of gravity on us does not decrease

Answer:

Explanation:

Since energy is conserved:

2

mu  

2

​

=  

2

mv  

2

​

+mgh

⇒u  

2

=v  

2

+2gh

⇒(3)  

2

=v  

2

+2(9.8)(0.5−0.5cos60)

⇒v=2m/s

Acceleration of the simple pendulum is 2.62 m/s².

When a point mass is suspended from a fixed support by a light, non-extensible string, the instrument is said to be a simple pendulum.

Here,

Let the mass of the bob be m. The simple pendulum is attached to the fixed support with a string having length l. The pendulum makes an angle of 15° with the vertical from the equilibrium point.

Let T be the tension acting on the string.

As, the bob passes through the angle,

The weight of the bob becomes equal to the vertical component of the tension.

mg = T cos15°

Also, the horizontal component of the tension,

T sin15° = ma

By solving these two equations, we get that,

Acceleration of the simple pendulum,

a = g tan15°

a = 9.8 x 0.267

a = 2.62 m/s²

Hence,

Acceleration of the simple pendulum is 2.62 m/s².

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Answer:

a) 0.46875

b) 8 km

Explanation:

Smokestack height ( H ) = 50 m

speed of pollutant / wind speed = 3 m/s

Half width = 2 [tex]\sqrt{2Dt }[/tex] = 50 m  ---- ( 1 )

a) If L = 2 km

value of turbulent diffusion coefficient D

back to equation 1

50 = 2 √ 2 * D * ( 2000/3 )

2500 = 4 * 2 * D * ( 2000/3 )

D = 2500 / ( 8 * ( 2000/3 )  )

   = 0.46875

where : time to travel ( t ) = Distance / speed = 2000 / 3

b) when the smoke stack = 50 * 2 = 100 m

L = 800 m = 8 km

attached below is the detailed solution