Two train 75 km apart approach each other on parallel tracks, each moving at
15km/h. A bird flies back and forth between the trains at 20km/h until the trains pass
each other. How far does the bird fly?

Answer:

a)   θ = 14.23º, b)   θ₂ = 75.77,  c) t = 0.6019 s

Explanation:

This is a missile throwing exercise.

a) the reach of the ball is the distance traveled for the same departure height

          R = [tex]\frac{v_o^2 \ sin 2 \theta }{g}[/tex]

          sin 2θ = [tex]\frac{Rg}{v_o^2}[/tex]

          sin 2θ = 7.00 9.8 / 12.0²

          2θ = sin⁻¹ (0.476389) = 28.45º

           θ = 14.23º

the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º

          θ ’= 90  -14.23

          θ’= 75.77º

b) the two angles that give the same range are

         θ₁ = 14.23

         θ₂ = 75.77

the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.

C) the time of the pass can be calculated with the expression

           x = v₀ₓ t

           t = x / v₀ₓ

           t = 7 / 11.63

           t = 0.6019 s

i. [tex]T = 36.8\:\text{N}[/tex]

ii. [tex]a = 2.45\:\text{m/s}^2[/tex]

iii. [tex]x = 1.23\:\text{m}[/tex]

Explanation:

Let's write Newton's 2nd law for each object. We will use the sign convention assigned for each as indicated in the figure. Let T be the tension on the string and assume that the string is inextensible so that the two tensions on the strings are equal. Also, let a be the acceleration of the two masses. And [tex]m_1 = 3\:\text{kg}[/tex] and [tex]m_2 = 5\:\text{kg}[/tex]

Forces acting on m1:

[tex]T - m_1g = m_1a\:\:\:\:\:\:\:(1)[/tex]

Forces acting on m2:

[tex]m_2g - T = m_2a\:\:\:\:\:\:\:(2)[/tex]

Combining Eqn(1) and Eqn(2) together, the tensions will cancel out, giving us

[tex]m_2g - m_1g = m_2a + m_1a[/tex]

or

[tex](m_2 - m_1)g = (m2 + m_1)a[/tex]

Solving for a,

[tex]a = \left(\dfrac{m_2 - m_1}{m_2 + m_1}\right)g[/tex]

[tex]\:\:\:\:= \left(\dfrac{5\:\text{kg} - 3\:\text{kg}}{5\:\text{kg} + 3\:\text{kg}}\right)(9.8\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:= 2.45\:\text{m/s}^2[/tex]

We can solve for the tension by using this value of acceleration on either Eqn(1) or Eqn(2). Let's use Eqn(1).

[tex]T - (3\:\text{kg})(9.8\:\text{m/s}^2) = (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]

[tex]T = (3\:\text{kg})(9.8\:\text{m/s}^2) + (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:= 29.4\:\text{m/s}^2 + 7.35\:\text{m/s}^2 = 36.8\:\text{N}[/tex]

Assuming that the two objects start from rest, the distance that they travel after one second is given by

[tex]x = \frac{1}{2}at^2 = \frac{1}{2}(2.45\:\text{m/s}^2)(1\:\text{s})^2 = 1.23\:\text{m}[/tex]

We know

[tex]\boxed{\large{\sf Velocity=\dfrac{Distance}{Time}}}[/tex]

[tex]\\ \Large\sf\longmapsto Velocity=\dfrac{6}{\dfrac{1}{2}}[/tex]

[tex]\\ \Large\sf\longmapsto Velocity=6\times 2[/tex]

[tex]\\ \Large\sf\longmapsto Velocity=12km/h[/tex]

Is this the right answer??

We should keep km and min in smallest SI unit

285.3 days

Explanation:

The centripetal force [tex]F_c[/tex] experienced by the planet is the same as the gravitational force [tex]F_G[/tex] so we can write

[tex]F_c = F_G[/tex]

or

[tex]m\dfrac{v^2}{R} = G\dfrac{mM}{R^2}[/tex]

where M is the mass of the star and R is the orbital radius around the star. We know that

[tex]v = \dfrac{C}{T} = \dfrac{2\pi R}{T}[/tex]

where C is the orbital circumference and T is orbital period. We can then write

[tex]\dfrac{4\pi^2R}{T^2} = G\dfrac{M}{R^2}[/tex]

Isolating [tex]T^2[/tex], we get

[tex]T^2 = \dfrac{4\pi^2R^3}{GM}[/tex]

Taking the square root of the expression above, we get

[tex]T = 2\pi \sqrt{\dfrac{R^3}{GM}}[/tex]

which turns out to be [tex]T = 2.47×10^7\:\text{s}[/tex]. We can convert this into earth days as

[tex]T = 2.47×10^7\:\text{s}×\dfrac{1\:\text{hr}}{3600\:\text{s}}×\dfrac{1\:\text{day}}{24\:\text{hr}}[/tex]

[tex]\:\:\:\:\:= 285.3\:\text{days}[/tex]

Answer:

The vibrating length of the air column is greater than the actual length of the organ pipe

Answer:

h = 27.17 m

Explanation:

First, we will calculate the total mechanical energy of the system at the bottom point of the third loop:

Mechanical Energy = Kinetic Energy + Potential Energy

[tex]E = \frac{1}{2}mv^2 + mgh[/tex]

where,

E = Total Mechanical Energy = ?

m = mass of the roller coaster = 1200 kg

v = velocity of the roller coaster = 22 m/s

g = acceleration due to gravity = 9.81 m/s²

h = height of roller coaster = 2.5 m

Therefore,

[tex]E = \frac{1}{2}(1200\ kg)(22\ m/s)^2+(1200\ kg)(9.81\ m/s^2)(2.5\ m)\\\\E = 290400 J +29430\ J\\\\E = 319830\ J = 319.83\ KJ[/tex]

Now, the total mechanical energy at the top position of the first hill must also be the same:

[tex]E = \frac{1}{2}mv^2 + mgh[/tex]

where,

v = 0 m/s

h = ?

Therefore,

[tex]319830\ J = \frac{1}{2}(1200\ kg)(0\ m/s)^2+(1200\ kg)(9.81\ m/s^2)(h)\\\\h = \frac{319830\ J}{11772\ N}\\\\[/tex]

h = 27.17 m

Answer:

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

Answer:

The results of the experiment indicated a shift consistent with zero, and certainly less than a twentieth of the shift expected if the Earth's velocity in orbit around the sun was the same as its velocity through the ether.

Explanation:

Answer:

The angular acceleration is 26.6 rad/s^2.

Explanation:

Force, F = 314 N

radius, r = 0.281 m

mass, m = 84.2 kg

The grindstone is a disc.

The torque is given by

torque = force x radius

Torque = 314 x 0.281 = 88.234 Nm

The torque is given by

Torque = Moment of inertia x angular acceleration

[tex]88.234 = 0.5 mr^2 \alpha \\\\88.234 = 0.5\times 84.2\times 0.281\times 0.281\times \alpha \\\\\alpha = 26.6 rad/s^2[/tex]

Explanation:

Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]

The sum of the two vectors is

[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]

[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]

The difference between the two vectors can be written as

[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]

[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]

(1) When the driver is at rest, the restoring force exerted by spring is equal in magnitude to the driver's weight, so that

∑ F = s - mg = 0   ==>   s = mg = 617.4 N

If the spring is compressed 0.015 m, then the spring constant k is such that

617.4 N = k (0.015 m)   ==>   k = 41,160 N/m ≈ 41 kN/m

(2) The total mass of the passengers is

24 (68 kg) + 3 (15 kg) + 5 (3 kg) + 25 kg = 1717 kg

so that if everyone is at rest, the spring is compressed a distance x such that

kx = (1717 kg) g   ==>   x ≈ 0.41 m

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

Answer:

due to production of heat through friction

Explanation:

because of the friction produce between her hands

Answer:

Explanation:

An impulse results in a change of momentum.

The impulse is the product of a force and a distance. This will be represented by the area under the curve

a) W = ½(4.00)(3.00) = 6.00 J

b) W = (11.0 - 4.00)(3.00) = 21.0 J

c) W = ½(17.0 - 11.0)(3.00) = 9.00 J

d) ASSUMING the speed at x = 0 is in the direction of applied force

½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00

v₄ = 2.05 m/s

½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00

v₁₇ = 4.92 m/s

If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.

Answer:

A change in pressure at one point in an incompressible fluid is felt at every other point in the fluid.

Explanation:

Pascal's principle states that ''pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.''(Science direct).

The implication of this law is; that a change in pressure at one point in an incompressible fluid is felt at every other point in the fluid. Hence the correct answer chosen above.

The Pascal's principle is applied in hydraulic jacks and automobile brakes.

Answer:

elevators

Theatre system

construction pulley

lifts

Answer:

elevator,cargo lift system

Answer:

to adjust from distant to the near objects

Explanation:

Answer:

The ability of eye lens to change the focal length of eye lens is called accommodation power of eye.

Explanation:

The human eye is the optical instrument which works on the refraction of light.

The ability of eye lens to change its focal length is called accommodation power of eye.

The focal length of eye lens is changed by the action of ciliary muscles.

When the ciliary muscles are relaxed then the thickness of lens is more and thus the focal length is small. When the ciliary muscles is stretched, the lens is thin and then the focal length is large.

Answer:

Sources:

-- https://intblog.onspot.com/en-us/why-do-bridges-become-icy-before-roads

and

-- https://www.accuweather.com/en/accuweather-ready/why-bridges-freeze-before-roads/687262

I hope this helps you! ^^

Answer:

[tex]N_f=248N[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=100kg[/tex]

Ladder Length [tex]l=4.0m[/tex]

Mass of Ladder [tex]m_l=25kg[/tex]

Angle [tex]\theta=56 \textdegree[/tex]

Generally the equation for Co planar forces is mathematically given by

[tex]mgcos \theta *2+Mgcos\theta*1 -N_fsin \theta*4=0[/tex]

Therefore

[tex]25*9.81cos 56 *2+100*9.81cos56*1 -N_fsin 56*4=0[/tex]

[tex]N_f=248N[/tex]

Answer:

The current in third branch is 11 A.

Explanation:

incoming current in one branch = 18 A

outgoing current in the other branch = 7 A

let the current in the third branch is i.

According to the Kirchoff's fist law in electricity

incoming current = out going current

18 = 7 + i

i = 11 A

The current in third branch is 11 A.

Answer:

300J

Explanation:

Work done = Force x the distance travelled in the direction of the force

=300 x 1

=300J

Answer:

The unit of the time constant RC is the second

Explanation:

The unit of resistance, R is the Ohm, Ω and resistance, R = V/I where V = voltage and I = current. The unit of voltage is the volt, V while the unit of current is the ampere. A.

Since R = V/I

Unit of R = unit of V/unit of I

Unit of R = V/A

Ω = V/A

Also, The unit of capacitance, C is the Farad, F and capacitance, F = Q/V where Q = charge and V = voltage. the unit of charge is the coulomb, C while the unit of voltage is the volt, V

Since C = Q/V

Unit of C = unit of Q/unit of V

Unit of C = C/V

F = C/V

Now the time constant equals RC.

So, the unit of the time constant = unit of R × unit of C = Ω × F = V/A × C/V = C/A

Also. we know that the  1 Ampere = 1 Coulomb per second

1 A = 1 C/s

So, substituting 1 A in the denominator, we have

unit of RC =  C/A = C ÷ C/s = s

So, the unit of RC = s = second

So, the unit of the time constant RC is the second

Answer:

433.15K

Explanation:

(320°F − 32) × 5/9 + 273.15 = 433.15K

Answer:

A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer.

Explanation:

This low resistance is called shunt resistance S. The scale is now calibrated in ampere and the range of the ammeter depends on the values of the shunt resistance.

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

Answer:

[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]

Explanation:

The pressure at a depth of a fluid is

       P = ρ g y

where ρ is the density of the fluid, y the depth of the gauge measured from the surface of the fluid.

In this case the pressure for fluid A is

      Pa = ρₐ g y

the pressure for fluid B is

      P_b = ρ_b g y

depth y not changes as the gauge is stationary

if we look for the relationship between these pressures

       [tex]\frac{P_a}{P_b} = \frac{ \rho_a}{\rho_b}[/tex]

        [tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]

therefore we see that the pressure measured for fluid B is different from the pressure of fluid A

if  ρₐ < ρ_b B the pressure P_b is greater than the initial reading

   ρₐ>  ρ_b the pressure in B decreases with respect to the reading in liquid A

Answer:

ygguguguhhihhihijijijjojojinjbgy

Answer:

the radius of the bigger loop is 5 cm.

Explanation:

Given;

current in the smaller loop, I₁ = 12 A

current in the larger loop, I₂ = 20 A

radius of the smaller loop, r₁ = 3 cm

let the radius of the larger loop, = r₂

Apply Biot-Savart's law to determine the magnetic field at the center of the circular loops.

[tex]B= \frac{\mu_0 I}{2r}[/tex]

The magnetic field at the center of the smaller loop;

[tex]B_1 = \frac{\mu_0 I_1}{2 r_1}[/tex]

The magnetic field at the center of the bigger loop;

[tex]B_2 = \frac{\mu_0 I_2}{2 r_2}[/tex]

If the magnetic field at the center is zero, then B₁ = B₂

[tex]B_1 = B_2 = \frac{\mu_0 I_1}{2 r_1} = \frac{\mu_0 I_2}{2 r_2} \\\\\frac{I_1}{ r_1} = \frac{ I_2}{r_2} \\\\r_2 = \frac{I_2 r_1}{ I_1} = \frac{(20 \ A) \times (3.0 \ cm)}{12 \ A} = 5 \ cm[/tex]

Therefore, the radius of the bigger loop is 5 cm.