David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.0 m/s2 at the instant when David passes. How far does Tina drive before passing David?

Answer:

(a)  209 Watt

(b) 4482.8 seconds

Explanation:

(a) P = 50×4.18

Where P = rate of heat loss in watt

    P = 209 Watt

Applying,

Q = cm(t₁-t₂)................ Equation 1

Where Q = amount of heat given off, c = specific heat capacity capacity of human, m = mass of the person, t₁ and t₂ = initial and final temperature.

From the question,

Given: m = 90 kg, t₁ = 40°C, t₂ = 37°C

Constant: c = 3470 J/kg.K

Substtut these values into equation 1

Q = 90×3470(40-37)

Q = 936900 J

But,

P = Q/t.............. Equation 2

Where t = time

t = Q/P............ Equation 3

Given: P = 209 Watt, Q = 936900

Substitute into equation 3

t = 936900/209

t = 4482.8 seconds

Answer:

There will be a phase change at the 1-1.38 interface and no phase change at the 1.38-1.33 interface.

At a thickness of lambda / 4       (y/4) one should get constructive interference for the reflected light.

y = 4 * 99.8 * 10E-9 m = 400 nm (about)  = 4 * 10E-7 m

The color of this light will be violet or blue

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:

a) The laser device

b) The tape

Explanation:

First, there is a need to understand what accuracy and precision mean.

Accuracy is the closeness of a measurement to its true (pre-determined) value.

Precision is the closeness of repeated measurements to each other.

Since 1 feet = 12 inches, then, 5 feet and 5.2 inches would be equivalent to 65.2 inches. This value represents the true value of my height.

The tape measured the height as 65 3/8, which is equivalent to 65.375 inches.

The laser device measured the height as 65.31.

Error = true value - measured value

Absolute error from the tape = 65.2 - 65.375

                                       = -0.175 inches

Absolute error from laser device = 65.2 - 65.31

                                 = -0.11

a) The magnitude of error from the tape is more than that of the laser device. Hence, the laser device is said to be more accurate.

b) Even though there were just single readings from both instruments, the tape can be read to the nearest 1/8 of an inch and as such, can give more precisive measurements than the laser device.

Answer:

I don't get it rewrite please

The ratio of foort dustance to load distance in a simple machine is called mechanical advantage or MA.

MA= Effort Distance / Load Distance

Answer:

hcbvdgsyyvjusvbxjxu usbsbhsi

Explanation:

ysggsghxuxgscsixigdvgsibxhdhshshjf

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let T represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, [tex]F_{f2}[/tex] = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = [tex]F_{f2}[/tex] × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

[tex]T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27[/tex]

[tex]Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92[/tex]

The frictional force on the block W₂, [tex]F_{f2}[/tex] ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ [tex]F_{w2}[/tex] = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + [tex]\mathbf{F_{w2}}[/tex]

The frictional force from the ground, [tex]\mathbf{F_{f1}}[/tex] = N×μ + [tex]\mathbf{F_{f2}}[/tex] = P

Where;

P = The horizontal force applied to the block

P = (W₁ + [tex]\mathbf{F_{w2}}[/tex]) × μ + [tex]\mathbf{F_{f2}}[/tex]

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

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Answer:

bcz we want to have fun there lolol

Answer:

for emergency, treatment, medicines,etc.....

Answer:

1231

Explanation:

∑ F (horizontal) = (150 N) cos(60°) - 15 N = (20 kg) a

==>   a = ((150 N) cos(60°) - 15 N)/(20 kg) = 3 m/s²

To calculate the acceleration of the box, we need to consider the net force acting on it. So, the acceleration of the box is 3 m/s².

The net force is the vector sum of the applied force and the force of friction. First, let's find the horizontal and vertical components of the applied force:

Horizontal component of the applied force (F[tex]_{horizontal}[/tex]) = F[tex]_{applied}[/tex] × cos(θ)

F[tex]_{horizontal}[/tex] = 150 N × cos(60°)

F[tex]_{horizontal}[/tex] = 150 N × 0.5

F[tex]_{horizontal}[/tex] = 75 N

Vertical component of the applied force (F[tex]_{vertical}[/tex]) = F[tex]_{applied}[/tex] × sin(θ)

F[tex]_{vertical}[/tex] = 150 N × sin(60°)

F[tex]_{vertical}[/tex] = 150 N × (√3 / 2)

F[tex]_{vertical}[/tex] ≈ 129.9 N

Now, let's calculate the net force in the horizontal direction:

Net Force in the horizontal direction (F[tex]_{net horizontal}[/tex]) = F[tex]_{horizontal}[/tex] - F[tex]_{friction}[/tex]

F[tex]_{net horizontal}[/tex] = 75 N - 15 N

F[tex]_{net horizontal}[/tex] = 60 N

Now, we can calculate the acceleration (a) using Newton's second law of motion, F = ma:

F[tex]_{net horizontal}[/tex] = m × a

60 N = 20 kg × a

Now, solve for acceleration (a):

a = 60 N / 20 kg

a = 3 m/s²

So, the acceleration of the box is 3 m/s².

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Answer:

The tempature of the matter increases.

Explanation:

The matter is heated by friction, weather that is through the air or if its touching the ground like a car.

please mark me as brainliest of this is right, thanks

Answer:

option A

Explanation:

simple harmonic motion

Answer:

random motion I think not sure

Answer:

Stress = F / A       force per unit area

A = 3.00 cm^2 = 3 E-4  m^2

F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N    max force applied

F/3 = 2.4E4 N  if force not to exceed limit   (= f)

f = M a

a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2      about 2 g

Answer:

The first law of motion describes the principle of law of inertia.

Answer:

Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.

Answer:

please follow Me

Explanation:

if the answer is right then like me otherwise you can share the right answer

Answer:

473.15

Explanation:

Answer:

(a) 5.46 C

(b) 3.4125 x 10^19

Explanation:

q' = 1.08 C, q'' = 6.74 C, q''' = 4.61 C, q'''' = 9.41 C

When the charges are in contact to each other.

(a) So, the net charge is

[tex]q = \frac{q' + q'' + q''' + q''''}{4}[/tex]

[tex]q = \frac{1.08+6.74+4.61+9.41}{4}\\\\q = 5.46 C[/tex]

(b) As the charge is positive in nature, so the protons are there. The number of protons is

[tex]n = \frac{q}{e}\\\\n = \frac{5.46}{1.6\times 10^{-19}}\\\\n = 3.4125\times 10^{19}[/tex]

Answer:

una ufx vek cintos kavres millianto

Explanation:

vekas moriqn5o siveria il a ola ola micanese ma n8 aou ko sevyaera

=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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Answer:

4.38 m/s

Explanation:

We have that the total distance covered by the runner is

[tex]d_t=20.723miles[/tex]

The total distance covered by the runner is a sum of all miles covered by the runner

Therefore

With

[tex]d_t[/tex]=Total distance

[tex]d_t=d_1+d_2+d_3\\\\d_t=9.5+8.89+2.333[/tex]

[tex]d_t=20.723miles[/tex]

in conclusion

The total distance covered by the runner is

[tex]d_t=20.723miles[/tex]

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We know

[tex]\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\Theta}}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{50^2+50^2+2(50)(50)cos35}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{2500+2500+2(2500)\times (-0.9)}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{5000+5000(-0.9)}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{5000+(-4500)}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{5000-4500}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{-500}[/tex]

[tex]\\ \sf\longmapsto R=22.4i[/tex]

Resultant using the Law of Sines and the Law of Cosines will be R=95 N

Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.

The Magnitude of two forces =50 N

Angle between the forces = 35

By using the resultant formula

[tex]\rm R=\sqrt{A^2+B^2+2ABCos\theta}[/tex]

[tex]\rm R=\sqrt{50^2+50^2+2(50)(50)Cos35}[/tex]

[tex]\rm R=\sqrt{5000+5000(0.81)}[/tex]

[tex]\rm R=\sqrt{5000+4500}[/tex]

[tex]\rm R=95\ N[/tex]

Hence the Resultant using the Law of Sines and the Law of Cosines will be R=95 N

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Answer:

Displacement current flows in the dielectric material(insulated region)

Explanation:

Firstly a capacitor stores charge when a capacitor is charging (or discharging), current flows in the circuit. Also, there is no charge transfer in the dielectric material in the capacitor which is contradictory to the flow of current. Hence, displacement current is the current in the insulated region due to the changing electric flux.

Answer:

Explanation:

Newton's Second Law is pretty much the standard for all motion that involves a force. It applies to gravitational force and torque and friction and weight on an elevator. The main formula for force is

F = ma. We have to adjust that to take into account that when the elevator is moving up, that "surge" of acceleration weighs down a bit on the scale, causing it to read higher than the actual weight until the acceleration evens out and there is no acceleration at all (no acceleration simply means that the velocity is constant; acceleration by definition is a change in velocity, and if there is no change in velocity, there is 0 acceleration). The force equation then becomes

[tex]F_n-w=ma[/tex]  where [tex]F_n[/tex] is normal force. This is what the scale will read, which is what we are looking for in this problem (our unknown). Since we are looking for [tex]F_n[/tex], that is what we will solve this literal equation for:

[tex]F_n=ma+w[/tex] .  m is the mass of the boy, a is the acceleration of the elevator (which is going up so we will call that acceleration positive), and w is weight. We have everything but the unknown and the weight of the boy. We find the weight:

w = mg so

w = 28(9.8) and

w = 274.4 N BUT rounding to the correct number of significance we have that the weight is actually

w = 270 N.

Filling in the elevator equation:

[tex]F_n=28(.50)+270[/tex] and according to the rules of significant digits, we have to multiply the 28(.50) {notice that I did add a 0 there for greater significance; if not that added 0 we are only looking at 1 significant digit which is pretty much useless}, round that to 2 sig fig's, and then add to 170:

[tex]F_n=14+270[/tex] and adding, by the rules, requires that we round to the tens place to get, finally:

[tex]F_n=280N[/tex]  So you see that the surge in acceleration did in fact add a tiny bit to the weight read by the scale; conversely, if he were to have moved down at that same rate, the scale would have read a bit less than his actual weight). Isn't physics like the coolest thing ever!?

Answer:

in which topic you need help