For a total trip of several hours duration, a ratio of the total distance traveled to the total time elapsed is a (an)?

A boat is stationary at  12 meters away from a dock. The boat then begins to move toward the dock with an acceleration of 50 m/s². How long will it take the boat to reach the dock?

Answer:

Explanation:

Given that

To find

So according the question

We have,

Now, for finding time travelled by boat to reach dock we can use the second equation of motion i.e.

         s = ut + [tex]\frac{1}{2}[/tex] at²

     where,

         s = Distance.

          u = Initial velocity.

          t = Time.

          a = Acceleration.          

Now, putting all given values in second equation of motion,

So, we get

          s = ut + [tex]\frac{1}{2}[/tex] at²

         12 = 0t + [tex]\frac{1}{2}[/tex] ×5× t²

         12×2 = 5t²

         t² = [tex]\frac{12 \times 2}{5}[/tex]

         t = [tex]\sqrt{\frac{24}{5}}[/tex]

         t = [tex]\sqrt{{4.8}}[/tex]

        t = [tex]2.19[/tex] s.

Answer:

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FALSE

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The magnitude of the electric field at point p is [tex]\frac{2kQ}{a\sqrt{4a^{2} + L^{2} } }[/tex].

An electric field is a physical field that surrounds electrically charged particles and acts as an attractor or repellent to all other charged particles in the vicinity. It can also refer to a system of charged particles' physical field.

The human body is affected by low-frequency electric fields in the same way that they are affected by other charged-particle-containing materials. The distribution of electric charges at conductive materials' surfaces is affected by electric fields acting on such materials. They make the body conduct current to the ground.

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In here E = 5.3 x 10^3 N/C, the potential difference (V) between the two plates is 116.6 V and if d is doubled, electric field E won't change but V gets doubled.

a) The electric field of infinite parallel plates can be formulated as

E= σ/2ε0 → eq1

Surface charge density is given as 47.0 nc/m2

So, the electric field is independent on d.

Hence the, so the net field is

E = E1 + E2 = σ/2ε0 + σ/2ε0 = σ/ε0

Substituting in eq 1,

E = [47 x 10^(-9) ] / 8.85 x 10^-12 = 5.3 x 10^3 N/C

b) Since the electric potential between them at any point and is given by.

V=Ed → eq 2

So, Substituting in eq 2 yields

V= 2.2 x 10^-2 x 5.3 x 10^3 = 116.6 V

c) If the distance between plates is doubled it will make no changes on the field because the field is independent on d.

For the voltage, according to equation 2 the voltage depends on d so if d is doubled then the potential between plates must be doubled.

Thus, magnitude of e in the region is 5.3 x 10^3 N/C.

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The steelhead lands on the top of waterfall when it is descending so the time taken is 1.36 seconds.

A waterfall is a factor in a river or movement in which water flows over a vertical drop or collection of steep drops.  

Waterfalls are formed in a variety of ways, but the most common  is for a river to flow over a top layer of elastic rock before dropping into soft rock. This rock will erode faster, resulting in a higher drop.  Waterfalls have been studied for their impact on the species that live in and around them. Over the years, people have developed a unique relationship with waterfalls by seeing them, exploring them, and naming them. They can pose a formidable obstacle to navigation along the river.

Using ,S= u*t + (1/2)at​2​​​​

         1.8 = 8t - [(1/2)*9.8*t​​​​​​2​​​​​]

By ,solving this quadratic equation we get, t= 0.27 s or t= 1.36 s

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Answer:

i cant see the picture clearly can u resend it?

Explanation:

Answer:

8.49673 minutes ≈ 8.5 minutes

Explanation:

Acceleration, a is defined as the rate of change in velocity divided by the change in time to attain that velocity

[tex]\mathsf a = \dfrac{\Delta v}{\Delta t} \\\\\textsf {where } \mathsf {\Delta v = v-v_0} \textsf{ v being final velocity and v_0 initial velocity}[/tex], v = final velocity and v₀ the initial velocity
and Δt is the time required to attain final velocity

Δt = number of seconds since launch = t since t₀ = 0

Therefore,
[tex]a = \dfrac{v-v_0}{t}[/tex]

The space shuttle's initial velocity is 0 m/s and its final velocity must be 7800 m/s to escape earth's gravity

So v - v₀ = 7800 - 0 = 7800 m/s, t - t₀ = t = 0 = t seconds
a = 15.3 m/s²

We have the equation

a = 7800/t

a is given as 15.3 m/s² so
15.3 = 7800/t

t  = 7800/15.3 = 509.80392 seconds

To convert to minutes, divide by 60
509.80392/60 = 8.49673 minutes ≈ 8.5 minutes

Answer: The Big Bang!

Answer: a. 6.5 m/s

b. s=10.83 seconds

Explanation:

a. 3+7(0.5)=

3+3.5=6.5 m/s

b. s=seconds

6.5-0.6s=0

6.5-0.6s+0.6s=0+0.6s

6.5=0.6s

6.5=3s/5

6.5*5=3s*5/5

3s=32.5

s=32.5/3

s=65/6

s=10.83 seconds

The apparent weight of the egg = 0.264N

Here the rocket is accelerating upward from the surface of the earth. The egg is on the rocket.

The acceleration due to gravity is 4.4m/[tex]s^{2}[/tex].

The mass of a chicken egg is 0.060 kg.

Form Newton's second law, the net force acting on a body is equal to the product of mass and acceleration due to gravity.

So form Newton's second law we know the formula i.e.,

                         F = m×a

                            = 0.06 × 4.4

                            = 0.264N

Therefore the apparent weight of the egg is 0.264N.

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Answer:

the process of solving a mystery.

Explanation:

While the flashlight light has no effect, the UV source's light discharges the electroscope.

According to Planck's formula, the energy of the particles that make up a beam of light is connected to their frequencies. The photons and atoms collide when that beam strikes a metal. The photoelectric effect is created by a collision when a photon's frequency is high enough to remove an electron.

One electron interacts with each photon. The incident photon's energy is employed to both liberate the surface-bound electrons and give the expelled electrons energy.

Because it offers insights into the properties of solids as well as the properties of atoms and molecules, the photoelectric effect is extensively researched in the fields of quantum chemistry, condensed matter physics, and solid-state chemistry.

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The speed of the rocket is 3000 km/h.

Distance is the sum of an object's movements, regardless of direction. Without respect to the object's starting or finishing positions, distance can be defined as the amount of space it has traveled.

A scalar quantity, speed is defined as the size of the change in an object's location over time or the size of the change in an object's position per unit of time.

The distance traveled by rocket is 6000 km.

The time is taken by the rocket is 2 hours.

The speed of an object is defined as the distance traveled by the object per unit of time.

Therefore,

Speed = distance / time

S = d / t

S = 6000 km / 2 hr

S = 3000 km/h

The speed of the rocket is 3000 kilometers per hour.

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The oscillation angular frequency of a drop with half of the first drop's radius is 4ω

Surface tension is the tension force exerted on an object by the surface of a liquid.

Angular frequency is the frequency of oscillation of a rotating object. It is given in rad/s.

Given that

Since the energy of the drop is conserved, using the law of conservation of angular momentum, we have

Iω = I'ω' where

So, Iω = I'ω'

Making ω' subject of the formula, we have

ω' = Iω/I'

ω' = mr²ω/mr'²

ω' = r²ω/r'²

Given that the drop is half of the first drop's radius, r' = r/2

So, ω' = r²ω/r'²

ω' = r²ω/(r/2)²

ω' = r²ω/r²/4

ω' = 4ω

So, the oscillation angular frequency of a drop with half of the first drop's radius is 4ω

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If the charge on object A is doubled then the electric field will remain unchanged in any of the cases. It is so because the question is asking us whether the electric field which the object is experiencing is changing or not, but in actuality, due to charge change, the magnitude of the electric field is changing only.

The electric field's magnitude may be easily determined by calculating the force per charge on the test charge. The electric field would exert a force on the item when its charge doubled since it is inversely proportional to the electric charge. so won't have any effect.

The field is supposed to have a direction corresponding to the force that it would exert on a positive test charge. The electric field emanates from a positive point charge in an outward direction, and from a negative point charge in an inward direction.

A charge is a characteristic that describes how many more or fewer electrons than protons a matter unit has.

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Electron transport chain use NADH and FADH2 molecules formed in Glycolysis and Krebs Cycle processes.

The third stage of cellular respiration is the electron transport chain. In cellular respiration, high-energy electrons are transported by NADH and FADH2 to the electron transport chain. In photosynthesis, NADPH transports energetic electrons to the electron transport chain.

Krebs cycle or the TCA cycle, the process of oxidizing acetyl-CoA, which is produced from proteins, carbs, and lipids, releases stored energy.

The process by which glucose is broken down to provide energy is known as glycolysis. It generates two pyruvate molecules, ATP, NADH, and water. There is no need for oxygen throughout the process, which occurs in the cytoplasm of a cell. Both aerobic and anaerobic respiration experience it.

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With the concept of Coulomb's Law, the net force acting on the 2μC charge is 2.06 × 10⁻² newton.

Coulomb's Law states that the force of attraction or repulsion exerted by a charged particle on another charged particle is given by,

F=kq₁q₂ / r²

Here q₁ and q₂ are the charges on the particles,

r is the separation between them

k is a constant known as Coulomb's constant, whose value is, k = 9 × 10⁹ N m²/C²

So here in the question, we have been given an equilateral triangle of side 2 meters with three-point charges at its vertices

q₁ = 1μC ,

q₂ = 2μC ,

q₃ = 3μC ,

r=2 m

From Coulomb's law,

F₂₁ = kq₁q₂/r²

F₂₁ = (9 × 10⁹ × 1 × 2 × 10⁻¹²) / 2²

F₂₁ =4.5 × 10⁻³ N

This is the force on charge q₂ because of q₁.

Similarly,

F₂₃ = kq₂q₃ / r²

F₂₃ = (9 × 10⁹ × 2 × 3 × 10⁻¹²) / 2²

F₂₃ = 18 × 10⁻³ N

Now, the resultant of the two forces will be,

F(net) = [tex]\sqrt{F_{21} ^2 + F_{23}^2 +2F_{21}F_{23}cos60}[/tex]

On substituting all the values the net force will be

F(net) = 2.06 × 10⁻² N

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Answer:

Q1: C

Q2: C

Q3: D

Q4: D

Code: Left, Down, Up, Right

Your welcome.

Answer: ctrl+U

ctrl+G

Explanation:

I don't know how else to help

The FitnessGram PACER Test is a multistage aerobic capacity test that progressively gets more difficult as it continues.

The test is used to measure a student's aerobic capacity as part of the FitnessGram assessment. Students run back and forth as many times as they can, each lap signaled by a beep sound. The test get progressively faster as it continues until the student reaches their max lap score.

The PACER Test score is combined in the FitnessGram software with scores for muscular strength, endurance, flexibility and body composition to determine whether a student is in the Healthy Fitness Zone™ or the Needs Improvement Zone™.

The vector tangential to the circle because the vector indicates the direction of the force on the object.

Centripetal force is the force acting on an object in circular motion directed towards the axis of rotation.

In a circular motion, the direction of the velocity vector is the same as the direction of the object's motion, the velocity vector is directed tangent to the circle as well.

It also indicate the direction of the centripetal force.

Thus, the vector tangential to the circle because the vector indicates the direction of the force on the object.

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A 60-kilogram individual ascends an 80 m-high building's roof by walking up from the ground. Once at the peak, the human has a potential energy of 44027.28 J.

This is our formula

Gravitational ability strength = mgh  

here (GPE = mgh)

mg = 60.0 kg

h = 80

mg = 60.0 kg

in view that we're searching for pressure and the SI Unit for force is "N," we must convert this to force (N).

consequently, we will multiply 60.0 via nine.81.

60.0 × 9.81 = 588.6

Now that will be

⇒ mg = 588.6 N

⇒ h = 80

GPE = (588.6) × (80) = 47088 J

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The correct option is (D) The iodine gets extracted into to the organic solvent, and the potassium iodide remains in the water.

During extraction, the iodine gets extracted into to the organic solvent, and the potassium iodide remains in the water.

For extraction, the principle of like dissolve like is followed.

There are two solvents - water which is polar and dichloromethane (or hexane) which are nonpolar.

We need to extract iodine [tex]I_{2}[/tex] and potassium iodide (KI).

[tex]I_{2}[/tex] is a non-polar molecule that is formed by covalent bonding. But KI is an ionic molecule that exhibits ionic behavior by giving [tex]K^{+}[/tex] and [tex]I^{-}[/tex] ions in aqueous media. Also, there is a difference in their electronegativity. Thus, [tex]I_{2}[/tex] will be extracted in an organic solvent and KI remains soluble in water.

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The complete question is:

"In this experiment, you start with plain dichloromethane (or hexane), and a solution of iodine and potassium iodide in water. what happens during the extraction process?

(a) Both the potassium iodide and the iodine remain in the water

(b) Both the potassium iodide and the iodine get extracted into the organic solvent

(c) The potassium iodide gets extracted into the organic solvent, and the iodine remains in the water

(d) The iodine gets extracted into to the organic solvent, and the potassium iodide remains in the water"

The angular frequency of the emf for the inductor and resistor to have the same peak voltage (i.e. for R and XL to be the same) is R/L.

The above situation represents Alternating Current(AC).

An alternating current is a type of current that reverses its direction after fixed intervals of time.

The following formulas are used in the case of AC,

XL = ωL

XC= 1/Cω

Here L and C represent inductance and capacitance respectively.

It is  given that XL = R

Hence, ωL = R

ω = R/L

Hence, the angular frequency(ω) of the emf for the inductor and resistor to have the same peak voltage (i.e. for R and XL to be the same) is R/L.

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The snapshot of light as the cart moves with constant velocity is represented by a graph with uniform displacement at each time interval.

The change in displacement with time is uniform at constant velocity. The displacement of the supplied moving item grows at the same pace.

The beginning velocity equals the ultimate velocity at constant velocity.

v₁ = v₂

The object's acceleration at constant velocity is zero since the velocity change with time is zero.

As a result, we may deduce that the graph with equal displacement at each time interval reflects a snapshot of light as the cart moves at a constant speed.

A moving object's displacement-time graph shows the distance traveled by a moving item as time passes. A vector quantity is displacement. The slope or gradient of this graph represents the velocity of the item. The displacement-time graph, also known as the position-time graph, describes an object's motion. In this graph, the displacement of the moving item is displayed on the y-axis as a dependent variable, while time is shown on the x-axis as an independent variable.

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The athlete who jumps a vertical distance of 0.8 m has a initial speed of: 3.959 m/s and his total time in the air is: 0.8078 s

The formulas for the vertical launch upward and the procedures we will use are:

Where:

Information about the problem:

Applying the maximum height formula and clearing the initial velocity we get:

y max = v₀²/(2*g)

v₀ = √(y max * (2*g))

v₀ = √( 0.8 m * (2 * 9.8 m/s²))

v₀ = √( 0.8 m * 19.6 m/s²)

v₀ = √15.68 m²/s²

v₀ = 3.959 m/s

Applying  the maximum time formula we get::

t max = v₀ / g

t max =(3.959 m/s)/(9.8 m/s²)

t max = 0.4039 s

Applying  the time of flight formula we get::

t(of) =2 * t max

t(of) =2 * 0.4039 s

t(of) = 0.8078 s

In physics vertical launch upwards is the motion described by an object that has been launched vertically upwards in which the height and the effect of the earth's gravitational force on the launched object are taken into account.

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The frequency of light emitted is 24.63 x [tex]10^1^5[/tex] Hz when an electron jumps from the 2nd to the 1st orbit in a hydrogen atom.

According to Bohr, the electrons in the atom's structure are situated around the nucleus at particular energy levels. An electron must either acquire or lose energy as it transitions from one of these energy levels to another. It is known as an absorption when an electron obtains energy and as an emittance when it loses energy. Both processes involve the absorption or emission of a PHOTON energy particle, which in turn causes the absorption or emission of light. Because various elements have varying energies, they each emit or absorb light in varying quantities (wavelengths).

The wavelength or wave number is used to compute the frequency of light emitted. 

The frequency of light emitted is inversely proportional to its wavelength.

Given:

Jump from 2nd orbit to 1st orbit

Planck’s constant, h = 6.626 x [tex]10^-^3^4[/tex]

To find:

Frequency of light emitted, υ = ?

Formula:

E = hυ

Calculations:

[tex]E = 13.6 * Z^2( \frac{1}{n_1^2} - \frac{1}{n_2^2} )[/tex]

E = hυ = 13.6 x 1 x [tex](1 - \frac{1}{4})[/tex]

υ x 6.626 x [tex]10^-^3^4[/tex] = 10.2 x 1.6 x [tex]10^-^1^9[/tex]

υ = 24.63 x [tex]10^1^5[/tex] Hz

Result:

The frequency of light emitted is 24.63 x [tex]10^1^5[/tex] Hz.

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(a)The net charge on the sphere is 3.38×[tex]10^{-9}[/tex] C.

(b) The charge inside sphere is positive and distributed symetrically inside the sphere .

(a) As we know the electric field is given as,

E = kq / [tex]r^{2}[/tex]

Plugging in the values we get,

575 = 8.99×[tex]10^{9}[/tex] q / [tex]0.230^{2}[/tex]

8.99×[tex]10^{9}[/tex]q = 30.41

q = 3.38×[tex]10^{-9}[/tex] C

So, the net charge on the sphere is 3.38×[tex]10^{-9}[/tex] C.

(b) We can conclude that the charge inside sphere is positive and distributed symetrically inside the sphere .

The complete question is: The electric field everywhere on the surface of a charged sphere of radius 0.230 m has a magnitude of 575 N/C and points radially outward from the center of the sphere. (a) What is the net charge on the sphere? (b) What can you conclude about the nature and distribution of charge inside the sphere?

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The distance travelled by the student is 1600m and the displacement is 0.

Displacement is the direction from the beginning point and the length of a straight path from the starting point to the finishing point, whereas distance is the length of a path that connects two places.

Displacement is a vector quantity that does not take into consideration the actual path taken; instead, it quantifies the distance between the beginning point and the finishing position.

Given:

Track length = 400m

No. of rounds = 4

To find:

Distance, d = ?

Displacement, D = ?

Formula:

Distance, d = Track length x No. of rounds

Displacement, D = End point – Start point

Calculations:

d = 400 x 4

d = 1600m

D = 0 since the start and end point are same.

Result:

The distance travelled is 1600m with 0 displacement.

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