Before we understood that objects have a tendency to maintain their velocity in a straight line unless acted upon by a net force, people thought that objects have a tendency to stop on their own. This happened because a specific force was not yet understood. What was that force?

Answer: 800N

Explanation:

Given :

Mass of ball =0.8kg

Contact time = 0.05 sec

Final velocity = initial velocity = 25m/s

Magnitude of the average force exerted on the wall by the ball is can be calculated using the relation;

Force(F) = mass(m) * average acceleration(a)

a= (initial velocity(u) + final velocity(v))/t

m = 0.8kg

u = v = 25m/s

t = contact time of the ball = 0.05s

Therefore,

a = (25 + 25) ÷ 0.05 = 1000m/s^2

Therefore,

Magnitude of average force (F)

F=ma

m = mass of ball = 0.8

a = 1000m/s^2

F = 0.8 * 1000

F = 800N

Answer:

Explanation:

velocity of ship with respect to water = 6.5 m/s due north

[tex]\overrightarrow{v}_{s,w}=6.5 \widehat{j}[/tex]

velocity of water with respect to earth = 1.5 m/s at 40° north of east

[tex]\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)[/tex]

velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth

[tex]\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}[/tex]

[tex]\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j} \right )[/tex]

[tex]\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}[/tex]

The magnitude of the velocity of ship relative to earth is [tex]\sqrt{1.15^{2}+5.54^{2}}[/tex] = 5.66 m/s

The most accurate weather for the next two days would be dry warm weather or severe thunderstorms if there is an occluded front over the area.

Explanation:

The weather front that is created during the cyclogenesis process is an occluded front in meteorology. Cyclogenesis process is the development of extra-tropical cyclone and its intensification.

During the occurrence of this, the warm air is occluded (separated) from the center of cyclone at the surface of the earth.

The cold front rotates the storm as it intensifies and holds up the warm front. This develops an occluded front, that is the boundary which separates the new cold air mass and the older cool air mass that is already in warm front's north.

Answer:

823.46 kgm/s

Explanation:

At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.

So, mgh = 1/2mv²

From here, his velocity just as he reaches the surface of the water is

v = √2gh

h = 9 m and g = 9.8 m/s²

v = √(2 × 9 × 9.8) m/s

v = √176.4 m/s

v₁ = 13.28 m/s

So his velocity just as he reaches the surface of the water is 13.28 m/s.

Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.

So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.

His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,

J = (62 × 0 + 62 × 13.28) kgm/s = 0 +  823.46 kgm/s = 823.46 kgm/s

So the magnitude of the impulse J of the water on him is 823.46 kgm/s

Answer: a

Explanation: ur mom

Answer:

A) an atom's ability to attract electrons that are shared in a chemical bond

Explanation:

I took the test and got the answer right.

Answer:

0.5 s

Explanation:

The discharge current in a series RC circuit is given by

[tex]i = \frac{V_{0} }{R} e^{-\frac{t}{RC} }[/tex] where i₀ = V₀/R

For the current i to decay to 5%i₀, i/i₀ = 0.05. Where R = resistance = 2.0 × 10⁶ Ω and C = capacitance = 3.0 μF = 3.0 10⁻⁶ F

So,

[tex]i = {i_{0}e^{-\frac{t}{RC} }[/tex]

[tex]{\frac{i}{i_{0} } = e^{-\frac{t}{RC}[/tex]

[tex]0.05 = e^{-\frac{t}{2 X 10^{6} X 3 X 10^{-6}}\\[/tex]

[tex]0.05 = e^{-\frac{t}{6}[/tex]

taking natural logarithm of both sides

-t /6 = ln(0.05)

t = -ln(0.05)/6 = 0.5 s

So it takes the current 0.5 s to reach 5% of its original value

In an RC circuit, the time required for the current to decay to 5% of its original value, or for the capacitor to reach 95% of its maximum charge, is approximately three times the circuit's time constant. Here, the time constant is 6 seconds, so the time required is approximately 3 * 6 = 18 seconds.

In an RC circuit, the time required for the current to decay to a certain percentage of its original value is connected to the circuit's time constant. The time constant is calculated by the formula τ = R*C. In your case, R equals 2.0 * 106 ohms and C equals 3.0 * 10-6 farads, so τ = 2.0 * 106 * 3 * 10-6 = 6 seconds.

The time required for the current to decay to 5% of its original value or for the capacitor to reach 95% of its maximum charge in an RC circuit is roughly three time constants, or 3τ.From the calculation above, τ equals 6 seconds, so the time required is approximately 3 * 6 = 18 seconds.

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Answer:

242.85 Hz

Explanation:

For maximum intensity of sound, the path difference,ΔL = (n + 1/2)λ/2 where n = 0,1,2...

Since Abby is standing perpendicular to one speaker, the path length for the sound from the other speaker to him is L₁ = √(2.00² + 5.50²) = √(4.00 + 30.25) = √34.25 = 5.85 m.

The path difference to him is thus ΔL = 5.85 m - 5.50 m = 0.35 m.

Since ΔL = (n + 1/2)λ/2 and for lowest frequency n = 0,

ΔL = (n + 1/2)λ/2 = (0 + 1/2)λ/2 = λ/4

ΔL = λ = v/f and f = v/4ΔL where f = frequency of wave and v = velocity of sound wave = 340 m/s.

f = 340/(4 × 0.35) = 242.85 Hz

The condition of constructive interference and the speed of a wave allows finding the frequency for the first constructive interference that Abby hears is:

         f = 971.4 Hz

The interference of the coherent sound wave occurs when two wave fronts are added at a point and the intensity depends on the path difference of these waves, we have two extreme cases:

          Δr = [tex]2n \ \frac{\lambda}{2}[/tex]

Where Δr is the path difference, λ is the wavelength and n is an integer.

In the attachment we can see the distance from Abby is to the speakers for the closest y = 5.50 m, the distance to the furthest speaker.

Let's use the Pythagoras' theorem,

             r =[tex]\sqrt{x^2 + y^2}[/tex]  

Let's calculate.

            r = [tex]\sqrt{2^2 + 5.5^2}[/tex]  

            r = 5.85 m

The difference in path is:

           Δr = r-y

           Δr = 5.85 - 5.5

           Δr = 0.35 m

let's find the wavelength.

          Δr = 2n [tex]\frac{\lambda}{2}[/tex]  

          λ = [tex]\frac{\Delta r}{n}[/tex]

The first constructive interference occurs for n = 1

           λ= Δr

           λ = 0.35 m

Wave speed is proportional to wavelength and frequency.

          v = λ f

          f = [tex]\frac{v}{\lambda }[/tex]

Let's calculate.

          f = [tex]\frac{340}{0.35}[/tex]  

          f = 971.4 Hz

In conclusion using the condition of constructive interference and the speed of a wave we can find the frequency for the first constructive interference that Abby hears is:

          f = 971.4 Hz

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Answer:

The required torque is 1.17 N-m.

Explanation:

The given data :-

The magnitude of force ( F ) = 4.5 N.

The length of arm ( r ) = 0.26 m.

Here given that force is applied at perpendicular means ( ∅ ) = 90°.

The torque ( T ) is given by

T = F * r * sin∅

T = 4.5 * 0.26 * sin 90°

T = 4.5 * 0.26 * 1

T = 1.17 N-m.

The magnitude of the torque produced by the given force perpendicular to the lever is 1.17N.m.

Given the data in the question;

Torque; [tex]T = \ ?[/tex]

Torque simply the measure of the force that can cause an object to rotate about an axis. It is expressed as:

[tex]T = rFsin\theta[/tex]

Where r is radius, F is force applied and θ is the angle between the force and the lever arm.

We substitute our given values into the equation;

[tex]T = 0.26m * 4.5N * sin90^o\\\\T = 0.26m * 4.5N * 1\\\\T = 1.17N.m[/tex]

Therefore, the magnitude of the torque produced by the given force perpendicular to the lever is 1.17N.m.

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Answer:

Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on

Limitations: yield only planet's mass and orbital properties

Explanation:

Strengths: yield planet's size

Limitations: few planets have the necessary orbital alignment to be detectable

The Doppler method is a valuable tool for detecting and characterizing exoplanets, particularly those with significant mass and short orbital periods. However, it has limitations in terms of its sensitivity to certain types of planets and the limited information it provides about exoplanet properties. So, option A is correct.

The Doppler method is particularly effective in detecting massive exoplanets that are relatively close to their host stars. It has been successful in identifying hundreds of exoplanets since its inception.The Doppler method is more sensitive to massive exoplanets that are closer to their host stars. This means that it may miss smaller, more distant planets or those with longer orbital periods. The method also has difficulty detecting exoplanets with orbits that are nearly aligned with the line of sight, resulting in underrepresentation of such planets in detection statistics.

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Complete question is below

What are the strengths and limitations of the doppler method?

A.it has limitations in terms of its sensitivity to certain types of planets and the limited information it provides about exoplanet properties.

B.it has NO limitations in terms of its sensitivity to certain types of planets and the limited information it provides about exoplanet properties

Answer:

Maximum altitude above the ground = 1,540,224 m = 1540.2 km

Explanation:

Using the equations of motion

u = initial velocity of the projectile = 5.5 km/s = 5500 m/s

v = final velocity of the projectile at maximum height reached = 0 m/s

g = acceleration due to gravity = (GM/R²) (from the gravitational law)

g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)

g = -9.82 m/s² (minus because of the direction in which it is directed)

y = vertical distance covered by the projectile = ?

v² = u² + 2gy

0² = 5500² + 2(-9.82)(y)

19.64y = 5500²

y = 1,540,224 m = 1540.2 km

Hope this Helps!!!

Since there's no answer choices available, I would say "reward mechanisms in the brain", "pleasure", "Schizophrenia" or "appetite".

Answer:

A neurotransmitter used in the parts of the brain involved in regulating movement and experiencing pleasure.

Explanation:

Dopamine is one of the chemicals in our brain which controls mood Interacting with the pleasure and reward center of our brain, dopamine, along with other chemicals like oxytocin and endorphins plays a very crucial role in how we feel

Answer:

Initial value problem is:

u'' + 10u' + 98u = (2 sin(t/2) N for u(0) = 0

u'(0) = 0.03m/s

Explanation:

The directions of Fd(t*) and U'(t*) are not specified in the question, so we'll take Fd(t*) to be negative and U'(t*) to be positive. This is due to the fact that the damping factor acts in the direction opposite the direction of the motion of the mass.

M = 5kg; L= 10cm or 0.1m;

F(t) = 10 sin(t/2) N ; Fd(t*) = - 2N

U'(t*) = 4cm/s or 0.04m/s

u(0) = 0

u'(0) = 3cm/s or 0.03m/s

Now, we know that W = KL.

Where K is the spring constant.

And L is the length of extension.

So, k = W/L

W= mg = 5 x 9.81 = 49.05N

So,k = 49.05/0.1 = 490.5kg/s^(2)

Now from spring damping, we know that; Fd(t*) = - γu'(t*)

Where,γ = damping coefficient

So, γ = - Fd(t*)/u'(t*)

So, γ = 2/0.04 = 50 Ns/m

Therefore, the initial value problem which describes the motion of the mass is;

5u'' + 50u' + 490u = (10 sin(t/2) N

Divide each term by 5 to give;

u'' + 10u' + 98u = (2 sin(t/2) N for u(0) = 0

u'(0) = 0.03m/s

Answer:

a) 242.5 m

Explanation:

Given :

Vi= 20m/s

t = 5 seconds

According to kinematic equation, the displacement is given by:

d = Vit + 1/2 gt^2

d = 20 × 6 ) + (1/2 × 9.8 × 5^2)

d = 120 + 122.5

d = 242.5m

Answer:

Explanation:

Given:

Mass = 124 kg

U = -20 m/s

t = 5 s

Using the equation of motion,

S = U×t + 1/2 × gt^2

S = -20 × 5 - 1/2 × 9.8 × 5^2

= -222.5 m

= 222.5 m (the sign is showing the direction of the motion from its origin)

Answer:

64.59kpa

Explanation:

See attachment

Answer:

Solid cylinder

Explanation:

Solid cylinder will reach first

If radius of each cylinder is r

mass is m

then,

Moment of inertia I= dm[tex]r^{2}[/tex]

Here d = measure as how close the mass is to the edge

Velocity of rolling cylinder is given by

v= [tex]\sqrt} \frac{2gh}{1+d}[/tex]

where,

g= 9.8 m/s2

h= height from ground

So from formula of  velocity we can say that velocity will be maximum if denominator is minimum i-e if k is of small value -- or in other word if mass is away from edge i-e if mass is closer to the center !

As all the mass in hollow cylinder is near the edge so k value will be higher for it and hence it will have low velocity value. So it will reach later  as compared to the solid cylinder in which mass is closer to the center and hence k is greater for solid cylinder.

Answer:

the equilibrium constant is equal to 1 (i.e., the reactant and product concentrations are always equal).

Explanation:

ΔG is a symbol related to Gibbs free energy, which is a physical quantity related to thermodynamics. ΔG refers to the difference between the change in enthalpy (and sometimes entropy) and the temperature of a chemical reaction.

Gibbs free energy is very useful for measuring the work done between the reactants in a reaction. It is calculated using the formula: ΔG = change in enthalpy - (temperature x change in entropy).

The ΔG of a reaction would have a minimum value (zero), if the equilibrium constant is equal to 1 (that is, the concentrations of the reagent and the product are always equal).

the equilibrium constant should be equal to 1 (i.e., the reactant and product concentrations are always equal).

ΔG refers to a symbol that should be related to Gibbs free energy, i.e. physical quantity related to thermodynamics. ΔG means the difference between the change in enthalpy and the temperature of a chemical reaction. Now it is determined using the formula:

ΔG = change in enthalpy - (temperature x change in entropy).

In the case when the ΔG of a reaction would have a minimum value (zero), so if the equilibrium constant is equivalent to 1 i.e. the concentrations of the reagent and the product should always equal.

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Answer:

THE ANSWER IS: TRUE

Explanation:

Strategic planning is indeed long-range and formulated by top management but not under the assumption that the company operates in a vacuum. Effective strategic planning acknowledges and incorporates external factors.

The statement is partially true and partially false. It is accurate that strategic planning is long-range and formulated by top management. These types of plans typically look several years into the future to set comprehensive goals for the company. However, the statement is false in suggesting that strategic planning is done as if a company operates in a vacuum. In truth, effective strategic planning must incorporate external influences like market trends, competitive activities, and regulatory changes.

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Answer:

The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.

Explanation:

As from the given data

the length of the rope is given as l=30 m

the stretched length is given as l'=41m

the stretched length required is give as  y=l'-l=41-30=11m

the mass is m=95 kg

the  force is  F=380 N

the gravitational acceleration is g=9.8 m/s2

The equation of  k is given by equating the energy at the equilibrium point which is given as

[tex]U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}[/tex]

Here

m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

[tex]k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m[/tex]

Now the force is

[tex]F=kx\\[/tex] or

[tex]x=\dfrac{F}{k}\\[/tex]

So here F=380 N, k=630.92 N/m

[tex]x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m[/tex]

So the distance is 0.602 m

The student requires help to calculate the stretch distance of a bungee cord when a force is applied, which is determined using Hooke's Law, F = kx, where F is the applied force and k is the spring constant.

The student is asking about the properties of a bungee cord that would be suitable for their father-in-law's jump so that he will fall safely without hitting the ground. Specifically, the question involves calculating the amount by which the bungee cord would stretch when a force of 380.0 N is applied to it. The restoring force that the bungee cord exerts is given as kx, where k is the spring constant and x is the stretch distance.

To find the correct bungee cord, one must solve for x in the equation F = kx given that F (the force applied) is 380.0 N. The spring constant k is the unknown in this problem and must be determined based on the requirement that the maximum fall before the cord stops the father-in-law is 41 m, factoring in his weight and the initial unstretched length of the bungee cord (30 m).

This problem requires knowledge in the area of Hooke's Law and the work-energy principle, as it deals with forces and the properties of elastic materials. The actual calculation is not provided because it would require additional information about the bungee cords the student has access to, such as their specific spring constants.

Answer:

The P site

Explanation:

After the formation of peptide linkage, the trna of P site becomes free, more to the E site and then slip away. The A site trna has dipeptide instead of a single aminoacid.

In the prescence of translocase and energy from GTP, the ribosome move in such a way the peptidyl bearing trna of A site come to lie with P sit. This exposes next codon to A site. A new aminoacid trna complex reaches the fresh codon exposed at A site.

An aminoacyl-tRNA molecule that enters the A site of the ribosome next moves to the P site during protein synthesis. This is part of the process known as translocation. After the P site, it moves to the E site, then exits the ribosome.

In protein synthesis, an aminoacyl-tRNA that enters the A site of the ribosome will next occupy the P site. This transition between the A site (aminoacyl site) and the P site (peptidyl site) happens during the process known as translocation, which is driven by elongation factors. Once in the P site, the tRNA molecule will be linked with the growing polypeptide chain. After this the tRNA will move to the E site (exit site) and leave the ribosome.

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Answer:

We can increase the strength by increasing current in the wire or increasing the number of turns of the coil around an iron core .

Explanation:

The magnetic strength due to current carrying conductor carrying current I and having N number of turns is given by

      B = [tex]\mu _{0}\times N\times I[/tex]

[tex]\mu_{O}[/tex] is vacuum permeability .and its value is equal to [tex]4\pi\times10^{-7} \frac{H}{m}[/tex]

.so from the above equation we can see that magnetic strength is directly proportional to the current through wire and no. of turns .

Final answer:

An electromagnet's strength can be increased by passing an electric current through a coil of wire wrapped around an iron core and using a ferromagnetic material to intensify the magnetic field.

Explanation:

An electromagnet increases its strength when an electric current passes through a coil of wire wrapped around an iron core. The strength of the magnetic field generated by the electromagnet is proportional to the amount of current flowing through the wire.

By using a ferromagnetic core within the coil, such as soft iron, the magnetic field can be intensified to thousands of times stronger than the field produced by the coil alone. This combination creates a ferromagnetic-core electromagnet with significantly increased magnetic strength.

Answer:

The current in R₁ is 0.0816 A.

The current at H point is 0.0243 A.

Explanation:

Given that,

Voltage = 64.5

Resistance is

[tex]R_{1}=711\ \Omega[/tex]

[tex]R_{2}=182\ \Omega[/tex]

[tex]R_{3}=663\ \Omega[/tex]

[tex]R_{4}=534\ \Omega[/tex]

[tex]R_{5}=265\ \Omega[/tex]

Suppose, The specified points are R₁ and H.

According to figure,

R₂,R₃,R₄ and R₅ are connected in parallel

We need to calculate the resistance

Using parallel formula

[tex]\dfrac{1}{R}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dfrac{1}{R_{4}}+\dfrac{1}{R_{5}}[/tex]

Put the value into the formula

[tex]\dfrac{1}{R}=\dfrac{1}{182}+\dfrac{1}{663}+\dfrac{1}{534}+\dfrac{1}{265}[/tex]

[tex]\dfrac{1}{R}=\dfrac{35501}{2806615}[/tex]

[tex]R=79.05\ \Omega[/tex]

R and R₁ are connected in series

We need to calculate the equilibrium resistance

Using series formula

[tex]R_{eq}=R_{1}+R[/tex]

[tex]R_{eq}=711+79.05[/tex]

[tex]R_{eq}=790.05\ \Omega[/tex]

We need to calculate the equivalent current

Using ohm's law

[tex]i_{eq}=\dfrac{V}{R_{eq}}[/tex]

Put the value into the formula

[tex]i_{eq}=\dfrac{64.5}{790.05}[/tex]

[tex]i_{eq}=0.0816\ A[/tex]

We know that,

In series combination current distribution in each resistor will be same.

So, Current in R and R₁ will be equal to [tex]i_{eq}[/tex].

The current at h point will be equal to current in R₅

We need to calculate the voltage in R

Using ohm's law

[tex]V=I_{eq}\timesR[/tex]

Put the value into the formula

[tex]V=0.0816\times79.05[/tex]

[tex]V=6.45\ Volt[/tex]

In resistors parallel combination voltage distribution in each part will be same.

So, [tex]V_{2}=V_{3}=V_{4}=V_{5}=6.45 V[/tex]

We need to calculate the current at H point  

Using ohm's law

[tex]i_{h}=\dfrac{V_{5}}{R_{5}}[/tex]

Put the value into the formula

[tex]i_{h}=\dfrac{6.45}{265}[/tex]

[tex]i_{h}=0.0243\ A[/tex]

Hence, The current in R₁ is 0.0816 A.

The current at H point is 0.0243 A.

Answer:

ΔPE= -4.8×10⁻²⁰J

Explanation:

Given data

Electric field of strength E=3.0 N/C

Charge of proton q=1.60×10⁻¹⁹C

Proton moves distance d=10 cm=0.10 m

To find

Change in electrical potential energy ΔPE

Solution

As we know that:

ΔPE= -qEd

[tex]=-(1.60*10^{-19}C )(3.0N/C)(0.10m)\\=-4.8*10^{-20}J[/tex]

ΔPE= -4.8×10⁻²⁰J

Answer:

Explanation:

Given:

D = 10 cm

= 0.1 m

E = 3.0 N/C

Qp = 1.602 × 10^-19 C

U = Q × V

But,

V = E × D

= 3 × 0.1

= 0.3 V

U = 1.602 × 10-19 × 0.3

= 4.806 × 10^-20 J.

Answer:735.46N, 626.34N

Explanation: when an elevator ascend, it means it moving against gravity

F= m(a+g)

M=66kg

a=v/t

g=9.81m/s^2

For the first a= 1.2/0.9

a=1.2/0.9

a=1.33m/s2

F=66(1.33+9.81)

F=66(11.143)

F=735.46N

F=m(a+g)

a=v/t

a=-1.6/5

a=-0.32m/s2

F=66(-0.32+9.8)

F=66(9.49)

F=626.34N

Answer:

1.9 s

Explanation:

We are given that

Length of cable=l=15 m

We have to find the time you have to move out of the way.

We know that

Time period,T=[tex]2\pi\sqrt{\frac{l}{g}}[/tex]

Where g=[tex]9.8m/s^2[/tex]

By using the formula

[tex]T=2\pi\sqrt{\frac{15}{9.8}}[/tex]

[tex]T=2\times 3.14\times \sqrt{\frac{15}{9.8}}=7.77 s[/tex]

Time you have to move out

[tex]t=\frac{T}{4}=\frac{7.77}{4}=1.9 s[/tex]

Hence,time you have to move out of the way=7.77 s

Answer:

Explanation:

The approach of Linear expansivity was used in solving the problem and the detailed steps and appropriate calculation is carefully shown in the attached file.

Answer:

a. 112.608 cm³ b. 2010 cm³ c. 0.032 cm

Explanation:

Given

Volume of aluminium cylinder = 2.0 L= 2.0 × 10³ cm³

Length of aluminium cylinder = 19.0 cm  

Volume of turpentine = 2.0 L= 2.0 10³ cm

Initial temperature of aluminium cylinder = 23.0 °C

Initial temperature of turpentine = 23.0 °C

Average linear expansion coefficient of aluminium α = 2.4 × 10⁻⁵ °C

Average volume expansion coefficient of turpentine γ₁ = 9.0 × 10⁻⁴ °C

(a) How much turpentine overflows in cm³?

We calculate the volume expansion of the aluminium cylinder and that of the turpentine and then subtract the difference.

The volume expansion of material  V = V₀(1 + γΔθ)

where V = final volume, V₀ = initial volume, γ = volume expansion and Δθ = temperature change.

For aluminium V₀ = 2000 cm³, γ = 3α = 3 × 2.4 × 10⁻⁵ /°C = 7.2 × 10⁻⁵ /°C

Δθ = θ₂ - θ₁ = 91 °C - 23 °C = 68 °C

V₁ = 2000(1 + 7.2 × 10⁻⁵ /°C × 68°C) = 2000(1 + 0.004896) = 2000(1.004896) = 2009.792 cm³

For turpentine V₀ = 2000 cm³, γ = 9.0 × 10⁻⁴ /°C

Δθ = θ₂ - θ₁ = 91 °C - 23 °C = 68 °C

V₂= 2000(1 + 9.0 × 10⁻⁴ /°C × 68°C) = 2000(1 + 0.0612) = 2000(1.0612) = 2122.4 cm³.

The volume of turpentine that overflows = V₂ - V₁ = 2122.4 cm³ - 2009.792 cm³ = 112.608 cm³

(b) What is the volume of turpentine remaining in the cylinder at 91.0°C? (Give you answer to four significant figures.)  in cm³?

The volume of turpentine remaining is the final volume of the turpentine minus overflow = 2122.4 cm³ - 112.608 cm³ = 2009.792 cm³ = 2010 cm³ to four significant figures.

(c) If the combination with this amount of turpentine is then cooled back to 23.0°C, how far below the cylinder's rim does the turpentine's surface recede in cm?

The linear expansion for the aluminium to 91 C from 23 C is L = L₀(1 + Δθ)

For aluminium L₀ = 19 cm, α = 2.4 × 10⁻⁵ /°C =

Δθ = θ₂ - θ₁ = 91 °C - 23 °C = 68 °C

L = 19(1 + 2.4 × 10⁻⁵ /°C × 68°C) = 19(1 + 0.001632) = 19(1.001632) = 19.031 cm

The linear expansion for the aluminium from 91 C to 23 C is L = L₀(1 + Δθ)

For aluminium L₀ = 19 cm, α = 2.4 × 10⁻⁵ /°C =

Δθ = θ₂ - θ₁ = 23 °C - 91 °C = -68 °C

L₁ = 19.031(1 + 2.4 × 10⁻⁵ /°C × -68°C) = 19.031(1 - 0.001632) = 19.031(0.998368) = 18.999 cm

How far below the rim it recedes is L - L₁ = 19.031 cm - 18.999 cm = 0.032 cm.

Answer:

Unusually large moons form in giant impacts, which are relatively rare events

Explanation:

Solution:

- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.

- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible. The transition from an undifferentiated cloud to a star system complete with planets and moons takes about 100 million years.

The HCP structure has the fewest slip directions, making HCP metals like Mg and Zn more difficult to deform at room temperature compared to those with BCC or FCC structures.

The crystal structure that has the fewest slip directions and is thus generally more difficult to deform at room temperature is the hexagonal close-packed (HCP) structure. Metals with the HCP structure include Cd, Co, Li, Mg, Na, and Zn. The nature of HCP's stacking arrangement, with alternating type A and type B close-packed layers (ABABAB...), results in fewer slip systems compared to body-centered cubic (BCC) and face-centered cubic (FCC) structures. BCC structures, found in metals like K, Ba, Cr, Mo, W, and Fe, have a coordination number of 8, whereas FCC structures have a coordination number of 12 and include metals like Ag, Al, Ca, Cu, Ni, Pb, and Pt. The closer atomic packing found in FCC, identified as CCP or cubic close-packed, is reflected in the ABCABCABC... stacking sequence, which affords a higher number of slip directions.

Ozone is a molecule that consists of three oxygen atoms. When high-energy ultraviolet rays strike ordinary oxygen molecules (O2), they split the molecule into two single oxygen atoms, known as atomic oxygen. A freed oxygen atom then combines with another oxygen molecule to form a molecule of ozone.  Hydrocarbons and nitrogen oxides come from great variety of industrial and combustion processes. Motor vehicle exhaust and industrial emissions, gasoline vapors, and chemical solvents are some of the major sources of NOx and VOC that acts as a precursor of ozone. In urban areas, the number of automobiles are more and therefore, more production of such harmful gases. These gases in presence of sunlight leads to the formation of bad ozone.

Answer:

There is no change in distance

Explanation:

Given that block slides a distance dd with initial velocity [tex]v_{ox}[/tex][tex]v_{ox}[/tex]

Work energy theorem states that Work done W is the difference between final kinetic energy [tex]KE_{f}[/tex] and initial kinetic energy [tex]KE_{i}[/tex]

W = [tex]KE_{f}[/tex] -  [tex]KE_{i}[/tex]  ...........(equation 1)

In the given problem, the work is done by frictional force

Hence the work done is given by

[tex]W_{Friction}[/tex] =  -[tex]F_{Friction}[/tex] x dd

[tex]F_{Friction}[/tex] is negative since it is resistive force and dd is the distance

[tex]W_{Friction}[/tex] = - μ mg X dd

where μ is coefficient of friction.

Also we know that Kinetic energy KE = [tex]\frac{1}{2}[/tex] [tex]mv^{2}[/tex]

[tex]KE_{f}[/tex] = 0 since final velocity is zero

[tex]KE_{i}[/tex] = [tex]\frac{1}{2}[/tex] [tex]m(v_{ox} v_{ox}) ^{2}[/tex]

Substituting the corresponding values equation 1 becomes

-μ mg x dd = 0 -  [tex]\frac{1}{2}[/tex] [tex]m(v_{ox} v_{ox}) ^{2}[/tex]

dd= [tex]\frac{(v_{ox}v_{ox} )^{2} }{2g}[/tex]x (1/μ) ........... (equation 2)

To find distance when mass of block is doubled and initial velocity is not changed:

Equation 2 shows that the distance dd is independent of mass, therefore there is no change in distance.

The block will slide only half the distance before stopping when its mass is doubled while the initial velocity remains the same.

When a block is pushed along the floor with an initial velocity [tex]\(v_{0x}\)[/tex] and slides a distance [tex]\(d\)[/tex] before stopping, its stopping distance is determined by the initial velocity and the frictional force opposing its motion.

The force of friction is given by [tex]\(F_{\text{friction}} = \mu \cdot N\)[/tex], where [tex]\(\mu\)[/tex] is the coefficient of friction and [tex]\(N\)[/tex] is the normal force. In this case, we are assuming a constant coefficient of friction.

The deceleration of the block can be calculated using Newton's second law: [tex]\(F = m \cdot a\)[/tex]. When the mass of the block is doubled, the force required to decelerate it to a stop is also doubled.

Let's assume the initial velocity [tex]\(v_{0x}\)[/tex] remains the same. The deceleration [tex]\(a\)[/tex] is proportional to the force, so [tex]\(a\)[/tex] is also doubled when the mass is doubled. This is because [tex]\(F\)[/tex] is directly proportional to [tex]\(m\)[/tex].

The stopping distance [tex]\(d'\)[/tex] can be calculated using the equation of motion:

[tex]\[v_f^2 = v_i^2 + 2as\][/tex]

Where:

[tex]\(v_f\)[/tex] is the final velocity (0 m/s, as the block stops),

[tex]\(v_i\)[/tex] is the initial velocity [tex](\(v_{0x}\)),[/tex]

[tex]\(a\)[/tex] is the deceleration (doubled),

[tex]\(s\)[/tex] is the stopping distance [tex](\(d'\)).[/tex]

Plugging in the values:

[tex]\[0 = v_{0x}^2 + 2(2a)d'\][/tex]

Now, we can solve for [tex]\(d'\):[/tex]

[tex]\[2ad' = -v_{0x}^2\][/tex]

[tex]\[d' = \frac{-v_{0x}^2}{2a}\][/tex]

Since [tex]\(a\)[/tex] is doubled, the stopping distance [tex]\(d'\)[/tex] will be halved compared to the initial distance [tex]\(d\).[/tex] So, the block will slide only half the distance before stopping when its mass is doubled while the initial velocity remains the same.

For more such questions on initial velocity:

https://brainly.com/question/29110645

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Answer:

At the closest point

Explanation:

We can simply answer this question by applying Kepler's 2nd law of planetary motion.

It states that:

"A line connecting the center of the Sun to any other object orbiting around it (e.g. a comet) sweeps out equal areas in equal time intervals"

In this problem, we have a comet orbiting around the Sun:

- Its closest distance  from the Sun is 0.6 AU

- Its farthest distance from the Sun is 35 AU

In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: [tex]A\propto vr[/tex], therefore if r is larger, then v (velocity) must be lower).

On the other hand, when the the comet is closer to the Sun the line must move faster ([tex]A\propto vr[/tex], if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.