The same explosive charge is used, so the total energy of the cannon plus cannonball system remains the same. Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all other parameters remained the same? Answer in units of m/s.

1) 12 cm

2) 3 N

Explanation:

1)

The relationship between force and elongation in a spring is given by Hooke's law:

[tex]F=kx[/tex]

where

F is the force applied

k is the spring constant

x is the elongation

For the spring in this problem, at the beginning we have:

[tex]F=2 N[/tex]

[tex]x=4 cm[/tex]

So the spring constant is

[tex]k=\frac{F}{x}=\frac{2N}{4 cm}=0.5 N/cm[/tex]

Later, the force is tripled, so the new force is

[tex]F'=3F=3(2)=6 N[/tex]

Therefore, the new elongation is

[tex]x'=\frac{F'}{k}=\frac{6}{0.5}=12 cm[/tex]

2)

In this second problem, we know that the elongation of the spring now is

[tex]x=6 cm[/tex]

From part a), we know that the spring constant is

[tex]k=0.5 N/cm[/tex]

Therefore, we can use the following equation to find the force:

[tex]F=kx[/tex]

And substituting k and x, we find:

[tex]F=(0.5)(6)=3 N[/tex]

So, the force to produce an elongation of 6 cm must be 3 N.

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:

t = 0.1 s

Explanation:

Given:

- The distance crushed (Stopping-Distance) s = 1.20 m

- The mass of the car m = 1,550 kg

-  The initial velocity vi = 24.0 m/s

Find:

- How long does the collision last t?

Solution:

- The stopping distance s is the average velocity v times time t as follows:

                                s = t*( vf + vi ) / 2

Where,

             vf = 0 m/s ( Stopped )

                                s = t*( vi ) / 2

                                t = 2*s / vi

                                t = 2*1.20 / 24

                                t = 0.1 s

Answer:

t=0.1seconds.

Explanation:

The mass of the car m = 1,550 kg  

We know the stopping distance, 1.20m,

we know the final velocity, 0m/s (its stopped),

the starting velocity, 24m/s.

d=t(v2+v1)/2

Rearange to solve for t, and remove the v2^2 as its zero

t=2d/v1  

t=2(1.20m)/(24m/s)

t=0.1seconds.

Answer:

The ride is above 22m in height for 1.33 minutes.

Explanation:

Let's first find the height required above the boarding platform for the ride to be 22 m above the ground:

Height required = 22 - 5 = 17 m

We can now, using a right angled triangle of height equal to the Ferris wheel radius, calculate the angle from the vertical axis to achieve this height:

Height of triangle = 15 - (17 - 15) = 13 m

Hypotenuse of triangle = radius = 15 m

Angle from the vertical:

Cos( Angle ) = base / hypotenuse = 13 / 15

Angle = 29.92 °

Multiplying this angle by 2 we get the total angle through which the ride is at the required height:

Total Angle = 29.92 * 2 = 59.85 °

To take out the time we can now simply multiply the ratio of this angle /360 by the time taken for one complete revolution:

Time = [tex]\frac{59.85}{360} * 8[/tex]

Time = 1.33 minutes

Answer:

11.714 kW

Explanation:

Here is the complete question

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 34.0∘ above the horizontal. The car accelerates uniformly to a speed of 2.25 m/s in 10.5 s and then continues at constant speed. What power must the winch motor provide when the car is moving at constant speed?

Solution

Since the loaded ore car moves along the mine shaft at an angle of θ = 34° to the horizontal, if F is the force exerted on the cable, then the net force on the laoded ore car is F - mgsinθ = ma where  mgsinθ = component of the car's weight along the incline, m = mass of loaded ore car = 950 kg and a = acceleration

F = m(a + gsinθ)  

When the car is moving at constant speed, a = 0

So F = m(a + gsinθ) = F = 950(0 + 9.8sin34) = 5206.1 N

Since it continues at a constant speed of v = 2.25 m/s, the power of the winch motor is P = Fv = 5206.1 N × 2.25 m/s = 11713.7 W = 11.714 kW

The force needed to pull a 950 kg ore car up a 34.0° inclined mine shaft is approximately 5518.493 N, involving both the acceleration force and the component of gravitational force parallel to the incline.

The question is about calculating the acceleration and force needed to pull a loaded ore car up a mine shaft inclined at 34.0° above the horizontal.

Step-by-Step Explanation:

Therefore, the total force needed to pull the ore car up the inclined shaft is approximately 5518.493 N.

Answer:

1.599 m.

Explanation:

Using

α = ΔL/(LΔT)............ Equation 1

Where

α = Coefficient of linear expansion of brass,

ΔL = Change in length,

ΔT = Change in temperature

L = Original length.

make ΔL the subject of the equation

ΔL = α(LΔT)................. Equation 2

Given: L = 1.6 m, ΔT = T₂-T₁ = 0-28 = -28 ⁰C

Constant: α = 1.8×10⁻⁵/°C

Substitute into equation 2

ΔL =  (1.8×10⁻⁵)(1.6)(-28)

ΔL = -8.064×10⁻⁴ m

But,

ΔL = L₁-L

Where L = final length of the pendulum rod.

make L₁ the subject of the equation,

L₁ = ΔL+L......... Equation 2

L₁ = 1.6+( -8.064×10⁻⁴ )

L₁ = 1.599 m.

Explanation:

Below is an attachment containing the solution.

The skydiver has more force acting downward due to gravity than upward from air resistance, and thus he's accelerating downwards. You calculate the acceleration by subtracting the air resistance from the skydiver's weight and dividing by his mass, in this case, yielding an acceleration of 7.36 m/s².

The skydiver's situation can be understood by taking into account the forces acting on him, mainly his weight and the air resistance. The weight of the skydiver is determined by multiplying his mass by the gravity constant, 9.81 m/s². This gives a weight of 101 kg × 9.81 m/s² = 991.81 N. The skydiver is said to be experiencing one-fourth that amount in upward force due to air resistance, which would be approximately 247.95 N.

In situations where opposing forces are equal, an object doesn't accelerate and we would expect the skydiver to be in a state of terminal velocity. However, in this case, there's more force acting downward (gravity) than upward (air resistance), so the skydiver is accelerating downwards. To find the amount of that acceleration, we subtract the air resistance from the skydiver's weight, then divide that by his mass. Therefore, (991.81 N - 247.95 N) / 101 kg = 7.36 m/s².

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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:

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: 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.

The number of orbitals in an atom can be determined using quantum numbers. (a) 2p can have 2 orbitals, (b) 4d can have 4 orbitals, and (c) 6s can have 1 orbital.

The number of orbitals in an atom with a set of quantum numbers can be determined using the following rules:

For each orbital, there can be a maximum of two electrons, each with opposite spins (+1/2 and -1/2).

Therefore, (a) 2p can have a maximum of 2 orbitals, (b) 4d can have a maximum of 4 orbitals, and (c) 6s can have a maximum of 1 orbital.

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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.

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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:

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.

Explanation:

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:

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

The satellite dish follows the geometry of a paraboloid. Given the dimensions of a paraboloid that is 12 feet wide and 2 feet deep, the receiver for the satellite signals should be placed 0.5 feet from the base, along the axis of symmetry.

This is a mathematical problem involving the geometry of paraboloids. A paraboloid must have a particular balance between its depth and its width that allows signals, such as those from satellites, to focus at a single point. This point is known as the focus of the paraboloid.

In this case, for a paraboloid that is 12 feet across and 2 feet deep, the focus will be at the position that is one-fourth the depth of the dish from its vertex. Therefore, 2 feet (the depth) divided by 4 equals 0.5 feet. Hence, the receiver should be placed 0.5 feet from the vertex of the paraboloid, along the axis of symmetry, in order for it to receive signals reflected off the dish.

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Answer:

Explanation:

- The atoms combine to form molecules and attain stability by completing their octet. The formation of compound can take place either by transfer of electron from one atom to other or by sharing of electrons between them.

- Resonance structure of a molecule is of two or more forms in which the distribution of electrons around the structure is different but the chemical connectivity is same.

- The total number of valence electrons VE in (N N O) is :

                                 VE = 2(5) + 6 = 16 electrons.

- Among the molecule, the electrons are distributed in atom in such a way that formation of triple bond will take place between two nitrogen atoms and a single bond will form between nitrogen and oxygen atom.

- The formal charge FC on each atom is determined as:

                                  FC = VE - NBE - BE/2

Where,

            NBE: Non-Bonding Electrons

            BE : Bonding Electrons.

- The formal charge on each atom is: the nitrogen atom in center will possess 1+ formal charge and oxygen will possess 1- charge (oxygen is electronegative atom). Thus, results in formation of neutral molecule.

- The structure of (N N O) is shown in attachment.

- The resonance form which is likely to contribute most to the correct structure of (N_2 O) is:

- Structure for (N N O)  showing one lone-pair of electrons on the terminal nitrogen atom, a triple bond between the two nitrogen atoms, a single bond between nitrogen and oxygen, and three lone-pairs of electrons on the terminal oxygen atom.

Answer:

1800/300 = 6ropes

Explanation:

The engine weighs 1800N and the person exerts a force of 300N, so for him to lift the engine and exerting a force of 300N all through we divide the weight of the engine by the force exerted to know how many ropes are used. Which makes it 6 thereby each rope uses 300N to lift the engine.

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:

For Potassium:

Frequency = 7.5 x 10¹⁴ Hz; E (energy) = 8.83 x 10⁻²¹ J

For Strontium:

Frequency = 4.3 x 0¹⁴ Hz

E (energy) = 2.85 x 10⁻¹⁹ J

Explanation:

Wavelength is represented by λ, and Frequency is represented by ν .

E (energy) = hν = hc/λ,  where ν = frequency; c = speed of light = 3 x 10⁸ m/s; 1 s-1 = 1 Hz

h = planck's constant = 6.62 x 10⁻³⁴ J.s; 1 nm = 10⁻⁹ m

1. Potassium λ (wavelength) = 400 nm, Frequency, ν  is given by :

ν = c/λ = (3 x 10⁸  m/s) / 400 nm

  = (3 x 10⁸ m/s) / 400 x 10⁻⁹ s-1

  = 0.0075 x 10¹⁷s-1

  = 7.5 x 10¹⁴  s-1

Frequency = 7.5 x 10¹⁴ Hz

E (energy) = hν = (6.62 x 10⁻³⁴ J.s) x (7.5 x 10¹⁴s-1)

E (energy) = 8.83 x 10⁻²¹ J

2. Strontium λ (wavelength) = 700 nm ,Frequency, ν  is given by :

ν = c/λ = (3 x 10⁸ m/s) / 700 nm

  = (3 x 10⁸ m/s) / 700 x 10⁻⁹s-1

  = 0.00428571428 x 10¹⁷s-1

  = 4.3 x 10¹⁴ s-1

Frequency = 4.3 x 0¹⁴ Hz

E (energy) = hν = (6.62 x 10⁻³⁴ J.s) x (4.3 x 0¹⁴s-1)

E (energy) = 2.85 x 10⁻¹⁹ J

Explanation:

Below is an attachment containing the solution.

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:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: [tex]F = 49.06N[/tex]

Pebble: [tex]F = 29.44N[/tex]

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: [tex]a =9.8m/s^{2}[/tex]

Pebble:  [tex]a =9.8m/s^{2}[/tex]

Explanation:

The universal law of gravitation is defined as:

[tex]F = G\frac{m1m2}{r^{2}}[/tex]  (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

Case for the rock [tex]m = 5.0 Kg[/tex]:

m1 will be equal to the mass of the Earth [tex]m1 = 5.972×10^{24} Kg[/tex] and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth [tex]r = 6371000m[/tex].

[tex]F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}[/tex]

[tex]F = 49.06N[/tex]

Newton's second law can be used to know the acceleration.

[tex]F = ma[/tex]

[tex]a =\frac{F}{m}[/tex] (2)

[tex]a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}[/tex]

[tex]a =9.8m/s^{2}[/tex]

Case for the pebble [tex]m = 3.0 Kg[/tex]:

[tex]F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}[/tex]

[tex]F = 29.44N[/tex]

[tex]a =\frac{F}{m}[/tex]

[tex]a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}[/tex]

[tex]a =9.8m/s^{2}[/tex]

(a) The magnitude of the gravitational force on the rock is 49 N, while the magnitude of the gravitational force on the pebble is 2.94 × 10^-3 N. (b) The magnitude of the acceleration for both objects when released is 9.8 m/s^2.

(a) Gravitational Force:

The magnitude of the gravitational force exerted on an object near the surface of the Earth can be calculated using the formula:

F = mg

where F is the gravitational force, m is the mass of the object, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

For the rock, m = 5.0 kg:

F = mg = (5.0 kg) * (9.8 m/s^2) = 49 N

For the pebble, m = 3.0 × 10-4 kg:

F = mg = (3.0 × 10-4 kg) * (9.8 m/s^2) = 2.94 × 10-3 N

(b) Acceleration:

The magnitude of the acceleration of an object when released can be calculated using Newton's second law:

F = ma

Rearranging the equation to solve for acceleration, we have:

a = F/m

Substituting the values we calculated in part (a) for the gravitational force exerted on each object, we can determine the magnitude of acceleration.

For the rock, F = 49 N and m = 5.0 kg:

a = (49 N) / (5.0 kg) = 9.8 m/s^2

For the pebble, F = 2.94 × 10-3 N and m = 3.0 × 10-4 kg:

a = (2.94 × 10-3 N) / (3.0 × 10-4 kg) = 9.8 m/s^2

Learn more about Gravitational force and acceleration here:

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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:

the membrane potential

Explanation:

typical values of membrane potential are in the range -40 mv to-70 mv

A

Explanation:

fred ball hit the pin the time for all just put a hit in his right knee and do not need to be on bed at the carbon footprint in his first year of practice at all but his teammates were also on bed and you were a good friend to be with you cute guys that he is the most likely and do you have the ability for you I don't want him on you and

The time it takes for Wilma's ball to hit the pins is equal to that for Barney's ball to hit. The time it takes for Barney's ball to hit the pins is greater than that for Fred's ball to hit. The time it takes for Betty's ball to hit the pins is greater than that for Fred's ball to hit.

The time it takes for Wilma's ball to hit the pins is equal to that for Barney's ball to hit.

The time it takes for Barney's ball to hit the pins is greater than that for Fred's ball to hit.

The time it takes for Betty's ball to hit the pins is greater than that for Fred's ball to hit.

The time it takes for Betty's ball to hit the pins is equal to that for Barney's ball to hit.

The time it takes for Fred's ball to hit the pins is equal to that for Wilma's ball to hit.

The time it takes for Wilma's ball to hit the pins is greater than that for Betty's ball to hit.

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Answer:

71.76 m

Explanation:

We will solve this question using the work energy theorem.

The theorem explains that, the change in kinetic energy of a particle between two points is equal to the workdone in moving the particle from the one point to the other.

ΔK.E = W

In the attached free body diagram for the question, the forces acting on the puck are given.

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Final kinetic energy = 0 J (since the puck comes to a stop)

Initial kinetic energy = (1/2)(m)(v²) = (1/2)(0.2)(26²) = 67.6 J

ΔK.E = 0 - 67.6 = - 67.6 J

W = Workdone between the starting and stopping points = (work done by the force of gravity) + (work done by frictional force)

Work done by the force of gravity = - mgh = - (0.2)(9.8)(h) = - 1.96 h

Workdone by the frictional force = F × d

F = μ N

μ = coefficient of kinetic friction = 0.30 (kinetic frictional force is the only frictional force that moves a distance of d, the static frictional force doesn't move any distance, so it does no work)

N = normal reaction of the plane surface on the puck = mg cos 30° = (0.2)(9.8)(0.866) = 1.697 N

F = μ N = 0.3 × 1.697 = 0.509 N

where d = distance along the incline that the puck travels.

d = h/sin 30° = 2h (from trigonometric relations)

Workdone by the frictional force = F × d = 0.509 × 2h = 1.02 h

ΔK.E = W = (work done by the force of gravity) + (work done by frictional force)

- 67.6 = - 1.96h + 1.02h

-0.942h = - 67.6

h = 71.76 m

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:

The lone pair of electrons occupy more space because the electrostatic force becomes weaker.

Explanation:

When there is a bond pair of electrons in the 2 positively charged the atomic nuclei draw the electron density towards them, thereby reducing the bond diameter.

In the case of the lone pair, only 1 nucleus is present, and the enticing electrostatic force becomes weaker and the intensity of the electrons will be increases. Therefore, the lone pair occupies more space than the pair of bonds.    

A lone pair of electrons occupies more space around the central atom than bonding pairs due to greater electrostatic repulsions, which are not mitigated by the sharing of electrons with another atom. This concept also applies to multiple bonds, with higher electron density leading to increased space occupancy compared to single bonds and influencing molecular geometry.

The reason why a lone pair of electrons occupies more space around the central atom than a bonding pair of electrons relates to electrostatic repulsions. Lone pairs are not shared with another atom, therefore they tend to occupy a larger region of space due to increased repulsions of their negatively charged electrons. As illustrated in the case of a molecule like ammonia (NH3), the lone pair of electrons on the nitrogen atom occupies more space than the bonding pairs.

Additionally, this concept extends to multiple bonds such as double or triple bonds, which occupy more space around a central atom than a single bond. The higher electron density in multiple bonds leads to greater repulsions, which can influence the bond angles in a molecule, causing deviations from ideal geometry.

The order of space occupied from largest to smallest is as follows: lone pair > triple bond > double bond > single bond. This hierarchy is essential for understanding molecular shape and bond angles, and is exemplified by deviations in expected bond angles due to these repulsions.