An electric vehicle starts from rest and accelerates at a rate of1.9m/s2in a straight line until it reaches a speed of23m/s. The vehicle then slows at a constant rate of 1.4m/s2 until it stops.(a)How much time elapses from start to stop

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

Answer:

Current circulating is 8.59×10^-6A

Explanation:

Magnetic field at the center of the loop is given by, B= μI/2R

I = 2RB/μ

given that B is 90x 10^-12 T

radius is 0.12 m /2 = 0.06m

we know that μ is 4π x 10^-7 T.A/m

Substituting the given values we get,

I= (((2(0.06m)(90x 10^-12 T))/(4π x 10^-7 T.A/m))

I is 8.59 x 10^-6 A

The current circulating around the heart is 8.59×10⁻⁶A

The magnetic field B at the center of the loop of radius R with current I is given by,

B= μ₀I/2R

Rearranging the terms we get:

I = 2RB/μ₀

Given that B is 90×10⁻¹² T

radius is R = 0.12 m /2 = 0.06m

Substituting the given values we get,

I = 2×0.06×90×10⁻¹² / 4π×10⁻⁷

I = 8.59 x 10⁻⁶ A

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Answer:

Explanation:

Given that,

Number of turn is 48

N=48

Radius is 4.8cm

r=0.048m

Magnetic Field

B=0.48T

Current in coil

i=23.3mA

i=0.233A

Maximum Torque?

Maximum torque occur at angle 90°

Torque is given as

τ = N•I•A•B•sinθ

Where N is number of turn =48

I is current in coil =0.233A

A is area of circular coil form

Area of a circle is given as

A=πr²

A=π×0.048²

A=0.007238m²

B is magnetic field =0.48T

Maximum torque occurs at 90°

τ = N•I•A•B•sinθ

τ=48×0.233×0.007238×0.48×Sin90

τ = 0.0389Nm

This torque is large enough to exert the coil

Given Information:

Magnetic field = B =  0.480 T

Current = I = 23.3 mA = 0.0233 A

Number of turns = N = 48 turns

Radius = r = 5.50 cm = 0.055 m

Required Information:

Maximum possible torque = τ = ?

Answer:

Maximum possible torque = 0.0051 N.m

Explanation:

We know that toque τ is given by

τ = NIABsin(θ)

Where N is the number of turns of the circular coil, I is the current flowing through the circular coil, A is the area of circular coil, B is the magnetic field induced in the circular coil.

The area of the circular coil is

A = πr²

A = π(0.055)²

A = 0.009503 m²

The maximum torque is possible when θ = 90°

τ = 48*0.0233*0.009503*0.480*sin(90°)

τ = 0.0051 N.m

Answer:

c, About the same for both balls.

Explanation:

If no air resistance is present, the rate of descent depends only on how far the object has fallen, no matter how heavy the object is. This means that two objects will reach the ground at the same time if they are dropped simultaneously from the same height.

Answer:

≅ 17000 years or 1.7 x 10⁴ years

Explanation:

time= total energy/power

=  (10⁸J/kg)(2x10³⁰ kg) / 3.8 x 10²⁶ J/s

 = 526,315,789,473 s

=  16689 years

≅ 17000 years or 1.7 x 10⁴ years

Final answer:

If the Sun were simply a large fire releasing chemical energy, it would have a lifetime of approximately 16,438 years, considering a mass of 2.0 x 10³⁰ kilograms and an energy release of 10^8 joules per kilogram. This is significantly shorter than the Sun's actual lifespan, confirming that nuclear reactions, not chemical combustion, power the Sun.

Explanation:

If the Sun were simply a large fire releasing chemical energy, we can estimate its lifetime using certain assumptions. Firstly, if the Sun's mass is predominantly hydrogen and its energy output is due to combustion with oxygen, the amount of energy released per kilogram would be about 108 joules. With this in mind, we recall that the Sun's mass is approximately 2.0 × 1030 kilograms. The resource provided illustrates that combustion produces about 45,000,000 calories, or approx. 1.88 × 108 joules, per second for a combusting mass of one kilogram (previously cited as 1.6 × 10-19 joules for a typical energy in a chemical bond). With a simple energy balance calculation, we find that the theoretical Sun-as-a-fire would consume its mass in:

Total energy available = mass × energy per kilogram = 2.0 × 1030 kg × 108 joules/kg = 2.0 × 1038 joules.

If the Sun's luminosity is 3.8 × 1026 watts and 1 watt is equivalent to 1 joule per second, the time in seconds for the Sun to deplete this energy at its current luminosity is:

Lifetime in seconds = Total energy / Luminosity = 2.0 × 1038 joules / 3.8 × 1026 watts = 5.26 × 1011 seconds.

Since there are roughly 32 million seconds in a year, the lifetime of the Sun, if it were burning chemically, can be estimated as:

Lifetime in years = Lifetime in seconds / seconds per year = 5.26 × 1011 seconds / 3.2 × 107 seconds/year = approximately 16,438 years.

This is a very short lifetime compared to the Sun's actual age of approximately 4.6 billion years, indicating the real Sun is sustained by nuclear reactions and not chemical combustion.

Complete Question:

Devon has several toy car bodies and motors. The motors have the same mass, but they provide different amounts of force, as shown in this table.  

The bodies have the masses shown in this table (refer attached figure).  

Which motor and body should Devon use to build the car with the greatest acceleration?

motor 1, with body 1

motor 1, with body 2

motor 2, with body 1

motor 2, with body 2

Answer:

Devon should build the car with motor 2 and body 1 for having the greatest acceleration.

Explanation:

As per Newton's second law of motion, the acceleration of any object is directly proportional to the force on the object and inversely proportional to the mass of the object.

It can be seen that motor 2 has greater force than the force provided by motor 1. Similarly, the mass of body 1 is found to be lesser compared to mass of body 2. So,

          [tex]acceleration =\frac{\text { Force }}{\text { mass }}[/tex]

It gives, the system with motor 2 and body 1 the maximum acceleration. So the car should be built with motor 2 and body 1.

The car with the greatest acceleration will be one that has a higher power-to-weight ratio, provided by a light yet powerful motor, and a lighter, aerodynamically efficient body. Both factors--lightweight and power-- are crucial for achieving high acceleration.

In deciding what motor and body to use, Devon must consider factors like the power to weight ratio, the torque of the motor, and the aerodynamics of the body. A higher power-to-weight ratio generally translates to greater acceleration. Therefore, Devon should choose a motor that is powerful yet light. When considering the body, Devon should go for a lighter body as heavy bodies slows down a car's acceleration. Besides weight, a body whose design is aerodynamically efficient will enhance acceleration because it reduces air resistance.

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Answer:

A) [tex]v_b=0.8\ m.s^{-1}[/tex] is the final velocity of the cart B after collision.

B) [tex]KE_A=0.176\ J[/tex]

C) [tex]KE_B=0\ J[/tex]

D) [tex]ke_a=0\ J[/tex]

E) [tex]ke_B=0.176\ J[/tex]

F) Yes, here the kinetic energy is conserved because the mass of both the bodies involved in the collision is same.

G) Yes, momentum is always conserved for an elastic collision.

Explanation:

Given:

A)

As given in the question that the cars undergo an elastic collision:

According to the conservation of momentum:

[tex]m_a.u_a+m_b.u_b=m_a.v_a+m_b.v_b[/tex]

[tex]0.55\times 0.8+0.55\times 0=0.55\times 0+0.55\times v_b[/tex]

[tex]v_b=0.8\ m.s^{-1}[/tex] is the final velocity of the cart B after collision.

B)

Initial kinetic energy of cart A:

[tex]KE_A=\frac{1}{2} m_a.u_a^2[/tex]

[tex]KE_A=0.5\times 0.55\times 0.8^2[/tex]

[tex]KE_A=0.176\ J[/tex]

C)

Initial kinetic energy of cart A:

[tex]KE_B=\frac{1}{2} \times m_b.u_b^2[/tex]

[tex]KE_B=0.5\times 0.55\times 0^2[/tex]

[tex]KE_B=0\ J[/tex]

D)

The final kinetic energy of cart A:

[tex]ke_A=\frac{1}{2} m_a.v_a^2[/tex]

[tex]ke_a=0.5\times 0.55\times 0^2[/tex]

[tex]ke_a=0\ J[/tex]

E)

The final kinetic energy of cart B:

[tex]ke_B=\frac{1}{2} m_b.v_b^2[/tex]

[tex]ke_B=0.5\times 0.55\times 0.8^2[/tex]

[tex]ke_B=0.176\ J[/tex]

F)

Yes, here the kinetic energy is conserved because the mass of both the bodies involved in the collision is same.

G)

Yes, momentum is always conserved for an elastic collision.

(a) The velocity of cart B is [tex]0.8m/s[/tex]

(b) Initial kinetic energy of cart A is [tex]0.176J[/tex]

(c) Initial kinetic energy of cart B is 0

(d) The final kinetic energy of cart A is 0

(e) The final kinetic energy of cart B is [tex]0.176J[/tex]

(f) The kinetic energy is conserved

(g) The momentum is conserved

(a) The momentum of the system must be conserved, which means momentum before the collision must be equal to momentum after the collision.

Momentum before collision = [tex]m_Au_A+m_bu_B=0.55\times0.8+0.55\times0=0.44kgm/s[/tex]

Momentum after collision:

[tex]m_A+m_A+m_Bv_B=0.55\times0+0.55v_B=0.44\\\\v_B=0.8m/s[/tex]

(b) Initial kinetic energy of cart A:

[tex]K_{iA}=\frac{1}{2}m_Au_A^2=0.5\times0.55\times0.64=0.176J[/tex]

(c) Initial kinetic energy of cart B:

[tex]K_{iB}=\frac{1}{2}m_Bu_B^2=0.5\times0.55\times0=0J[/tex]

(d) The final kinetic energy of cart A:

[tex]K_{fA}=\frac{1}{2}m_Av_A^2=0.5\times0.55\times0=0J[/tex]

(e) Initial kinetic energy of cart B:

[tex]K_{iB}=\frac{1}{2}m_B_v_B^2=0.5\times0.55\times0.64=0.176J[/tex]

(f) yes the kinetic energy of the system is conserved for elastic collision

(g) The momentum is conserved in any case including the elastic collision

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Final answer:

Doubling the distance between a pair of oppositely charged parallel plates, without altering the charge on them, does not change the potential difference. The potential difference remains at 402 V because it is directly influenced by the electric field and the charge on the plates, both of which remain constant in this scenario.

Explanation:

The question involves understanding how the potential difference between a pair of oppositely charged parallel plates changes when the distance between them is altered, without changing the charge on the plates. The original potential difference is given as 402 V when the spacing between the plates is a certain value.

If the spacing between the plates is doubled, the new potential difference remains the same, 402 V, because the potential difference between two plates is determined by the equation V = Ed, where 'E' is the electric field strength and 'd' is the distance between the plates. Since the charge on the plates remains constant, and assuming the electric field remains uniform, the electric field (E) does not change.

Therefore, doubling the distance without altering the charge or the electric field does not affect the potential difference. This is a key concept in electrostatics, highlighting how changes in geometry and charge distribution affect electric fields and potential differences.

Final answer:

Total internal reflection is possible for light incident from air on water.

Explanation:

Total internal reflection occurs when light traveling from a medium with a higher refractive index to a medium with a lower refractive index is incident at an angle greater than the critical angle.

In the case of light incident from air on water, the critical angle is 48.6°. If the angle of incidence is greater than this critical angle, the light will undergo total internal reflection and not refract out of the water. Therefore, total internal reflection is possible for light incident from air on water.

Answer:

The cable would stretch 14 cm when loaded with 1000 kg ore.

Explanation:

The question is incomplete.

The complete question would be.

A mine shaft has an ore elevator hung from a single braided cable of diameter 2.5 cm. Young's modulus of elasticity of the cable is [tex]10\times 10^{10}\ N/m^2[/tex]. When the cable is fully extended, the end of the cable is 700 m below the support.

How much does the fully extended cable stretch when 1000 kg of ore is loaded into the elevator?

Given the diameter of the cable is [tex]2.5\ cm[/tex]. The length of the cable is [tex]700\ m[/tex].

And the mass of the ore is [tex]1000\ kg[/tex]. Also the Young's modulus of  elasticity of the cable is [tex]10\times 10^{10}\ N/m^2[/tex].

We will use Hook's law

[tex]\sigma=E\epsilon[/tex]

Where [tex]\sigma[/tex] is stress. E is Young's modulus of elasticity. And [tex]\epsilon[/tex] is strain.

We can rewrite .

[tex]\frac{P}{A}=E\times \frac{\delta l}{l}[/tex]

Where [tex]P[/tex] is the applied force, [tex]A[/tex] is the area of the cross-section. [tex]\delta l[/tex] is the change in length. [tex]l[/tex] is the initial length of the cable.

Also, the applied force [tex]P[/tex] is due to mass of the ore. That would be [tex]P=mg\\P=1000\times 9.81\ N[/tex]

Given diameter of the cable [tex](d)[/tex] [tex]2.5\ cm[/tex].

[tex]d=\frac{2.5}{100}=0.025\ m\\ A=\frac{\pi}{4}d^2\\ \\A=\frac{\pi}{4}(0.025)^2=4.91\times 10^{-4}\ m^2[/tex]

[tex]E=10\times 10^{10}\ N/m^2[/tex]

[tex]l=700\ m[/tex]

Plugging these values

[tex]\frac{P}{A}=E\times \frac{\delta l}{l}[/tex]

[tex]\frac{P}{A}\times \frac{l}{E}=\delta l \\\\ \delta l =\frac{1000\times 9.81\times 700}{4.91\times 10^{-4}\times 10\times 10^{10}} \\\delta l=.139\ m\\\delta l=14\ cm[/tex].

So, the cable would stretch 14 cm when loaded with 1000 kg ore.

Final answer:

The peak emf from a generator can be maximized by increasing the number of turns, increasing the area of the wired loop, and increasing the frequency. Both the one-turn and two-turn square coils can produce the same peak emf for a given frequency and magnetic field strength.

Explanation:

The peak emf from a generator can be maximized by maximizing the number of turns, maximizing the area of the wired loop, or maximizing the frequency. The one-turn and two-turn square coils are both options that can be used, as they both give the same peak emf for a given frequency and magnetic field strength. However, the other factors, such as the number of turns and area of the loop, may differ between the two options and can affect the overall performance of the generator.

Answer:

A=0

Explanation:

I got 100% on this assignment

This question involves the basic concept of displacement.

The frog's displacement is "0 cm".

In order to calculate the displacement of the frog, we must consider both the magnitude and direction of the movements made by the frog. Here, we will take the forward direction as positive and the backward direction as negative. Hence, the displacement of the frog will be as follows:

Displacement = 18 cm - 6 cm - 12 cm

Displacement = 0 cm

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The attached picture shows the difference between the displacement and the distance.

Answer:

The hang time is 0.95 s

Explanation:

Spud Webb has a height of 5 feet 8 inches and is one of the shortest basketball players to play in the NBA.

He could make an amazing vertical leap of 1.1 m off the ground. For such a leap to calculate the hang time(the time spent in the air after leaving the ground and before touching down again), we use the formula:

[tex]S=ut -\frac{1}{2}gt^{2}[/tex]

Where S is the distance traveled, u is the initial velocity, t is the time taken and g is the acceleration due to gravity.

Given that:

g = 9.8 m/s²

S = [tex]y_{f} -y_{i} =0-1.1=-1.1[/tex]

u = 0

[tex]S=ut -\frac{1}{2}gt^{2}[/tex]

Substituting values:

[tex]-1.1=(0)t -\frac{1}{2}(9.8)t^{2}\\-1.1=0-4.9t^{2} \\-1.1=-4.9t^{2}[/tex]

Dividing through by -4.9 we get:

[tex]\frac{-1.1}{-4.9} =\frac{-4.9}{-4.9}t^{2}[/tex]

[tex]t^{2}=0.2245\\ t=\sqrt{0.2245}=0.474[/tex]

t = 0.474 s

The hang time = 2t = 2 × 0.474 = 0.95 s

The hang time is 0.95 s

The average speed of Frank's car is 2 m/s and the average speed of Noah's car is 1.8 m/s.

Explanation:

Average speed is the measure of total distance covered at different time period. Since, the formula used for calculating the average speed is the ratio of total distance covered by each car to the time taken to cover that distance.

As there is only one set of data i.e., distance and time, the average speed will be equal to the speed of the car.

So in this case, the total distance covered by franks car is 300 cm = 3 m in 1.5 s. Then, the average speed will be

[tex]Avg. S_{frank} = \frac{Total\ distance\ rolled\ by\ his\ car}{Time\ taken}[/tex]

[tex]Avg.S_{frank} =\frac{300 \times 10^{-2} }{1.5}=2\ m/s[/tex]

Similarly, the average speed of Noahs car which rolled a distance of 360 cm = 3.6 m in time 2 s, will be

[tex]Avg.S_{Noah} =\frac{360 \times 10^{-2} }{2}=1.8\ m/s[/tex]

Thus, the average speed of Frank's car is 2 m/s and the average speed of Noah's car is 1.8 m/s.

The average speed for Frank's car and Noah's car is 200 cm/s and 180 cm/s respectively calculated using the formula speed = distance / time.

Average speed is a measure of how quickly an object moves over a certain distance in a specific amount of time. It's calculated using the formula:

Average Speed (v) = Total Distance Traveled (d) / Total Time Taken (t)

To calculate the average speed of any moving object, use the formula: speed = distance / time. For Frank's car which rolled 300 cm in 1.5 seconds, the speed would be 300 cm / 1.5 s = 200 cm/s. Similarly, for Noah's car which rolled 360 cm in 2 seconds, the speed would be 360 cm / 2 s = 180 cm/s.

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Answer:

0 J

Explanation:

The work done by an ideal gas is given by the equation:

[tex]W=p\Delta V[/tex]

where

W is the work done by the gas

p is the pressure of the gas

[tex]\Delta V[/tex] is the change in volume of the gas

In this problem, we have a gas kept at a constant volume of

V = 2.00 L

This is an isochoric process (constant volume). Since the volume is constant, the change in volume is zero:

[tex]\Delta V=0[/tex]

And therefore, this means that the work done by the gas is zero:

W = 0

Answer: The distance is 723.4km

Explanation:

The velocity of the transverse waves is 8.9km/s

The velocity of the longitudinal wave is 5.1 km/s

The transverse one reaches 68 seconds before the longitudinal.

if the distance is X, we know that:

X/(9.8km/s) = T1

X/(5.1km/s) = T2

T2 = T1 + 68s

Where T1 and T2 are the time that each wave needs to reach the sesmograph.

We replace the third equation into the second and get:

X/(9.8km/s) = T1

X/(5.1km/s) = T1 + 68s

Now, we can replace T1 from the first equation into the second one:

X/(5.1km/s) = X/(9.8km/s) + 68s

Now we can solve it for X and find the distance.

X/(5.1km/s) - X/(9.8km/s) = 68s

X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s

X = 68s/0.094s/km = 723.4 km

Answer:

10.56 ft/s.

Explanation:

given,

Speed of side walk = 7 ft/s

time taken to travel 74 ft on the moving sidewalk = time taken to walk 15 ft on non moving side walk.

We know,

[tex]time = \dfrac{distance}{speed}[/tex]

[tex]\dfrac{74}{w+7}=\dfrac{15}{w-7}[/tex]

[tex]59 w = 7\times 74 + 15\times 7[/tex]

[tex] 59 w = 623[/tex]

[tex]w = 10.56\ ft/s[/tex]

Rate at which person walk on non moving side walk is equal to 10.56 ft/s.

The person's walking speed on a non-moving sidewalk is approximately 1.78 ft/sec, determined by setting up and solving the equation based on equal times on the moving and non-moving sidewalks.

To find the rate at which a person would walk on a non-moving sidewalk, we need to set up an equation based on the given information.

We know the speed of the moving sidewalk is 7 ft/sec. Let's denote the walking speed of the person (on a non-moving sidewalk) as v ft/sec.

From the problem, the time taken by the person to walk 74 ft on the moving sidewalk should be equal to the time taken to walk 15 ft on a non-moving sidewalk in the opposite direction:

       [tex]\frac{74}{7 + v} = \frac{15}{v}[/tex]

   74v = 15(v+7)
   74v = 15v + 105

  74v − 15v = 105

 59v = 105

  [tex]v = \frac{105}{59}[/tex]

 v ≈ 1.78 ft/sec

Therefore, the person's walking speed on a non-moving sidewalk is approximately 1.78 ft/sec.

Answer:

45 C.

Explanation:

Given:

Time, t = 5.0 h

= 18000 s

Current, I = 2.5 mA

= 0.0025 A

Potential difference, V = 9 V

Q = I × t

= 18000 × 0.0025

= 45 C.

Explanation:

Below is an attachment containing the solution.

Final answer:

Johnnie's move from automotive manufacturing to the IT sector due to changes in the economy is an example of structural mobility, which reflects broad societal shifts affecting whole groups within the social class system. Option a

Explanation:

Johnnie's career change from the declining auto industry that his father was a part of, to the burgeoning information technology sector, is an example of structural mobility. This type of mobility occurs due to large-scale economic and societal shifts rather than individual actions. For instance, in the past, industrialization led to the general upliftment of the population, causing widespread upward structural mobility. In more recent times, factors such as recessions and outsourcing have resulted in many experiencing downward structural mobility, as traditional jobs have become obsolete or have moved overseas.

Moreover, technological advancements have altered the kinds of skills employers seek in their workforce, leading to a situation where some individuals may find themselves structurally unemployed if they lack the necessary analytical and communication skills, despite having traits like reliability and physical strength that were formerly highly valued.

Therefore, Johnnie's entrance into a different industry than his father's, brought upon by changes in the economic landscape, is a reflection of the broader trend where entire groups in society move up or down the social class ladder due to changes in the social structure itself.

Answer:

a)

The expression for the electric potential in this problem is:

[tex]V=5x-3x^2y+2yz[/tex]

where

x, y, z are the three spatial coordinates

The relationship between components of the electric field and electric potential is:

[tex]E_x=-\frac{dV}{dx}\\E_y=-\frac{dV}{dy}\\E_z=-\frac{dV}{dz}[/tex]

Therefore, we have to calculate the derivatives of the potential over the three variables.

Doing so, we find:

[tex]E_x=-\frac{d}{dx}(5x-3x^2y+2yz)=-(5-6xy)=6xy-5[/tex]

[tex]E_y=-\frac{d}{dy}(5x-3x^2y+2yz)=-(-3x^2+2z)=3x^2-2z[/tex]

[tex]E_z=-\frac{d}{dz}(5x-3x^2y+2yz)=-(2y)=-2y[/tex]

b)

Here we want to find the magnitude of the electric field at the point P that has coordinates

P (1.00, 0, 22.00) m

First of all, we find the components of the electric field at that point by substituting

x = 1.00

y = 0

z = 22.0

We find:

[tex]E_x=6xy-5=6(1)(0)-5=-5 N/C\\E_y=3x^2-2z=3(1)^2-2(22)=-41 N/C\\E_z=-2y=-2(0)=0[/tex]

Now, the magnitude of the electric field is given by

[tex]E=\sqrt{E_x^2+E_y^2+E_z^2}[/tex]

And by substituting,

[tex]E=\sqrt{(-5)^2+(-41)^2+0}=41.3 N/C[/tex]

Answer:

5174.4 J

Explanation:

Parameters given:

Mass of calorimeter, m = 1.4 kg = 1400 g

Specific heat capacity, c = 3.52 J/g°C

Temperature difference, ΔT = 28.5 - 27.45 = 1.05 °C

Heat absorbed by reaction, Q = m * c * ΔT

Q = 1400 * 3.52 * 1.05

Q = 5174.4 J

Answer:

5174.4Joules

Explanation:

Heat capacity is defined as the quantity of heat required to raise the temperature of total mass of a substance by 1Kelvin. Mathematically,

Q = mc∆t where;

Q is the amount of heat absorbed (in Joules)

m is the mass of the substance (bomb calorimeter) in g or kg

c is the specific heat capacity of the bomb calorimeter in J/g°C

∆t is the change in temperature in °C

Given m = 1.4kg

Since 1kg = 1000g

1.4kg = (1.4×1000)g

m = 1.4kg = 1400g

c = 3.52J/g°C

∆t = final temperature - initial temperature

Since heat was absorb (heat gained), final temperature will be 28.5°C

Initial temperature = 27.45°C

Substituting the data given into the heat capacity formula will give us;

Q = 1400×3.52×(28.5-27.45)

Q = 1400×3.52× 1.05

Q = 5174.4Joules

Amount of heat absorbed by the reaction is 5174.4Joules

Explanation:

Answer:

A customer is about to buy a limited edition sports car from Torque. It is most likely that the customer will have  d. limited problem solving, in which consumers decision rules to purchase are simple, and are not motivated to search for information about other optons, deciding to buy the car.

Explanation:

a. advertising clutter is the big amount of ad-messages that the consumer is exposed to everyday.

b. high involvement  product or purchases features many variables to be considered by the customer before getting to a decision.

c. cognitive dissonance  or conflicting attitudes causing mental discomfort.

Answer:

Dispersion

Explanation:

Chromatic aberration refers to the failure of a lenses to focus all the wavelengths or colours in white light at the same point. This is due to dispersion of light in glass. The refractive index of the lenses elements varies with different wavelengths of light. The different colours of white light will travel at different speeds through the lens hence they cannot all be focused at the same point. Dispersion results from the refraction of the various wavelengths of light to varying degrees.

Chromatic aberration in lenses is a result of the property of light, wavelength. It leads to dispersion of different colors of light due to the dependence of lens's refractive index on color or wavelength, causing different colors to have different focal points. This can be partially corrected in optical systems using multiple lenses of different materials.

Chromatic aberration in lenses is caused by the property of light known as wavelength. This concept is related to the wave nature of light, where the wavelength is the distance between successive peaks or troughs of a light wave. Chromatic aberration occurs because the index of refraction in a lens is dependent on the light's wavelength, which is related to its color. Because the index of refraction varies with wavelength, different colors (wavelengths) of light are bent by different amounts when they pass through a lens and hence have different focal points.

For example, in a single convex lens, violet light, which has a shorter wavelength, is bent more than red light, which has a longer wavelength. Therefore, violet light converges to a focal point closer to the lens compared to red light, which results in the dispersion of colors, as depicted in Figure 26.28. However, this chromatic aberration can be partially corrected by employing a two-lens system, often consisting of a converging and a diverging lens made of different materials with different dispersions, referred to as an achromatic doublet. However, completely rectifying chromatic aberration is typically not possible, which is why systems such as telescopes often prefer mirrors as they do not depend on wavelength for reflection.

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Answer:

[tex]t_t=4.131\ s[/tex]

Explanation:

Given:

height above the horizontal form where the ball is hit, [tex]y=1\ m[/tex]

angle of projectile above the horizontal, [tex]\theta=30^{\circ}[/tex]

initial speed of the projectile, [tex]u=40\ m.s^{-1}[/tex]

Firstly we find the vertical component of the initial velocity:

[tex]u_y=u.\sin\theta[/tex]

[tex]u_y=40\times \sin30^{\circ}[/tex]

[tex]u_y=20\ m.s^{-1}[/tex]

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

Using eq. of motion:

[tex]v_y^2=u_y^2-2g.h[/tex] (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

[tex]0^2=20^2-2\times 9.8\times h[/tex]

[tex]h=20.4082\ m[/tex] (from the height where it is thrown)

Now we find the time taken to ascend to this height:

[tex]v_y=u_y-g.t[/tex]

[tex]0=20-9.8t[/tex]

[tex]t=2.041\ s[/tex]

Time taken to descent the total height:

[tex]h'=u_y'.t'+\frac{1}{2} g.t'^2[/tex]

[tex]21.4082=0+4.9t'^2[/tex]

[tex]t'=2.09\ s[/tex]

Now the total time taken by the ball to hit the ground:

[tex]t_t=t'+t[/tex]

[tex]t_t=2.09+2.041[/tex]

[tex]t_t=4.131\ s[/tex]

Explanation:

Answer:

D) The element is most likely from Group 6A or 7A and in period 2 or 3.

Explanation:

Electronegativity of an atom is the tendency of an atom to attract shared paired of electron to itself. Electronegativity increase across the period from left to right.The ability of an atom to attract electron to itself is electronegativity. Group 7A and 6A elements can easily attract atoms to itself so they are highly electronegative. The most electronegative element in the periodic table is fluorine.Group 6A and 7A is likely to have high electronegativity.

Electron affinity of an atom is the amount of energy release when an atom gains electron . Generally, when atom gains electron they become negatively charged. Group 6A and 7A elements have high electron affinity.  

Ionization energy is the energy required to remove one or more electron from a neutral atom to form cations.  ionization energy of group 7A and 6A are usually high because the energy required to remove these electron is usually very high . The elements in this groups usually gain electron easily so the energy to remove electron is very high.

Answer:

The smallest separation distance between the speakers is 0.71 m.

Explanation:

Given that,

Two speakers, one directly behind the other, are each generating a 240-Hz sound wave, f = 240 Hz

Let the speed of sound is 343 m/s in air. The speed of sound is given by the formula as :

[tex]v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{343\ m/s}{240\ Hz}\\\\\lambda=1.42\ m[/tex]

To produce destructive interference at a listener standing in front of them,

[tex]d=\dfrac{\lambda}{2}\\\\d=\dfrac{1.42}{2}\\\\d=0.71\ m[/tex]

So, the smallest separation distance between the speakers is 0.71 m. Hence, this is the required solution.

According to the question,

The wavelength will be:

→ [tex]\lambda = \frac{v}{f}[/tex]

By substituting the values, we get

     [tex]= \frac{343}{240}[/tex]

     [tex]= 1.43 \ m[/tex]

hence,

The smallest separation distance will be:

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

= [tex]\frac{1.43}{2}[/tex]

= [tex]0.715 \ m[/tex]

Thus the above solution is correct.

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Answer:

The answer to the question is;

1637.769 grams of water will need to be perspired in order to maintain his original temperature.

Explanation:

Energy intake of the person = 4000 kJ

Energy required to vaporize 1 mole of water = 44.0 kJ

That is 44.0 kJ/mole

Therefore

The number of moles of water that can be vaporized by 4000 kJ is given by

(4000 kJ)/ (44.0 kJ/mole) = 90.91 moles.

Mass of one mole of water = Molar mass of water = 18.01528 g/mol

Since number of moles of water = ([tex]\frac{Mass .of. water}{Molar. mass. of. water}[/tex])

We therefore have

Mass of water = (Number of moles of water)× (Molar mass of water)

Mass of water = 90.91 moles× 18.01528 g/mol = 1637.769 g

The mass (in grams) of water that he  would need to perspire in order to maintain his original temperature is 1637.769 g.

The person would need to perspire approximately [tex]\( 1638.86 \, \text{g} \)[/tex] (or about 1.64 kg) of water to maintain their original temperature after consuming 0.50 pounds of cheese with an energy intake of 4000 kJ.

To determine the mass of water the person needs to perspire to maintain their original temperature, we need to calculate the amount of water required to dissipate the energy intake through perspiration. Given:

Energy intake: [tex]\( 4000 \, \text{kJ} \)[/tex]

Heat of vaporization of water: [tex]\( 44.0 \, \text{kJ/mol} \)[/tex]

First, we need to find out how many moles of water need to be vaporized to dissipate 4000 kJ.

[tex]\[ \text{Moles of water} = \frac{\text{Total energy intake}}{\text{Heat of vaporization per mole}} = \frac{4000 \, \text{kJ}}{44.0 \, \text{kJ/mol}} \][/tex]

[tex]\[ \text{Moles of water} = \frac{4000}{44.0} \approx 90.91 \, \text{moles} \][/tex]

Next, we convert the number of moles to mass. The molar mass of water [tex](\( \text{H}_2\text{O} \))[/tex] is approximately [tex]\( 18.015 \, \text{g/mol} \).[/tex]

[tex]\[ \text{Mass of water} = \text{Moles of water} \times \text{Molar mass of water} = 90.91 \, \text{moles} \times 18.015 \, \text{g/mol} \][/tex]

[tex]\[ \text{Mass of water} = 1638.86 \, \text{g} \][/tex]