Urban cities like Atlanta have to contend with a serious problem like pollution. Drivers in California are testing out a car that is fueled by hyrodgen; therefore, the emissions produced are water vapor. Read the article from the link below and answer the following questions:

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1. What does the author mean by the statement:
“If we’re really going to make a significant reduction in carbon emissions, you can only do that with fuel-cell vehicles in the mix"?

2. What are some of the challenges with these type of vehicles? How can these challenges be overcome?

3. Do you think fuel celled cars are a viable answer to decrease pollution why or why not?

Answer:

Option-(D): 10¹¹ waves per second.

Explanation:

The electromagnetic waves is such form of a energy transfer or wave propagation through any space with or without having any medium(particles).As, the medium or particles inside a space are able to transfer the amount of energy from the origin towards the receiver, which makes it very easy for the wave propagation through a medium.Now, the electromagnetic waves are generated from the sensor on a traffic light when the frequency,f level of the wave generation is about 10¹¹ Hertz(Hz).

Answer:

answer is A

Explanation:

Answer: If it has ions, it is an electrolyte

Explanation:

Let's start by explaining that electrolytes are compounds that contain charged particles or ions, which can be cations (positive ions) or anions (negative ions).

So, it is this composition that makes an electrolytic material conduct electricity.

In this sense, the way to identify if a material is an electrolyte or not, is knowing whether it is composed of ions or not.

Answer:

Rate of change of area is [tex]75.398ft^{2}/sec[/tex]

Explanation:

[tex]Area=\pi r^{2}\\\\\frac{d(Area)}{dt}=\frac{\pi r^{2}}{dt}=2\pi r\frac{dr}{dt}[/tex]

Applying values we get [tex]\frac{d(Area)}{dt}=2\pi 4\times 3=75.398ft^{2}/sec[/tex]

Answer:

Boat speed = 8 miles/hr

Explanation:

Let the speed of the boat be U

Let the speed of the current be V.

Therefore, the downstream speed is U+V

and the upstream speed is U-V

Now we know that Speed = distance / time

Therefore, downstream speed, U+V = 63 / 7

                                                    U+V = 9 miles/hr   ------(1)

                  Upstream speed, U-V = 63 / 9

                                                U-V = 7 miles/hr      --------(2)

Therefore subtracting (2) from (1), we get

( U+V) - ( U-V ) = 9-7

2V = 2

V = 1 miles/hr

Therefore the speed of the current is V = 1 mile/hr

Now from (1) we get

U+V = 9

U+1 = 9

U = 8

Therefore, the speed of the boat is U = 8 miles/hr

Answer: 3 seconds

Explanation:

Since they provide the same impulse,

Impulse= Ft

F1 = 6N

t1 = 2s

F2= 4N

t2= ?

F1= Force of first engine

t1= time elapsed by first engine

F2= force of second engine

t2= time elapsed by second engine

F1t1 = F2t2

6 × 2 = 4t2

t2= 12/4

t2 = 3 seconds

This second engine fires for 3 s

[tex]\texttt{ }[/tex]

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

thrust of first engine = F₁ = 6 N

elapsed time of first engine = t₁ = 2 s

thrust of second engine = F₂ = 4 N

Asked:

elapsed time of second engine = t₂ = 2 s

Solution:

We will use Newton's Law of Motion to solve this problem as follows:

[tex]\Sigma F = ma[/tex]

[tex]\Sigma F = m \Delta v \div t[/tex]

[tex]\Sigma F = I \div t[/tex]

[tex]\boxed {I = \Sigma F \times t}[/tex] → Impulse Formula

[tex]\texttt{ }[/tex]

Two rocket engines provide the same impulse :

[tex]I_1 = I_2[/tex]

[tex]\Sigma F_1 \times t_1 = \Sigma F_2 \times t_2[/tex]

[tex]6 \times 2 = 4 \times t_2[/tex]

[tex]12 = 4 \times t_2[/tex]

[tex]t_2 = 12 \div 4[/tex]

[tex]\boxed {t_2 = 3 \texttt{ s}}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

Answer:

Ffs = 251 N

Fn = 691 N

Explanation:

Take the y direction to be normal to the ramp and the x direction to be parallel to the ramp.

The angle of the ramp is 20°, so the angle that the weight vector makes with the normal is also 20°.  Therefore:

Fgx = Fg sin 20°

Fgy = Fg cos 20°

Sum of the forces in the x direction (parallel to the ramp):

∑F = ma

Ffs − Fgx = 0

Ffs = Fgx

Ffs = Fg sin 20°

Ffs = 735 sin 20°

Ffs ≈ 251

Sum of the forces in the y direction (normal to the ramp):

∑F = ma

Fn − Fgy = 0

Fn = Fgy

Fn = Fg cos 20°

Fn = 735 cos 20°

Fn ≈ 691

To answer that question we need to apply equations of movement ( from Newton´s laws )

In equilibrium:

∑F  = 0         or    ∑Fx = 0     ;  ∑Fy = 0

Solution is:

a) F(sf) = 251 [N]

b) Fn = 691 [N]

From the attached drawings we can see:  ( Body free diagram)

∑ Fₓ  = F(sf)  - Pₓ  = 0           where P = m×g  = 735 [N] ( the weigth)

and Pₓ = P× cos20°

Then     F(sf)  = Pₓ × sin 20°

F(sf) = m×g×cos20°  =  735× 0.34202 [N]

F(sf) = 251.3847

F(sf) = 251 [N]

rounding to the nearest number

F(sf) = 691 [N]

∑ Fy = 0

∑ Fy = Fn - Py  = 0                     Py = P×cos20°      Py = m×g×cos20°

Py = 735×0.939693 [N]     Py =   [N]

Fn = Py = 690.674 [N]

rounding to the nearest whole number

Fn = 691 [N]

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Answer:given below

Explanation:

Cart is pulled 14 cm from mean position

and its position is given by

[tex]x=14cos\left ( \pi t\right )[/tex]

therefore its velocity is acceleration is given by

[tex]v=-14\pi sin\left ( \pi t\right )[/tex]

[tex]a=-14\pi ^2cos\left ( \pi t\right )[/tex]

[tex]\left ( a\right ) cart\ max.\ speed\ is[/tex]

[tex]v_{max}=14\pi at\ t=0.5sec[/tex]

and its position is x=0

acceleration at t=0.5sec

a=0

[tex]\left ( b\right )[/tex]

[tex]a_{max}=14\pi ^2 cm/s^2[/tex]

at t=0,1,2 sec

at t=0

x=14 cm

v at t=0

v=0 cm/s

for t=1 sec

x=-14 cm i.e. 14 cm behind mean position

v=0 m/s

The cart's maximum speed is 14π m/s and occurs at the equilibrium point in its oscillation at integer multiples of the half period. At these moments, the magnitude of the acceleration is 14π² m/s². The magnitude of the acceleration is greatest when the cart is at the extreme points of its oscillation, where its speed is zero.

The cart, undergoing simple harmonic motion, has a maximum speed when it passes through equilibrium - the midpoint of its oscillating path. From the equation for the straightforward harmonic motion, we know that the speed v of the cart is given by the derivative of s(t) = 14cos(πt). The derivative of this function is v(t) = -14πsin(πt), and the maximum speed occurs when sin(πt) = ±1, which gives a maximum speed of ±14π m/s.

The maximum speed is attained at integer multiples of the half period (1/2, 3/2, 5/2 seconds, and so on). At this moment, the cart is at the equilibrium position, s=0. Acceleration a(t) is given by the second derivative of the position function s(t), which is a(t) = -14π²cos(πt). The magnitude of the acceleration at the time of maximum speed is 14π² m/s².

In the case of the maximum magnitude acceleration, this is attained at the turning points of the oscillation when cos(πt) = ±1. At these moments, the speed of the cart is zero.

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Applying the work-energy theorem, which states that the work done by nonconservative forces equates to the change in kinetic plus potential energy, the skateboarder's change in potential energy results in -18.4J, and the absolute change in height becomes 0.032 m.

The student is asking about the concept of conservation of energy in physics, particularly in the context of the work-energy theorem. This theorem states that work done by nonconservative forces (like friction) is equal to the change in the kinetic and potential energy of the system. According to the given problem, the initial kinetic energy of a 55.6-kg skateboarder is KE0 = 0.5 * m * v², and after performing 80.4J of work on himself, while friction does -244J of work on him, the final kinetic energy becomes KEf = m * v² / 2 where 'v' is the final speed of 7.24 m/s.

(a) Considering that there are no conservative forces, the work-energy theorem states that Wnc = ΔKE + ΔPE. From here, we can calculate that ΔPE or PEf - PE0 results in -18.4J.

(b)The change in vertical height can be obtained by Δh = ΔPE / (m*g), where 'g' is the gravitational constant. Thus, Δh =-18.4J divided by (55.6 * 9.8), which results in -0.032 m, but since we are asked for the absolute value, the height change is 0.032 m.

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Answer:

F = 32.2m

Explanation:

The force of gravity on an object is given by:

[tex]F=mg[/tex]

where

m is the mass of the object

g is the acceleration due to gravity

Here we have:

- An object of mass m

- The acceleration of gravity is expressed as [tex]g=32.2 ft/s^2[/tex]

Therefore, substituting into the formula above, we find that the force of gravity on the object is

[tex]F=m\cdot 32.2 = 32.2m[/tex]

Answer:

F = 32.2m

Explanation:

Answer:

Speed of A = 891 km/h

Speed of B = 804 km/h

Explanation:

Let the speed of aeroplane B is v, the speed of aeroplane is A is 87 km/h faster than B.

So, the speed of aeroplane A is 87 + v.

Distance traveled by A after 7 hours, d1 = (87 + v) x 7

Distance traveled by B after 7 hours, d2 = v x 7

Total distance traveled = 11865 km

So, d1 + d2 = 11865

(87 + v) x 7 + v x 7 = 11865

609 + 14 v = 11865

14 v = 11256

v = 804 km/h

So, the speed of A = 87 + 804 = 891 km/h

Speed of B = 804 km/h

Answer:

   V₂  =  225 m^{3}

so no exact option is match with the calculated answer.

Explanation:

According to Boyle's Law, we have

                                     P_1 V_1  =  P_2 V_2    ----------- (1)

Data Given;

                  P_1  =  1.0 atm

                  V_1  =  180 m³

                  P_2  =  0.80 atm

                  V_2  =  ?

Solving equation 1 for V₂,

                  [tex]V_2= \frac{P_1 V_1}{P_1}[/tex]

Putting values,

                   [tex]V_2 = \frac{1.0*180}{0.80}[/tex]

                  [tex]V_2 = 225 m^{3}[/tex]

so no exact option is match with the calculated answer.

Answer:

E

Explanation:

Answer:

C. Interference from the sun causes data to be collected inaccurately.

Explanation:

Snow predictions by meteorologists are sometimes incorrect because from the sun causes data to be collected inaccurately.

Explanation:

The equation of motion of an object is given by :

[tex]h(t)=-16t^2+112t+128[/tex]

Where

t is the time in seconds

We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0

So,

[tex]-16t^2+112t+128=0[/tex]

[tex]-t^2+7t+8=0[/tex]

On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.

Answer:

Part a)

[tex]EMF = 5.6 \times 10^{-3} V[/tex]

Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

Explanation:

As we know that magnetic flux linked with the coil is given as

[tex]\phi = \pi r^2 B[/tex]

now the rate of change in flux is given as

[tex]\frac{d\phi}{dt} = 2\pi r \frac{dr}{dt} B[/tex]

now we know that circumference is decreasing at rate of 15 cm/s

so here we know the length of circumference as

[tex]C = 2\pi r[/tex]

So rate of change in circumference is

[tex]\frac{dC}{dt} = 2\pi \frac{dr}{dt}[/tex]

[tex]\frac{1}{2\pi}(15 cm) = \frac{dr}{dt}[/tex]

final length of circumference at t = 8 s

[tex]C = 167 - (15)(8) = 47[/tex]

Part a)

Now the induced EMF is given as

[tex]EMF = (2\pi r)(\frac{1}{2\pi})(0.15)(0.5)[/tex]

[tex]EMF = (0.47)(\frac{1}{2\pi})(0.15)(0.5)[/tex]

[tex]EMF = 5.6 \times 10^{-3} V[/tex]

Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

The problem is solved using Faraday's Law of Electromagnetic Induction: the induced emf equals to the change in magnetic flux divided by change in time. Then, Lenz's law is used to determine the direction of the induced current.

The problem you're dealing with is an application of Faraday's Law of Electromagnetic Induction. This law states that the electromotive force (or emf) induced in a circuit is equivalent to the rate of change of magnetic flux through that circuit.

For Part A, you know that the rate of change of the circumference is constant and you can convert that to a rate of change of radius (since circumference = 2πr). Using these formulas, you can determine the rate of change of area as πr(dr/dt), where dr/dt is the rate of change of radius. The induced emf is proportional to the rate of change of magnetic flux, which is the product of magnetic field strength and area. Thus, induced emf equals to (change in flux/change in time), or -Bπr(dr/dt).

For part B, you apply Lenz's law, which states that the direction of induced current is such that the magnetic field due to it opposes the change in the initial magnetic field. Since the field is getting smaller (since the loop is contracting), then the current will act in such a way as to try to keep the size of the field the same. This means going in the counterclockwise direction.

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Answer:

Required mass of sand is 20 kg

Explanation:

Given:

Mass of the plank = 25 kg

Distance of the Center of gravity of the Plank from the fulcrum = [tex]\frac{2}{2}-0.50 = 0.5m[/tex]

Distance of the Center of gravity of the sand box from the fulcrum = [tex]\frac{2}{2}-\frac{0.75}{2}= 0.625m[/tex]

Balancing the torque due to the plank and the sand box with respect to the fulcrum

Torque = Force × perpendicular distance

thus, we get

(25 × g) × 0.5 = weight of sand × 0.625

where, g is the acceleration due to gravity

or

(25 × g) × 0.5 = (mass of sand × g) × 0.625

or

mass of sand = 20 kg

Hence, the required mass of the sand is 20 kg

Answer:

20 Kg mass of sand should be put into the box so that the plank balances horizontally on a fulcrum placed horizontally on a fulcrum placed just below its midpoint.

Explanation:

Use the second condition of equilibrium.

[tex]$\sum} \tau=0$[/tex]

[tex]MgL-$M g x_{c m}=0$[/tex]

[tex]$M=\frac{m x_{c m}}{L}[/tex]

[tex]=\frac{25(0.50)}{0.625}[/tex]

[tex]=20 \mathrm{~kg}$[/tex]

Learn more about mass, refer:

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Answer:

50°

Explanation:

It is given that the both scales are the same at 18˚

Now,

at 30˚ on the Cantor scale, the temperature is 30˚ − 18˚ = 12˚ above

the 18˚ mark.

also, it is given that with every 3˚ increase in the temperature on the Cantor scale is there is an 8˚ increase on the Frobenius scale,

mathematically, we can write it as (by unitary method)

3˚ increase in the temperature on the Cantor =  8˚ increase on the Frobenius scale

or

1˚ increase in the temperature on the Cantor =  (8/3)˚ increase on the Frobenius scale

thus, for x˚ increase in the temperature on the Cantor =  ((8/3)˚ × x) increase on the Frobenius scale

hence, for 12° increase we have

12˚ increase in the temperature on the Cantor =  ((8/3)˚ × 12) increase on the Frobenius scale

or

12˚ increase in the temperature on the Cantor =  32° increase on the Frobenius scale

hence, the final reading on the Frobenius scale will be, 18˚ + 32˚ = 50˚.

Answer:

intensity.

Explanation:

when the light collected by the lens is focused into a small spot it tends to increase the intensity of the light.

as different path of light with different intensity combines from passing through the lens it tends to make the light path and intensity coherent and after being coherent there intensity increases.

Answer:

[tex]\lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]

Explanation:

the velocity as a function of time is

[tex]v(t)=\frac{mg}{c}(1-e^{\frac{-ct}{m}})[/tex]

[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})[/tex]

[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-\frac{1}{\infty })\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-0)\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]

The limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance considered, is the terminal velocity mg/c. The exponential term in the equation approaches zero as time becomes very large, leading to the object reaching a constant terminal velocity.

To calculate the limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance taken into account, we use the provided equation v = mg/c (1 − e^{-ct/m}), where g is the acceleration due to gravity, which averages 9.80 m/s², and c is a positive constant representing air resistance. The limit represents the object's terminal velocity, which is the constant speed an object reaches when the force of gravity is balanced by the drag force of air resistance.

As t → ∞ (increases towards infinity), the exponential term e^{-ct/m} approaches zero. Hence, the limit of the speed v becomes:

∑ lim t→∞ v = lim t→∞ mg/c (1 − e^{-ct/m}) = mg/c.

The value mg/c is known as the terminal velocity, which is the maximum speed that the object will reach as it continues to fall.

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Answer:

Part a)

2)  >  1)  >  3)

Part b)

1)  =  2)  =  3)

Explanation:

Due to charge induction the magnitude of charge on the inner surface of the outer shell is having same charge as that of the small sphere inside but the sign of charge must be opposite.

So here we can say

1)+ 4q, 0

so inner surface has charge - 4q and outer surface charge is +4q

2) -6q , +10q

so inner surface charge is +6q, outer surface charge is +4q

3) +16q , -12q

so inner surface charge is -16q, outer surface charge is +4q

Part a)

situations in which inner surface charge is arranged in decreasing order is given as

2)  >  1)  >  3)

Part b)

Situations in which outer surface charge is arranged in decreasing order is given as

1)  =  2)  =  3)

Situation (1) has the most positive charge on the inner surface of the shell, while situation (2) has the most positive charge on the outer surface.

To rank the situations according to their charge on the inner and outer surfaces of the metallic spherical shell, we need to consider the net charges on the small charged ball and the shell. Let's analyze each situation:

Therefore, ranking the situations according to the charge on the inner surface would be: (1), (3), (2) - from most positive to least positive. For the outer surface, the ranking would be: (2), (1), (3) - from most positive to least positive.

Answer:

Kinetic Energy

Explanation:

Heat energy is another name for thermal energy. Kinetic energy is the energy of a moving object. As thermal energy comes from moving particles, it is a form of kinetic energy.

Answer:

a) Increase by a factor of more than 2

Explanation:

The compression process is an adiabatic one. The opposite of this kind of process is the isothermal process, where the heat flow makes the temperature to be constant.

Let's consider this kind of process by means of an equation of state; for example, the ideal gas law:

[tex]P=\frac{RT}{v}[/tex]

Be [tex]v_{0}[/tex] the initial volume. And [tex]v_{f}=2v_{0}[/tex] the final volume.

If we consider the ideal gas law, it is evident that if the temperature remains constant (isothermal process), the pressure increases by a factor of 2; but in an adiabatic process the temperature of a gas tends to increase its temperature, so the pressure will be a higher than the resultant for the isothermal process.

Answer: 768000km

Explanation:

Velocity is given by the relation between the distance [tex]d[/tex] and the time it takes to travel that distance [tex]t[/tex]:

[tex]V=\frac{d}{t}[/tex]   (1)

In this problem we are told the time it takes for radio wave to travel from the Earth to the Moon and back is the "echo":

[tex]t=2.56s[/tex]  (2)

In addition, we know radio waves are electromagnetic waves (light), and its velocity is:

[tex]V=3(10)^{8}m/s[/tex]   (3)

Substituting (2) and (3) in (1):

[tex]3(10)^{8}m/s=\frac{d}{2.56s}[/tex]   (4)

And finding [tex]d[/tex]:

[tex]d=(3(10)^{8}m/s)(2.56s)[/tex]   (5)

Finally we can obtain the distance:

[tex]d=768000000m=768000km[/tex]  

Answer:

384,000 km

Explanation:

Given

Velocity of radio wave [tex]v = 3.00 \times 10^{8}m/s[/tex]

Duration of echo T = 2.56 s

Solution

Time taken for the radio wave to travel to moon and to travel back to earth as it was picked up by the astronaut's microphone is 2.56 s

Since any time delays in the electronic equipment can be ignored

time taken for the radio wave to reach moon

[tex]t = \frac{T}{2}\\\\t = \frac{2.56}{2} \\\\t = 1.28 s[/tex]

[tex]v = \frac{d}{t}\\\\d = vt\\\\d = 3 \times 10^8 \times 1.28\\\\d = 3.84 \times 10^8 m\\\\d = 384,000 km[/tex]

Answer:

Density of unknown liquid is [tex]2047 kg/m^3[/tex]

Explanation:

When rock is suspended in air then the weight of the rock is counter balanced by the tension force in the string

So here we have

[tex]T = mg = 43.8 N[/tex]

now when the rock is immersed in water then the tension in the string is and buoyancy force due to water is counter balanced by the weight of the object

so here we have

[tex]T_1 + F_b = mg[/tex]

[tex]31.3 + F_b = 43.8[/tex]

[tex]F_b = 43.8 - 31.3 = 12.5 N[/tex]

now we have

[tex](1000)V(9.81) = 12.5[/tex]

[tex]V = 1.27 \times 10^{-3} m^3[/tex]

now when the rock is immersed into other liquid then we have

[tex]T_2 + F_b' = mg[/tex]

[tex]18.3 + F_b' = 43.8[/tex]

[tex]F_b' = 25.5 N[/tex]

now we have

[tex]\rho(1.27 \times 10^{-3})(9.81) = 25.5[/tex]

[tex]\rho = 2047 kg/m^3[/tex]

Answer: .2) The final temperature will be exactly midway between the initial temperatures of substances A and B.

Explanation:

[tex]heat_{absorbed}=heat_{released}[/tex]

As we know that,  

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_A\times c_A\times (T_{final}-T_A)=-[m_B\times c_B\times (T_{final}-T_2)][/tex]         .................(1)

where,  

q = heat absorbed or released

[tex]m_A[/tex] = mass of A = 2x

[tex]m_B[/tex] = mass of B = x

[tex]T_{final}[/tex] = final temperature = z

[tex]T_A[/tex] = temperature of A

[tex]T_2[/tex] = temperature of B

[tex]c_A[/tex] = specific heat capacity of A = y

[tex]c_B[/tex] = specific heat capacity of B = 2y

Now put all the given values in equation (1), we get

[tex]2x\times y\times (z-T_A)=-[x\times 2y\times (z-T_B)][/tex]

[tex]2z=T_B+T_A[/tex]

[tex]z=\frac{T_B+T_A}{2}[/tex]

Therefore, the final temperature of the mixture will be exactly midway between the initial temperatures of substances A and B.

When two substances with different temperatures come into contact and reach thermal equilibrium, heat flows until they reach the same final temperature. In this case, the final temperature will be closer to the initial temperature of substance B because substance B requires more heat to raise its temperature and has a smaller mass. The specific heat capacity ratio determines the proportion of heat absorbed or released by the substances.

When two substances at different temperatures come into contact with each other and reach thermal equilibrium, heat flows from the hotter substance to the cooler substance until they reach the same final temperature. In this case, substance A has twice the mass of substance B but substance B has twice the specific heat capacity of substance A. The specific heat capacity determines how much heat is needed to raise the temperature of a substance. Since substance B requires more heat to raise its temperature and has a smaller mass, it will be able to absorb more heat from substance A than substance A can absorb from substance B.

This means that the final temperature will be closer to the initial temperature of substance B than substance A. The specific heat capacity ratio determines the proportion of heat absorbed or released by the substances.

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Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

[tex]E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2[/tex]

[tex]E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2[/tex]

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

Answer:

B. False

Explanation:

Acceleration is the slope of a velocity vs. time graph.

Displacement is the area under a velocity vs. time graph.

Answer:

1.034 m above the floor

Explanation:

The location of center of body for a compound body, when the weights are given is calculated as:

[tex]\bar x = \frac{W_1x_1+W_2x_2+W_3x_3+W_4x_4+W_5x_5+.......+Wnx_n}{W_1+W_2W_3W_4W_5+.....+W_n}[/tex]

where,

[tex]\bar x[/tex] is the center of gravity of the entire body

W = weight of the individual body

x = center of gravity of the individual body

Thus on substituting the values we get,

[tex]\bar x = \frac{438\times 1.28+144\times 0.760+87\times 0.250}{438+144+87}[/tex]

or

[tex]\bar x = \frac{691.83}{669}[/tex]

or

[tex]\bar x =1.034m[/tex]

Hence, the center of gravity of the entire body lies 1.034 m above the floor

Answer:

It is called single transformation

Answer:

Single transformation

Explanation:

Energy can neither be created nor be destroyed, it can only covert from one form to another. Sometimes, to get a work done one form of energy only require to be transformed into another form then it is known as single transformation.

For Example, cell phone transforming electrical energy into electromagnetic energy that makes it's working feasible.

Moving vehicle converting chemical energy to kinetic energy.

Based on the provided information, the correct statement is:

b. The bottom of your front bumper must not be more than 28 inches above the pavement.

The modification to increase ground clearance in the pick-up truck involves specific regulations for the front bumper height.

The statement "The bottom of your front bumper must not be more than 28 inches above the pavement" indicates a legal restriction to maintain a certain elevation for safety and compliance reasons.

This limitation aims to prevent potential hazards, such as underride collisions, and ensures that the modified truck adheres to regulatory standards. By specifying the maximum height of the front bumper, authorities aim to strike a balance between vehicle customization and road safety.

This regulation emphasizes the importance of maintaining a reasonable height for front bumpers to mitigate risks and uphold public safety standards, reflecting the broader considerations in vehicular modifications within the legal framework.

The probable question may be:

You have a pick-up truck that weighed 4,000 pounds when it was new. You are modifying it to increase its ground clearance. When you are finished

a. The bottom of your front bumper can be up to 32 inches above the pavement.

b. The bottom of your front bumper must not be more than 28 inches above the pavement.

c. You will no longer be required to have a rear bumper.

Answer:

i = 7.777 × 10⁻⁴ A = 0.77 mA

Explanation:

Given:

loop radius, r = 3.0 cm = 0.03 m

Area, A = π x r² = π x 0.03² = 0.0028 m²

Magnetic Field, B = 0.75 T

Loop resistance, R = 18 Ω

time, t = 0.15 seconds

Now,

the induced emf is given as:

EMF = [tex]-\frac{BA}{t}[/tex]

also

EMF = i x R

Where, i is the current flowing

equating both the formulas for EMF, we get

[tex]{i}{R}=-\frac{BA}{t}[/tex]

or

[tex]{i}=-\frac{BA}{tR}[/tex]

substituting the values in the above equation we get

[tex]{i}=-\frac{0.75\times 0.0028}{0.15\times 18}[/tex]

or

the magnitude of the current, i = 7.777 × 10⁻⁴ A = 0.77 mA

The electric current produced in a loop due to a changing magnetic field can be calculated using Faraday's law of electromagnetic induction. Based on the given data, the magnitude of the induced current in the loop is approximately 1.18 mA.

The scenario described in the question depicts a situation where a magnetic field decreases uniformly in magnitude, thus inducing an electric current within a circular loop according to Faraday's law of electromagnetic induction.

The electric current produced in the loop as a result of the change in the magnetic field can be calculated using the expression, I = ΔΦ/ Δt * R, where ΔΦ is the change in magnetic flux, Δt is the time interval, and R is the resistance of the loop.

In this case, the initial magnetic flux, Φ1 = B1 * A, where B1 is the initial magnetic field and A is the area of the loop. Since the magnetic field decreases to zero, the final magnetic flux, Φ2, is zero. So, ΔΦ = Φ2 - Φ1 = - B1 * A. Here, A = πr², where r is the radius of the loop.

So, I = - [ (B1* πr²) / ( Δt * R ) ]. Substituting the given values into the equation, we get I = (0.75 T * π * (0.03 m)²) / ( 0.15 s * 18 Ω ) = 0.00118 A or 1.18 mA.

So, the magnitude of the induced current is approximately 1.18 mA.

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