A car with a total mass of 1800 kg (including passengers) is driving down a washboard road with bumps spaced 4.9 m apart. The ride is roughest—that is, the car bounces up and down with the maximum amplitude—when the car is traveling at 5.7 m/s. What is the spring constant of the car's springs? Express your answer to two significant figures and include the appropriate units.

Io experiences tidal heating primarily because of the intense gravitational pull of Jupiter which creates friction inside Io and ultimately results in heating. This process is enhanced by Io's elliptical orbit.

Io, one of Jupiter's moons, experiences tidal heating primarily due to the intense gravitational pull of Jupiter. These gravitational forces create friction within Io, resulting in heat. This process is also facilitated by Io's elliptical orbit which varies the gravitational forces acting on it during its orbit around Jupiter, thereby enhancing the frictional heating. Therefore, the main contributor to Io's tidal heating is its interaction with Jupiter's gravitational field.

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Answer:

1. True

2. False

3. True

Explanation:

Semiconductors are the materials whose conduction is intermediary to that of the conductor and insulator and can be made more conducting by mixing impurities or we call it as doping.

In semiconductors, the concentration of the holes in the valence band and the electrons in the conduction band is usually very small and can be varied noticeably by doping.

(1) After being excited by a photon, an electron can move to the conduction band as a hole is created in the valence band band and this hole moves to the p-side of the p-n junction.

(2) It is not the n-type but the p-type semiconductor with missing electrons or holes in the valence band of lower energy.

(3) Electrons are free to move in a partially filled conduction band and valence band.

There is no movement of electrons along the band which are completely full or totally vacant.

Answer:

[tex]v_r=13.68\ m.s^{-1}[/tex] is the final velocity of the racket.

[tex]F=18.86\ N[/tex]

Explanation:

Given:

By the law of conservation of momentum:

[tex]m_r.u_r+m_b.u_b=m_r.v_r+m_b.v_b[/tex]

where: [tex]v_r=[/tex] final velocity of the racket

[tex]1000\times 15+60\times 18=1000\times v_r+60\times 40[/tex]

[tex]v_r=13.68\ m.s^{-1}[/tex] is the final velocity of the racket.

By the Newton's second law of motion:

[tex]F=\frac{dp}{dt}[/tex] ............................(1)

where:

dp = change in momentum

dt = change in time

Change in momentum of ball:

[tex]\Delta p_b=m_b.v_b-m_b.u_b[/tex]

[tex]\Delta p_b=60\times 10^{-3}\times (40-18)[/tex]

[tex]\Delta p_b=1.32\ kg.m.s^{-1}[/tex]

Now, using eq.(1):

[tex]F=\frac{1.32}{0.07}[/tex]

[tex]F=18.86\ N[/tex]

The tennis racket will be moving at 11.52 m/s immediately after the impact. The average force exerted on the ball by the racket, during the brief 7 ms contact time, is approximately 497 N.

The problem can be solved using the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision if no external forces act on the system. The formula for momentum is p = mv, where p is momentum, m is mass, and v is velocity.

The momentum before the collision (pi) for both the racket and the ball is:

pi,racket = mracket \\times vi,racket = 1.0 kg \\times 15.0 m/s = 15 kg\\cdot m/s

pi,ball = mball \\times vi,ball = 0.060 kg \\times (-18.0 m/s) = -1.08 kg\\cdot m/s

The negative sign indicats that the ball is moving in the opposite direction to the racket. Total initial momentum Pi,total = pi,racket + pi,ball.

The momentum after the collision (pf) for the ball is:

pf,ball = mball \\times vf,ball = 0.060 kg \\times 40.0 m/s = 2.4 kg\\cdot m/s

Using the conservation of momentum, Pi,total = Pf,total, i.e.,

15 kg\\cdot m/s - 1.08 kg\\cdot m/s = pf,racket + 2.4 kg\\cdot m/s

Solving for pf,racket, the momentum of the racket after collision:

pf,racket = 15 - 1.08 - 2.4 = 11.52 kg\\cdot m/s

The velocity of the racket after the collision (vf,racket) can be found by dividing pf,racket by the mass of the racket:

vf,racket = pf,racket / mracket = 11.52 kg\\cdot m/s / 1.0 kg = 11.52 m/s

To calculate the average force exerted on the ball by the racket, we can use the change in momentum (impulse) and divide it by the time of contact. The change in momentum of the ball (\\Delta p) is the momentum after minus the momentum before:

\\Delta p = pf,ball - pi,ball = 2.4 kg\\cdot m/s - (-1.08 kg\\cdot m/s) = 3.48 kg\\cdot m/s

The average force (F) exerted on the ball is:

F = \\Delta p / \\Delta t = 3.48 kg\\cdot m/s / 0.007 s = 497.14 N

The racket will be moving at 11.52 m/s immediately after the impact, and the average force that the racket exerts on the ball is approximately 497 N.

The angular speed of the merry-go-round after the child gets on is approximately 1.28 rev/s.

To find the angular speed of the merry-go-round after the child gets onto it, we can use the principle of conservation of angular momentum. The total angular momentum before the child gets on should be equal to the total angular momentum after.

The angular momentum (L) of the merry-go-round before the child gets on is given by:

L_initial = I_initial * ω_initial

Where:

I_initial = Moment of inertia of the merry-go-round

ω_initial = Initial angular speed

The angular momentum (L) of the child after getting on is given by:

L_child = I_child * ω_final

Where:

I_child = Moment of inertia of the child

ω_final = Final angular speed

Since angular momentum is conserved, L_initial = L_child.

The moment of inertia (I) of a solid disk like a merry-go-round can be calculated using the formula:

I = (1/2) * m * r^2

Where:

m = Mass

r = Radius

For the merry-go-round:

I_initial = (1/2) * 120 kg * (1.80 m)^2

I_child = (1/2) * 37.5 kg * (1.80 m)^2

Now, we can set up the equation for conservation of angular momentum:

I_initial * ω_initial = I_child * ω_final

Solve for ω_final:

(1/2) * 120 kg * (1.80 m)^2 * ω_initial = (1/2) * 37.5 kg * (1.80 m)^2 * ω_final

Now, we can cancel out some terms:

120 * ω_initial = 37.5 * ω_final

Now, solve for ω_final:

ω_final = (120 * ω_initial) / 37.5

Substitute the given value for ω_initial (0.400 rev/s):

ω_final = (120 * 0.400 rev/s) / 37.5

ω_final ≈ 1.28 rev/s

So, the angular speed of the merry-go-round after the child gets on is approximately 1.28 rev/s.

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The centripetal acceleration is [tex]0.63 m/s^2[/tex]

Explanation:

The centripetal acceleration for an object in circular motion is given by

[tex]a=\frac{v^2}{r}[/tex]

where

v is the linear speed

r is the radius of the circle

The body in the problem cover half of revolution in

t = 10 s

And the corresponding linear distance covered is

L = 10 m

which corresponds to half of the circumference, so

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

From this equation we find the radius of the circle:

[tex]r=\frac{L}{2\pi}=\frac{10}{2\pi}=1.59 m[/tex] [m]

While the linear speed is:

[tex]v=\frac{L}{t}=\frac{10}{10}=1 m/s[/tex]

Therefore, the centripetal acceleration is

[tex]a=\frac{1^2}{1.59}=0.63 m/s^2[/tex]

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Answers:

a) [tex]65.075 kgm/s[/tex]

b) [tex]10.526 s[/tex]

c) [tex]61.82 N[/tex]

Explanation:

According to the Impulse-Momentum theorem we have the following:

[tex]I=\Delta p=p_{2}-p_{1}[/tex] (1)

Where:

[tex]I[/tex] is the impulse

[tex]\Delta p[/tex] is the change in momentum

[tex]p_{2}=mV_{2}[/tex] is the final momentum of the ball with mass [tex]m=0.685 kg[/tex] and final velocity (to the right) [tex]V_{2}=57 m/s[/tex]

[tex]p_{1}=mV_{1}[/tex] is the initial momentum of the ball with initial velocity (to the left) [tex]V_{1}=-38 m/s[/tex]

So:

[tex]I=\Delta p=mV_{2}-mV_{1}[/tex] (2)

[tex]I=\Delta p=m(V_{2}-V_{1})[/tex] (3)

[tex]I=\Delta p=0.685 kg (57 m/s-(-38 m/s))[/tex] (4)

[tex]I=\Delta p=65.075 kg m/s[/tex] (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately [tex]1.0 cm=0.01 m[/tex]:

[tex]V_{2}=V_{1}+at[/tex] (6)

[tex]V_{2}^{2}=V_{1}^{2}+2ad[/tex] (7)

Where:

[tex]a[/tex] is the acceleration

[tex]d=0.01 m[/tex] is the length the ball was compressed

[tex]t[/tex] is the time

Finding [tex]a[/tex] from (7):

[tex]a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}[/tex] (8)

[tex]a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}[/tex] (9)

[tex]a=90.25 m/s^{2}[/tex] (10)

Substituting (10) in (6):

[tex]57 m/s=-38 m/s+(90.25 m/s^{2})t[/tex] (11)

Finding [tex]t[/tex]:

[tex]t=1.052 s[/tex] (12)

According to Newton's second law of motion, the force [tex]F[/tex] is proportional to the variation of momentum  [tex]\Delta p[/tex] in time  [tex]\Delta t[/tex]:

[tex]F=\frac{\Delta p}{\Delta t}[/tex] (13)

[tex]F=\frac{65.075 kgm/s}{1.052 s}[/tex] (14)

Finally:

[tex]F=61.82 N[/tex]

Answers:

a) 65.125 Ns

b) 5.263 * 10^(-4) s

c) 123737.5 N

Explanation:

a) Impulse delivered to the ball F.dt

According to the Impulse-Momentum we have the following:

[tex]F*dt = m*(V_{2} - V_{1})[/tex]

Using the given data we insert in equation above:

[tex]Impulse = 0.685 kg (57 - (-38))\\\\Impulse = 65.1225 Ns[/tex]

Answer: 65.125 Ns

b)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 0.01m :

Using the kinematic equations for constant acceleration:

[tex](v_{f})^2 = (v_{i})^2 + 2*a*s[/tex]

Where:

vf = 0 (ball stops on the bat)

vi = 38 m/s

s = compression = 0.01 m

Using the equation above we compute acceleration:

[tex]a = \frac{(v_{f})^2 - (v_{i})^2}{2*s} \\\\a = \frac{0^2 - 38^2}{2*0.01} \\\\a = -72,200 m/s^2[/tex]

Using the acceleration to compute time:

[tex]v_{f} = v_{i} + a*t\\\\t = \frac{v_{f} - v_{i}}{a}\\\\t = \frac{0 - 38}{-72,200}\\\\t = 5.263*10^(-4) s[/tex]

Answer: 5.263*10^(-4) s

c)

According to Newton's second law of motion:

[tex]F_{avg} * dt = Impulse[/tex]

Using answer from part a and b

[tex]F_{avg} = \frac{Impulse}{dt} \\\\F_{avg} = \frac{65.125}{5.263*10^(-4)} \\\\F_{avg} = 123737.5 N[/tex]

Answer: 123737.5 N

Answer:

Three times

Explanation:

The amount of heat added to a substance when its temperature is increased from T_1 to T_2 is given by

[tex]Q=mc(T_2-T_1)[/tex]

c= specific heat capacity

m= mass of the substance

⇒Q∝c

that is if c is increased three times the amount of heat added is also increased three times. Therefore, the amount of heat added to the copper is three times the heat added to lead.

The given parameters;

The potential energy of the hydrogen atoms is calculated as follows;

[tex]V = \frac{kq}{r}[/tex]

where;

The potential energy when the distance = 0.74 x 10⁻¹⁸ m

[tex]V = \frac{9\times 10^{9} \times 1.602\times 10^{-19}}{0.74\times 10^{-18}} = 1.95\times 10^9 \ N.m[/tex]

The potential energy when the distance = 0.5 x 10⁻¹⁸ m

[tex]V = \frac{9\times 10^{9} \times 1.602\times 10^{-19}}{0.5\times 10^{-18}} = 2.88 \times 10^9 \ N.m[/tex]

Thus, we can conclude that the following;

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When hydrogen atoms approach each other, their potential energy decreases and they begin to form a bond. At an optimal distance of 0.74 å, a stable covalent bond is formed. The potential energy being zero or less does not signify a bond formation.

When two hydrogen atoms are far apart, the potential energy indeed approaches zero because there is no overlap of their atomic orbitals. As they approach each other, their atomic orbitals start overlapping, the energy of the system decreases, initiating the process of bond formation. When the distance between the two hydrogen atoms reaches 0.74 å (Angstroms), their energy is at its lowest point which signifies the formation of a covalent bond. This is because at this optimal distance, both repulsive and attractive forces combine to achieve the lowest possible energy configuration, leading to a stable covalent bond.

However, if the distance were to reduce to 0.50 å, the atoms would be too close and the energy of the system would increase due to stronger nuclear repulsions and electron confinement. In essence, a scenario where the potential energy is either less than or equals to zero does not correlate to the formation of a covalent bond. Instead, it's the optimal balance of forces at a specific distance that gives rise to a stable covalent bond in hydrogen atoms.

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Answer:

Explanation:

One therm = 1.055 x 10⁸J

= 1.055 X 10⁵ kJ

= [tex]\frac{1.055\times10^5}{60\times60}[/tex]kWh

29.3055 kWh

cost of 29.3055 kWh is 86 cent

or cost of imput= [tex]\frac{86}{29.3055}[/tex]

= 2.93 cent /]kWh

compare it with cost rate quoted earlier which each was

= 10

Answer:

Angle of reflection  at plane surface of prism = 30°

Explanation:

Angle of reflection = angle of incidence at any surface plane or curved .

Here angle of incidence at plane surface of prism = 30°

Angle of reflection  at plane surface of prism = 30°

If i be the angle of incidence and r be angle of refraction

sin i / sin r = refractive index

= 1.52

sin r = sin i / 1.52

=  sin 30 / 1.52

= .5 / 1.52

= .3289

r = 19.2 °

Answer:

Q=116.37 W

Explanation:

Given that

Emissivity ,ε= 0.200

Surface temperature ,T₁ = 10⁰ C = 283 K

Surrounding T₂ = - 15⁰ C = 258 K

Area ,A= 1.6 m²

The net heat transfer given as

[tex]Q=\sigma A \varepsilon (T_1^4-T_2^2)[/tex]

σ = 5.67 x 10⁻⁸

The temperature should be in Kelvin.

Now by putting the values

[tex]Q=5.67\times 10^{-8}\times 1.6\times 0.2\times (283^4-258^2)\ W[/tex]

Q=116.37 W

Therefore answer will be 116.37 W.

Answer:

72.25 g

Explanation:

mass of 1.7 L water = 1.7 x 10⁻³ x 10³ kg

= 1.7 kg

heat required to raise its temperature from 25 degree to 100 degree

= mass x specific heat x rise in temperature

= 1.7 x 4.18 x 10³ x 75 J

= 532.95 kJ

Now calorific value of LP gas = 46.1 x 10⁶J / kg

Let required mass of LP gas be m kg

heat evolved from this amount of gas

=  46.1 x 10⁶ m

Heat utilized in warming water  

= 46.1 x 10⁶ m  x .16 J

So

46.1 x 10⁶ m  x .16  =  532.95 x 10³

7376 x 10³ m = 532.95 x 10³

m = 532.95 / 7376 kg

= 72.25 g

In order to heat 1.7 L of water from room temperature to boiling, approximately 66.7 kg of LP gas, assuming that during heating only 16% of the heat emitted by LP gas combustion goes to heat the water, would be required.

To calculate the mass of LP gas needed to heat the water, we firstly need to determine the amount of energy required to heat the water from room temperature to the boiling point. This can be determined using the formula for heat Q=m*c* ΔT, where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. Considering water's specific heat capacity is 4.184 J/g °C and the change in temperature is 75 °C, the heat needed to warm the water would be (1.7 kg * 4.184 kJ/kg °C * 75 °C) = 532.722 kJ.

Since only 16% of the heat from the LP gas goes to heat the water, the total energy provided by the combustion of LP gas would be obtained by dividing the heat to warm water (532.722 kJ) by 0.16, which equals to 3.33 * 10^3 kJ.

Considering that the heat of combustion of LP gas is about 50 kJ/g approx, the mass of LP gas needed would finally be found by dividing the total energy provided by the LP gas (3.33 * 10^3 kJ) by the heat of combustion of LP gas, which results in approximately 66.7 kg.

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Answer:6.0×10^5m/s

Explanation:

According to the law of conservation of momentum, sum of the momenta of the bodies before collision is equal to the sum of their momenta after collision.

After their collision, the two bodies will move with a common velocity (v)

Momentum = mass × velocity

Let m1 be the mass of the proton = m

Let m2 be the mass of the alpha particle = m2

Let v1 be the velocity of the proton = 3.0×10^6m/s

Let v2 be the velocity of the alpha particle = 0m/s (since the body is at rest).

Using the law,

m1v1 + m2v2 = (m1 + m2)v

m(3.0×10^6) + 4m(0) = (m + 4m)v

m(3.0×10^6) = 5mv

Canceling 'm' at both sides,

3.0×10^6 = 5v

v = 3.0×10^6/5

The common velocity v = 6.0×10^5m/s

Answer:

One of the most important difference between the two galaxies" is that the sbb galaxy has a lot of material and bar structures passing thourgh the center and the sb galaxy does not have this".

Explanation:

Basically we have 3 types of galaxies (ellipticals, spirals, and irregulars)

Sb galaxies

On this type of galaxies the "intermediate type of spiral typically has a medium-sized nucleus. They can contain stars, star clouds, and interstellar gas and dust. Usually show wide dispersions in terms of their shape. Other members of this subclass have arms that begin tangent to a bright, while still others reveal a small, bright spiral pattern inset into the nuclear bulge. In any of these cases, the spiral arms may be set at different pitch angles. is important to mention that a pitch angle is defined as the angle between an arm and a circle centred on the nucleus and intersecting the arm".

SB galaxies

"For these types of galaxies the luminosities, dimensions, spectra, and distributions of the barred spirals tend to be indistinguishable from those of normal spirals".

One special case occuers at the SBb systems, that "have a smooth bar as well as relatively smooth and continuous arms. In some galaxies of this type, the arms start at or near the ends of the bar, with conspicuous dust lanes along the inside of the bar that can be traced right up to the nucleus"

One of the most important difference between the two galaxies" is that the sbb galaxy has a lot of material and bar structures passing thourgh the center and the sb galaxy does not have this".

Answer:

F = 84.61 N

Explanation:

As in the figure, since there is no friction so if component of Force applied along the incline is greater than the component of weight along the incline, then the object will move up the incline.

component of Force along the incline = F cos(23° - 15°) = F cos(8°)

component of weight along the incline = 33*g*sin(15°) = 33*9.81*sin(15°)

Equating the above two components of forces will give the minimum Force required.

F cos(8°) = 33*9.81*sin(15°)

F = 33*9.81*sin(15°) / cos(8°)                  (calculate the value using a scientific calculator)

F = 84.61 N

Answer:

20%

Explanation:

Relative Humidity (%) = (water vapor content÷water vapor capacity) × 100

=(7÷35)×100

=(0.2)×100

=20%

According to the Temperature-Water Vapor Capacity Table, the water capacity at 35 °C is 35 grams.

Water Vapor Capacity: The amount of water (grams) which air can hold at a given temperature.

Water Vapor Content: The amount of water vapor actually present in the air.

It takes the same amount of work to move up the stairs whether running or walking, as work is related to the change in potential energy, which does not depend on the speed of climbing. However, running up the stairs requires additional work to be done against gravity due to the higher kinetic energy, leading to a feeling of greater exhaustion.

The question concerns whether it takes more work to run up stairs than to walk up. In physics, the term 'work' refers to the amount of energy transferred by a force through a distance. In the context of climbing stairs, work is done against gravity to elevate a person's body to a higher potential energy level. According to the work-energy principle, work is equal to the change in kinetic energy plus the change in potential energy.

Using the given information, the person does 1764 J of work to move up the stairs, regardless of whether they run or walk.

This is the work done to overcome gravity. Only 120 J of work is used to increase the person's kinetic energy when running, which is additional work that occurs due to the higher speed.

Therefore, the total work done when running will be the sum of these two amounts.

However, the amount of work done to raise the person's body to a higher elevation (the potential energy) does not change whether the person runs or walks.

The difference is in the kinetic energy.

The feeling of exhaustion may be greater when running due to the higher pace requiring more power output (rate of doing work) and possibly higher energy expenditure related to physiological effects, but the work done against gravity, which depends on the change in elevation and mass of the person, remains the same in both cases.

Answer:

D) shrivel up, since the atmosphere exerts more force on the can as it cools.

Explanation:

As the water in the can is boiled the can gets heated up and contains hot vapour and gases which are rare in density and are in their expanded state. In this state when the can is sealed tightly such that no air leaks in or out of the can. When the temperature of the can drops, the gases shrink in volume and the pressure inside the can become less than the pressure of the atmosphere which leads to shriveling of the can.

Answer:

a) Kinetic energies

K₁ = 1.2 J

K₂ = 7.5 J

b) The bullet that has the highest kinetic energy is the one with the highest speed , v = 50 m/s , K₂ = 7.5 J

c) K₂ -K₁  = 6.3 J

Explanation:

The kinetic energy (K) is that due to the movement of a body and is calculated as follows:

K = (1/2) m*v²  (J)

Where :

m : the mass of the body ( kg)

v is the speed of the body (m/s)

Data

m₁ = m₂ = 0.006 Kg

v₁ = 20 m/s

v₂ = 50 m/s

a)Calculation of the kinetic energy

K₁ = (1/2) (m₁)*(v₁)²

K₁ = (1/2) (0.006)*(20)²

K₁ = 1.2 J

K₂= (1/2) (m₂)*(v₂)²

K₂ = (1/2) (0.006)*(50)²

K₂ = 7.5 J

b) K₂ ˃ K₁

The bullet that has the highest kinetic energy is the one with the highest speed , v = 50 m/s, K₂ = 7.5 J

c) Difference of their kinetic energies (K₂ -K₁)

K₂ -K₁  = 7.5 J - 1.2 J = 6,3 J

Answer:

The work done by the man is 260 J.

(E) is correct option.

Explanation:

Given that,

Force = 80 N

Distance = 5.0 m

Angle = 30°

Acceleration = 1.5 m/s²

Let F be the man's force

We need to calculate the force

Using balance equation

[tex]F-mg\sin\theta=ma[/tex]

[tex]F-80\sin30=\dfrac{80}{9.8}\times1.5[/tex]

[tex]F=\dfrac{80}{9.8}\times1.5+80\sin30[/tex]

[tex]F=52\ N[/tex]

We need to calculate the work done by the man

Using formula of work done

[tex]W=F\dotc d[/tex]

[tex]W=52\times 5.0[/tex]

[tex]W=260\ J[/tex]

Hence, The work done by the man is 260 J.

If the speed of the crate decreases at a rate of 1.5 m/s2, then the work done by the man is E) 260 J

Work is the quantity of energy transferred by displacement. The following is formula of work done

[tex]\Delta KE = F*d[/tex]

Whereas the following is formula for  balance equation:

[tex]F-mg\sin\theta=ma[/tex]

The equation above is come from Newton's Second Law.  It applies to a wide range of physical phenomena, but it is not a fundamental principle like the Conservation Laws and also it applies only if the force is the net external force.

For the Case of fictionless slope, the speed at the bottom of a frictionless incline does not depend upon the angle of the incline

Then the work done by the man is...

[tex]w = mgd* sin\theta + \Delta KE\\w = 80 N *5.0 m *sin 30 + (F*d)\\w = 80 N *5.0 m *\frac{1}{2}  + (F*d)\\w = 40 N *5.0 m+ (F*d)\\w = 40 N *5.0 m+ (m*a*d)\\w = 200 Nm+ (8*1.5*5)\\w = 200 Nm+ 60\\w = 260 J\\[/tex]

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Answer:

Answered

Explanation:

Terrestrial Planets are the planets that have solid surface and are smaller in size. Mercury, Venus, Earth and Mars are the terrestrial planets of our solar system.

The main greenhouse gases in the atmospheres of the terrestrial planets are, Carbon dioxide( the atmosphere of Venus has mainly CO_2 in it making it hottest planet of the solar system because of greenhouse effect). Water vapor and Cloroflorocarbons in the Earth atmosphere. Traces of methane and nitrous oxide are also present.

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

     U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

The amplitude (A) of the oscillation is 0.650 m, and the frequency (f) is approximately 1.34 Hz. To determine the total energy and the kinetic and potential energy at the point when x = 0.360 m, the spring constant is needed.

In order to answer your question about a 1.15-kg mass oscillating according to the equation x = 0.650 cos(8.40t), we can use the facts about simple harmonic motion.

(a) The amplitude is the coefficient of the cosine function, which is 0.650 m.

(b) The frequency (f) can be determined from the coefficient of the t inside the cosine function. In simple harmonic motion, the angular frequency (ω) equals 2πf. Therefore f = ω / 2π = 8.40 / 2π = 1.34 Hz.

(c) The total energy (E) of an oscillator is given by the formula E = 1/2 kA². However, without the spring constant (k) given in the question, this cannot be determined.

(d) When x = 0.360 m, the potential energy (PE) is given by PE = 1/2 kx², and the kinetic energy (KE) can be calculated by subtracting the potential energy from the total energy. But again, without the spring constant, these cannot be determined.

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Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = [tex]11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}[/tex]

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = [tex]390\ W/m.^{\circ}C[/tex]

Conduction of heat from the rod per second is given by:

[tex]q = \frac{KA\Delta T}{L}[/tex]

where

[tex]\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C[/tex] = temperature difference between the two ends of the rod.

Thus

[tex]q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s[/tex]

Now,

To calculate the mass, M of the ice melted per sec:

[tex]M = \frac{q}{L_{w}}[/tex]

where

[tex]L_{w}[/tex] = Latent heat of fusion of water = 333 kJ/kg

[tex]M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g[/tex]

To find the number of grams of ice melted per second, we can calculate the rate of heat transfer through the copper rod from the boiling water to the ice-water mixture using Fourier's Law of Heat Conduction. The formula is:

[tex]\[ Q = kA \frac{\Delta T}{L} \][/tex]

Where:

- Q is the rate of heat transfer (in watts, W)

- k  is the thermal conductivity of the material (in W/m°C)

-  A  is the cross-sectional area of the rod (in m²)

- [tex]\( \Delta T \)[/tex] is the temperature difference between the hot and cold ends (in °C)

- L is the length of the rod (in meters)

First, we need to find the temperature difference [tex]\( \Delta T \)[/tex]between the hot end (boiling water) and the cold end (ice-water mixture). We know water boils at 100°C and ice-water mixture is at 0°C. Therefore,[tex]\( \Delta T = 100°C - 0°C = 100°C \).[/tex]

Given:

- Cross-sectional area [tex]\( A = 11.6 \, \text{cm}^2 = 0.00116 \, \text{m}^2 \)[/tex]

- Length of the rod [tex]\( L = 19.6 \, \text{cm} = 0.196 \, \text{m} \)[/tex]

- Thermal conductivity of copper [tex]\( k = 390 \, \text{W/m°C} \)[/tex]

- Latent heat of fusion of ice [tex]\( L_f = 334 \, \text{J/g} \)[/tex]

Now, let's calculate the rate of heat transfer  Q :

[tex]\[ Q = (390 \times 0.00116) \times \frac{100}{0.196} \][/tex]

[tex]\[ Q \approx 236.8 \, \text{W} \][/tex]

Next, we need to convert this rate of heat transfer into grams of ice melted per second using the latent heat of fusion of ice:

[tex]\[ \text{Grams of ice melted per second} = \frac{Q}{L_f} \][/tex]

[tex]\[ \text{Grams of ice melted per second} = \frac{236.8 \, \text{J/s}}{334 \, \text{J/g}} \][/tex]

[tex]\[ \text{Grams of ice melted per second} \approx 0.708 \, \text{g/s} \][/tex]

Therefore, approximately [tex]\( 0.708 \)[/tex] grams of ice melt each second. This means that [tex]\( 0.708 \)[/tex] grams of ice in the ice-water mixture will melt every second due to the heat transferred from the boiling water through the copper rod.

Answer:

2 seconds

5.74165 seconds

-32 ft/s²

119.73 ft/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32 ft/s²

[tex]s=-16t^2+64t+160[/tex]

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-64}{-32}\\\Rightarrow t=2\ s[/tex]

The maximum height will be reached in 2 seconds

From the given equation

[tex]s=-16(2)^2+64\times 2+160\\\Rightarrow s=224\ ft[/tex]

The maximum height is 224 ft from the ground

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 224=0t+\frac{1}{2}\times 32\times t^2\\\Rightarrow t=\sqrt{\frac{224\times 2}{32}}\\\Rightarrow t=3.74165\ s[/tex]

Time taken to fall from the maximum height is 3.74165 seconds

Time taken from the point in time the ball is thrown is 2+3.74165=5.74165 seconds

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 32\times 224+0^2}\\\Rightarrow v=119.73\ ft/s[/tex]

Velocity with which the ball will hit the ground is 119.73 ft/s

At t = 1

Acceleration of a body thrown is always i.e., a body in free fall is always 32 ft/s². Here as the ball is thrown up the it will be negative so -32 ft/s²

Answer:

0.882 m/s²

Explanation:

m = Mass of car = 643 kg

a = Acceleration of the car = 2 m/s²

[tex]a_2[/tex] = Acceleration of the truck

Force exerted on the car

[tex]F=ma\\\Rightarrow F=643\times 2\\\Rightarrow F=1286\ N[/tex]

Mass of the truck with the woman = [tex]M=1415+43=1458\ kg[/tex]

From Newton's second law

[tex]F=Ma_2\\\Rightarrow a_2=\frac{1286}{1458}\\\Rightarrow a=0.882\ m/s^2[/tex]

The acceleration of the truck is 0.882 m/s²

Final answer:

The acceleration of the truck can be found by using Newton's second and third laws. Equal and opposite forces act on the car and the truck, and with the mass of each known, calculations can provide the truck's acceleration.

Explanation:

To find the acceleration of the truck when the woman pushes the car, we use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. When the woman pushes the car with a certain force, the car pushes back on the woman (and hence the truck she is sitting on) with an equal and opposite force. This means the force causing the car to accelerate is the same force causing the truck to accelerate, but in the opposite direction.

Let's denote F as the force with which the woman pushes the car. Given the mass of the car (mcar = 643 kg) and its acceleration (acar = 2 m/s²), we can find the force using Newton's second law of motion: F = mcar × acar.

The same force is responsible for the acceleration of the truck in the opposite direction. Therefore, for the truck with mass mtruck = 1415 kg and acceleration atruck, Newton's second law gives us the relationship: F = mtruck × atruck. By substituting the force F from the previous calculation, we can solve for atruck, which is the acceleration of the truck.

a) The magnetic flux is 0.0005Tm².

b) 1. The magnitude of the magnetic field due to the coil is uniform over the area of the loop.3. The magnetic field due to the coil is uniform in direction over the area of the loop.

c) The direction of the electric field inside the wire of the loop at time t1 is zero

d) The direction of the "curly" electric field in the loop at time  when the current in the coil t₂ is counter clockwise

e)  The absolute value of the rate of change of the magnetic flux through the loop is [tex]0.000045T^2[/tex]

f) The absolute value of the emf in the loop at previous instant is 0.000045V

g) The magnitude of the electric field at location P, which is inside the wire is 0.0036V/m

h) Now the wire loop is removed.

The question is investigating the interactions of current, magnetic fields, magnetic flux, and electromagnetic forces.

(a) The magnetic flux is the magnetic field times the area it is flowing through. So we have

[tex]\phi_{mag}= B * A = 0.4T * \pi*(0.02m)^2 = 0.0005 T m^2[/tex]

(b) The assumptions made are: 1. The magnetic field is uniform over the loop's area, 2. Magnetic field outside the loop is zero, 3. The magnetic field is uniform in direction over the loop's area.

(c) The direction of the electric field inside the wire of the loop at time t1 is zero, as the magnetic field and the current are constant and hence there’s no change in magnetic flux.

(d) If the current in the coil begins to decrease, Lenz's law tells us that an induced current will always counteract the change in magnetic flux that caused it. Therefore, the direction of the "curly" electric field in the loop will be counter-clockwise.

(e) At the later time the rate of change of the magnetic flux, |dΦmag/dt|, would be equal to the rate of change of the magnetic field times the area of the loop [tex]= |0.36 T/s * \pi*(0.02m)^2| = 0.000045 T m^2/s.[/tex]

(f) The absolute value of the emf in the loop would be equal to the absolute value of the rate of change of magnetic flux. Based on Faraday's law this means [tex]|emf| = |d\phi_{mag}/dt| = 0.000045 V.[/tex]

(g) [tex]\|E| = \frac{|\text{emf}|}{2 \pi r} = \frac{0.000452 \, \text{V}}{2 \pi (0.02 \, \text{m})} \approx 0.0036 \, \text{V/m}[/tex]

(h ) The magnitude of the electric field at location P, which is inside the wire, would depend on specifics not given in the question.

Answer:

Circuit breaker

Explanation:

Circuit breaker is the devise designed to protect the circuit from over current by opening the circuit automatically. Breaker can also be off manually by toggle switch. Earlier fuses were used but circuit breakers have replaced them. Fuse and circuit breakers operates differently. in case of overloading fuses blown off and opens the circuit while circuit breaker opens the circuit automatically without being blown off.

Final answer:

A circuit breaker is a safety device in a circuit that opens the circuit when an overcurrent occurs. It is faster than fuses and can be reset.

Explanation:

A circuit breaker is a device designed to open and close a circuit by non-automatic means and to open the circuit automatically on a predetermined overcurrent without damage to itself when properly applied within its rating. It acts as a safety device that switches off an appliance if the current in the circuit is too strong.

Circuit breakers are rated for a maximum current and can be reset, reacting much faster than fuses. They consist of components like bimetallic strips that respond to heat to break the electrical connection in the circuit.

Answer:

1. A only

Explanation:

First law of thermodynamics it sates that energy of the universe is constant.It is also known as energy conservation law.

Therefore according to the first law of thermodynamics ,the thermal energy of the system is unchanged.

Second law of thermodynamics states that ,it is impossible to make a device which take heat from higher temperature energy resource and give 100 % work without heat rejecting.

Therefore only option A is correct.

1. A only

Answer:

Explanation:

The photons travel faster faster through space because photons always travel through space faster than electrons, in fact when an electron gets hitted by a photon this boost its speed

Answer:

The z-component of the net torque is: 2.58 (N*m)

Explanation:

We need to apply the torque equation ([tex]T=RXF[/tex]), where T is the torque, R is the distance at the point where the force is applied and F is the force, and remembering the cross product ([tex]AXB=abSin(\beta)(n)[/tex]), where A and B are the vectors and ab are the magnitudes, (beta) the angle between A and B in the plane containing them, and (n) is the direction of the vector given by the right-hand rule. Knowing those things, we can get:[tex]T=RXF=4.1(i)+5.4(j)X1.8(i)+3(j)=0+4.1*3(k)+5.4*1.8(-k)+0=2.58(k)(N*m)[/tex] that is the z-component of the torque.

The torque on the object is -2.58Nm

Data;

The net torque is calculated as the product of the force and distance

[tex]t = r * F[/tex]

Let's substitute the values and solve

Since this is a dot product, we can easily multiply through

[tex]\tau = r * F\\\tau = (1.8*5.4) - (3 * 4.1) = -2.58Nm[/tex]

Using 2 by 2 matrix method, the torque on the object is -2.58Nm

Learn more on torque here;

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