A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 7950 kcal of heat is added to the gas, the volume is observed to increase slowly from 12.8 m^3 to 19.4 m^3.

Required:
a. Calculate the work done by the gas. Express your answer with the appropriate units.
b. Calculate the change in internal energy of the gas, Express your answer with the appropriate units.

Answer:

21.73 kg

Explanation:

Applying,

T = 2π√(m/k)............... Equation 1

Where T = period, m = mass on the spring, k = spring constant, π = pie.

make m the subject of the equation

m = T²k/4π²................. Equation 2

From the question,

Given: T = 3.45 s, k = 72.0 N/m, π = 3.14

Substitute these values into equation 2

m = (3.45²×72)/(4×3.14²)

m = 21.73 kg.

Hence the mass should be 21.73 kg

Answer:

1210 ohm

Explanation:

Given :

P=40 W

V=220 V

Now,

[tex]P=\frac{V^{2} }{R} \\40=\frac{(220)^{2} }{R} \\40R=48400\\R=\frac{48400}{40} \\R=1210 ohm[/tex]

Therefore, resistance of bulb will be 1210 ohm

Explanation:

1. First, let's find the total resistance of the circuit. We begin by combining [tex]R_{4}[/tex], [tex]R_{5}[/tex] and [tex]R_{6}[/tex]:

[tex]R_{456}=R_{4} + \dfrac{R_{5}R_{6}}{R_{5} + R_{6}}[/tex]

[tex]= 6\:Ω + \dfrac{(3\:Ω)(5\:Ω)}{3\:Ω+5\:Ω} = 7.9\:Ω[/tex]

Now time to combine [tex]R_{2}[/tex] and [tex]R_{3}[/tex] and they are connected in series so

[tex]R_{23} =R_{2} + R_{3} = 17\:Ω[/tex]

Note that [tex]R_{23}[/tex] and [tex]R_{456}[/tex] are connected in parallel so

[tex]R_{23456} = \dfrac{R_{23}R_{456}}{R_{23}+R_{456}}=5.4\:Ω[/tex]

Finally, [tex]R_{23456}[/tex] is connected in series with [tex]R_{1}[/tex] so the total resistance [tex]R_{T}[/tex] is

[tex]R_{T} = R_{1} + R_{23456} = 10\:Ω + 5.4\:Ω = 15.4\:Ω[/tex]

2. The total current in the circuit is

[tex]I_{T} = \dfrac{V}{R_{T}} = \dfrac{20\:V}{15.4\:Ω} = 1.3\:A[/tex]

3. The voltage drop across [tex]R_{1},\:V_{1}[/tex] is

[tex]V_{1} = I_{T}R_{1} = (1.3\:A)(10\:Ω) = 13\:V[/tex]

4. We can see that [tex]I_{T} = I_{1} + I_{2}[/tex]. To solve for [tex]I_{1}[/tex], we need [tex]V_{23}[/tex], which is just [tex]V_{T} - V_{1} = 20\:V - 13\:V = 7\:V[/tex] , which gives us

[tex]I_{1} = \dfrac{V_{23}}{R_{23}} = \dfrac{7\:V}{17\:Ω} = 0.4\:A[/tex]

5. From #2 & #4, [tex]I_{2} = 1.3\:A - 0.4\:A = 0.9\:A[/tex] and we also know that the voltage drop across [tex]R_{456}[/tex] is 7 V, the same as that of [tex]R_{23}[/tex]. The voltage drop across [tex]R_{4}[/tex] is

[tex]V_{4} = I_{2}R_{4} =(0.9\:A)(6\:Ω) = 5.4\:V[/tex]

This means that the voltage drop across [tex]R_{6}[/tex] is 7 V - 5.4 V = 1.6 V. Knowing this, the current through [tex]R_{6}[/tex] is

[tex]I_{6} = \dfrac{1.6\:V}{5\:Ω} = 0.3\:A[/tex]

Answer:

If air resistance is taken as negligible, then the ball is in freefall the moment it is thrown so gravity is the only force acting on the object. If air resistance is not negligible then gravity will be the greatest force acting on the ball while it is going up and coming down, because Fair has to be less than gravity at all times otherwise the atmosphere would wither away.

Answer:

a = ?

u = 0

v = 30

by using v = u + at equation we can find " a "

30 = 0 + 5a

6 m/s = a

by using f = ma equation we can find force produce by engine ,

f = ?

a = 6

m = 1200

f = 1200 × 6

f = 7200 N

so the answer is "B"

Answer:

9.6352× 10²⁰ C atoms

Explanation:

From the given information,

The molar mass of Carbon = 12 g/mol

number of moles = 0.020g/ 12 g/mol

number of moles = 0.0016 mol

If 1 mole of C = 6.022 × 10²³ C atoms

∴

0.0016 mol of C = (6.022 × 10²³ C atoms/ 1 mol of C)×0.0016 mol of C

= 9.6352× 10²⁰ C atoms

Hence, the number of carbon atoms present in 0.020 g of carbon = 9.6352× 10²⁰ C atoms

Answer:

336.96m/s

Explanation:

answer is in photo above

The speed of sound on that day is  336.96 m/s.

Speed is distance travelled by the object per unit time. Due to having no direction and only having magnitude, speed is a scalar quantity With SI unit meter/second.

given that:

frequency of the sound = 312 Hz.

2nd resonant length = 81.0 cm.

If the wavelength of sound is λ; the 2nd resonant length of  plastic pipe with an adjustable closed end = 3λ/4

Hence, wavelength of sound is = 81×(4/3) cm = 0.81 ×(4/3) m.

So,  the speed of sound on that day is = frequency × wavelength

= 312 Hz ×  0.81 ×(4/3) m.

= 336.96 m/s.

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Answer:

The wave speed is 2.8 m.

Explanation:

Wavelength = 1.4 m

frequency, f = 2 Hz

the wave speed is given by

wave speed = wavelength x frequency

wave speed = 1.4 x 2 = 2.8 m

option (D) is correct.  

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g=[tex]9.50\times 10^{3} kg[/tex]

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

[tex]mu+ MU=mv+ MV[/tex]

Substitute the values

[tex]9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V[/tex]

[tex]3.61i=2.375j+0.150V[/tex]

[tex]3.61 i-2.375j=0.150V[/tex]

[tex]V=\frac{1}{0.150}(3.61 i-2.375j)[/tex]

[tex]V=24.07i-15.83j[/tex]

Magnitude of velocity of stone

=[tex]\sqrt{(24.07)^2+(-15.83)^2}[/tex]

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]

=[tex]tan^{-1}(\frac{-15.83}{24.07})[/tex]

[tex]\theta=tan^{-1}(-0.657)[/tex]

=33.3 degree below the horizontal

(b)

Initial kinetic energy

[tex]K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2[/tex]

[tex]K_i=685.9 J[/tex]

Final kinetic energy

[tex]K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2[/tex]

=[tex]\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2[/tex]

[tex]K_f=359.12 J[/tex]

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

Explanation:

Given that,

The radius of rain drop, r = 0.12 cm = 0.0012 m

The viscocity of air is, [tex]\eta=18\times 10^{-5}\ poise[/tex]

Let the viscous force is, [tex]F = 0.010173\ N[/tex]

The viscous force is given by :

[tex]F=6\pi \eta rv\\\\v=\dfrac{F}{6\pi \eta r}[/tex]

Put all the values,

[tex]v=\dfrac{0.010173}{6\pi 18\times 10^{-5}\times 0.0012 }\\\\v=2498.58\ m/s[/tex]

Answer:

effort arm mean the use of any work by using your hand force motion or by hand power

Answer:

C

Explanation:

horizintal speed stays same

only vertical speed changes

so H speed will stay 30 m/s

Answer:

The second statement.

Answer:

After 15 seconds, the green car will catch up with the blue car

Explanation:

Let the time for the green car to catch up with the blue car be T

When the green car catches up to the blue car, the distances covered by each car after time T will be equal. Also, their velocities at that instant will be equal

Distance covered by blue car after time T is given by: s = ut + 0.5 at²

Where u = 0, a = 0.2 m/s², t = T

S = 0.5 × 0.2 × T² = 0.1 T²

Velocity of blue car, v = u+ at

v = 0.2T

Distance covered by green car at T is given as: S = Velocity × time

Where v = 0.2T, t = T - 7.5 (since the blue car started 7.5 seconds earlier)

S = 0.2T (T - 7.5)

S = 0.2 T² - 1.5T

Equating the distance covered by the two cars

0.2T² - 1.5T = 0.1T²

0.1T² - 1.5T = 0

T(0.1T - 1.5) = 0

T = 0 or

T = 1.5/0.1 = 15 secs

Therefore, after 15 seconds, the green car will catch up with the blue car

Answer:

less than the weight of the block.

Explanation:

From the free body diagram, we get.

The normal force is N = Mg cosθ

The tension in the string is T = Mg sinθ

Wight of the block when the block is static, W = Mg

Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,

therefore, the normal force is less than the weight of the static block.

Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0​, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.

Answer:

the answer is c which is a+2 charge

Explanation:

Beryllium is in group 2A. It's nearest noble gas is Helium, which is 2 elements behind Beryllium. ThBeryllium wants to lose two electrons. When it does that, Beryllium will have a positive chargeof two, and it will be stated as B-e two plus.

The Beryllium (Be) has an atomic number of 4 and belongs to Group-2 elements. The Beryllium will form a divalent cation (+2). Thus, option C is correct.

In an atom, the number of electrons equals the number of protons. If the electrons are removed from the atom or the electrons are added to the atom, the atom has an excessive positive or negative charge.

This excessive of electrons or lack of electrons forms Ions. The excess of electrons has a negative charge or anions and the lack of electrons has a positive charge or cations.

Beryllium has 4 electrons. Two electrons are occupied in the valence shell of beryllium. Group 2 elements always form the positive ions or cations, to become stable ions.

The outermost shell of beryllium has two electrons. In order to form a stable ion, beryllium should lose its two electrons or gain six electrons. Beryllium belongs to the Group-2 element, it always loses two electrons and forms Be²⁺, to form a stable ion.

Hence, Beryllium forms an ion with a +2 charge. Thus, the correct option is C.

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Answer:

the correct answer is C   v = 60 cm / s

Explanation:

The speed of a wave is related to the frequency and the wavelength

         v = λ f

They indicate that the object performs 20 oscillations every second, this is the frequency

         f = 20 Hz

the wavelength is the distance until the wave repeats, the distance between two consecutive peaks corresponds to the wavelength

         λ = 3 cm = 0.03 m

let's calculate

       v = 20 0.03

       v = 0.6 m / s

       v = 60 cm / s

the correct answer is C

Answer:

[tex]{ \tt{n _{w} = \frac{real \: depth}{aparent \: depth} }} \\ \\ \frac{4}{3} = \frac{4}{d} \\ { \tt{d = 3 \: meters}}[/tex]

A boy observes a fish at the depth of 4 meters under the water's surface. The refractive index of the water is 4/3 then the apparent depth of the fish as seen by the boy is 3 meters.

Refractive index is also known as the refraction, is a measure of the way a light beam bends when it goes through different media. The refractive index n is computed by dividing the sine of the angle of incidence by the sine of the angle of refraction, i.e., n = sin i/sin r, if I will be the angle of incidence of a light source in a void (angle between the inbound ray and the normal, or perpendicular to the surface of a medium).

The Index of refraction is also equal to c/v, or the ratio of a wavelength of light's velocity in a substance to its velocity in a void.

As per the given data in the question,

n(w) = real depth/apparent depth

4/3 = 4/d

d = 3 meters.

Therefore, the apparent depth will be equal to 3 meters.

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Answer:

E = 2.01 J

Explanation:

The energy lost by the car due to friction through the entire trip must be equal to the potential energy of the car at the starting point.

Energy Lost = E = Potential Energy

E = mgh

where,

m = mass of car = 150 g = 0.15 kg

g = acceleration due to gravity = 9.81 m/s²

h = height of the track = 137 cm = 1.37 m

Therefore,

E = (0.15 kg)(9.81 m/s²)(1.37 m)

E = 2.01 J

Answer:

Time, t = 8 seconds.

Explanation:

Given the following data;

Initial velocity = 24 m/s

Final velocity = 0 m/s (since the object is coming to a stop).

Acceleration = -4 m/s²

To find how long it will take the object to stop, we would use the first equation of motion;

[tex] V = U + at[/tex]

Where;

Making time, t the subject of formula, we have;

[tex] t = \frac{V - U}{a}[/tex]

Substituting into the equation, we have;

[tex] t = \frac{0 - 24}{-4}[/tex]

[tex] t = \frac{-24}{-4}[/tex]

Time, t = 8 seconds.

Answer:

v₀ = 2.67 m / s

Explanation:

This problem can be solved using the Kinetic Enemy Work Theorem

         W = ΔK

Work is defined by the relation

         W = fr. d

The bold letters indicate vectors, in this case the blow is in the direction of the slope of the ramp and the displacement is also in the direction of the ramp, therefore the angle between the force and the displacement is zero.

the friction force opposes the displacement therefore its angle is 180º

          W = - fr d

Let's use Newton's second law, we define a reference frame with the horizontal axis parallel to the plane

Y axis  

           N- Wy = 0

           N - W cos tea = 0

the friction force has the expression

          fr = μ N

          fr = μ W cos θ

we substitute

           W = - μ W cos θ d

let's look for kinetic energy

the minimum velocity at the highest point is zero

           K_f = 0

the initial kinetic energy is

            K₀ = ½ m v₀²

we substitute energy in the work relationship

         - μ W cos θ d = 0 - ½ m v₀²

           v₀² = - μ W cos θ  2d / m

Let's use trigonometry to find distance d

         sin θ=  y / d

         d = y /sin θ

         d = 3.50 / sin 30

         d = 7 m

let's calculate

           v₀² = (6 10⁻² 2.50 9.8 cos 30) 2 7 / 2.50

           v₀ = √7.129

           v₀ = 2.67 m / s

Answer:

The efficiency of Carnot's heat engine is 26.8 %.

Explanation:

Temperature of hot reservoir, TH = 100 degree C = 373 K

temperature of cold reservoir, Tc = 0 degree C = 273 K

The efficiency of Carnot's heat engine is

[tex]\eta = 1-\frac{Tc}{T_H}\\\\\eta = 1 -\frac{273}{373}\\\\\eta = 0.268 =26.8 %[/tex]

The efficiency of Carnot's heat engine is 26.8 %.

Answer:

if we want to find force by using newton's law equation ( f = ma ) we have to use mass and acceleration not velocity ,but in this question they did not mention about acceleration but speed so the answer is D

Answer:

C) 300 Ohm.

Explanation:

In a series circuit, total resistance is just adding all the resistance together. So R (total) = 75 +75 +75+ 75 = 300 ohms

Parallel circuit are different because you add the inverses of resistance and you flip the final answer.

You can confirm your answers using the tools below:

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Answer:

Explanation:

The mass of that science book...wow. In pounds that would be 35.2! Yikes!

Anyway, we need final velocity here, and the mass of the book has nothing to do with how fast it falls. Everything is pulled by the same gravity. A feather falls at 9.8 m/s/s and so does an elephant. Mass is useless information. The equation we will use is

[tex]v^2=v_0^2+2a[/tex]Δx  where

v is the final velocity, our unknown,

v₀ is the initial velocity which is 0 since someone had to be holding the book before dropping it,

a is the pull of gravity which is always -9.8 m/s/s, and

Δx = -120 which is the displacement (it's negative because the book falls below the point from which it was dropped). Filling in:

[tex]v^2=0^2+2(-9.8)(-120)[/tex] so

[tex]v=\sqrt{2(-9.8)(-120)}[/tex] and

v = 48 m/s

As far as how far above the bottom of the cliff the object is when it is moving at 12 m/s we will use the same equation, but the velocity will be 12:

[tex]12^2=0^2+2(-9.8)[/tex]Δx and

144 = -19.6Δx so

Δx = -7.3 m. That's how far from the top of the cliff it is. We subtract then t find out how far it is from the bottom:

120 - 7.3 = 112.7 m off the ground.