Could someone explain to me how to got the answer B, thank you very much​

Answer:

1.671L

Explanation:

In the given case the pressure is constant therefore it is an isobaric process, the process in which the pressure remains constant is called as isobaric process.

Work done= external pressure× change in volume.

288= 2×( final volume-intial volume)×101.32

144÷101.32=(finalvolume-0.250)

1.421=final volume-0.250

final volume=1.671L

Answer:

If you know the current and voltage across the whole circuit, you can find total resistance using Ohm's Law: R = V / I.

Explanation:

Answer:

F = 118 N

Explanation:

Assume Ann and Bob lift at their respective ends of the table

Sum moments about Bob's position to zero.

Let F be Ann's upward force

F[2.25] - 18.5(9.80)[2.25 / 2] - 8.33(9.80)[0.750] = 0

F = 117.86133333...  

Given:

The weight of table will be:

→ [tex]W_t = m_t g[/tex]

        [tex]= 18.5\times 9.8[/tex]

        [tex]= 181.3 \ N[/tex]

Now,

→ [tex]\sum M_A_{nn} = -81.634\times 0.750-181.3\times 1.125+R_{bob}\times 1.125[/tex]

or,

→ [tex]R_{bob} = \frac{61.2255+203.9625}{2.25}[/tex]

          [tex]= \frac{265.188}{2.25}[/tex]

          [tex]= 117.9 \ N[/tex]

Thus the above answer is right.

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Answer:

The acceleration is in 2 D as in between east and south.

Explanation:

mass, m = 50 kg

acceleration, a = 0.25 m/s^2 horizontal

acceleration of elevator, a' = 1 m/s^2 downwards

When a person on the ground the resultant acceleration of the person with respect to the ground is between east and south direction so the path os parabolic in nature. It graph is shown below:

Answer:

[tex]r=200m[/tex]

Explanation:

From the question we are told that:

Charges:

[tex]Q_1=8.0mC[/tex]

[tex]Q_2=2.0mC[/tex]

[tex]Q_3=8.mC[/tex]

Distance [tex]d=300m[/tex]

Generally the equation for Force is mathematically given by

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

Therefore

[tex]F_{32}=F_{31}[/tex]

[tex]\frac{q_2}{(300-r)^2}=\frac{q_1}{r^2}[/tex]

[tex]\frac{2*10^{-3}}{(300-r)^2}=\frac{8*10^{-3}}{r^2}[/tex]

[tex]r=2(300-r)[/tex]

[tex]r=200m[/tex]

Answer:

[tex]F=202650N[/tex]

Explanation:

From the question we are told that:

Area [tex]a=50m^2[/tex]

Difference in air Pressure [tex]dP=4.0\% atm=>0.04*101325=>4035Pa[/tex]

Generally the equation for Force is mathematically given by

[tex]F=dP*A[/tex]

[tex]F=4053*50[/tex]

[tex]F=202650N[/tex]

Answer:

12.74 V

Explanation:

We are given that

Current, I1=54 A

Potential difference, V1=9.18V

I2=2.10 A

V2=12.6 V

We have to find the battery's emf.

[tex]E=V+Ir[/tex]

Using the formula

[tex]E=9.18+54r[/tex] ....(1)

[tex]E=12.6+2.10r[/tex]  .....(2)

Subtract equation (1) from (2)

[tex]0=3.42-51.9r[/tex]

[tex]3.42=51.9r[/tex]

[tex]r=\frac{3.42}{51.9}=0.0659ohm[/tex]

Using the value of r in equation (1)

[tex]E=9.18+54(0.0659)[/tex]

[tex]E=12.74 V[/tex]

Answer:

75

Explanation:

i am not sure but if 200N boy is being held back then the force that's holding him back must be equal to or greater than his weight. if 125N is already exerted then the tension will be:

T=200-125

= 75

Answer:

The final speed of electron=[tex]5.93\times 10^6m/s[/tex]

Explanation:

We are given that

Initial velocity, u=0

Potential, V=100 V

We have to find the final speed.

Mass of electron, [tex]m=9.1\times 10^{-31} kg[/tex]

Charge on electron, q=[tex]1.6\times 10^{-19}C[/tex]

We know that

[tex]qV=\frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]

Using the formula

[tex]1.6\times 10^{-19}\times 100=\frac{1}{2}\times 9.1\times 10^{-31} v^2-0[/tex]

[tex]v^2=\frac{2\times 1.6\times 10^{-19}\times 100}{9.1\times 10^{-31}}[/tex]

[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 100}{9.1\times 10^{-31}}}[/tex]

[tex]v=5.93\times 10^6m/s[/tex]

Hence, the final speed of electron=[tex]5.93\times 10^6m/s[/tex]

Answer:

In order to lift off the ground, the air in the balloon must be heated to 710.26 K

Explanation:

Given the data in the question;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

let F represent the force acting upward.

Now in a condition where the hot air balloon is just about to take off;

F - Mg - m[tex]_g[/tex]g = 0

where M is the mass of the balloon and its occupants, m[tex]_g[/tex] is the mass of the hot gas inside the balloon.

the force acting upward F = Vρg

so

Vρg - Mg - m[tex]_g[/tex]g = 0

solve for m[tex]_g[/tex]

m[tex]_g[/tex] = ( Vρg - Mg ) / g

m[tex]_g[/tex] =  Vρg/g - Mg/g

m[tex]_g[/tex] =  ρV - M ------- let this be equation 1

Now, from the ideal gas law, PV = nRT

we know that number of moles n = m[tex]_g[/tex] / μ

where μ is the molecular mass of air

so

PV = (m[tex]_g[/tex]/μ)RT

solve for T

μPV = m[tex]_g[/tex]RT

T = μPV / m[tex]_g[/tex]R -------- let this be equation 2

from equation 1 and 2

T = μPV / (ρV - M)R

so we substitute in our values;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

T = [ (29 × 10⁻³) × (1.01 × 10⁵) × 480 ] / [ (( 1.29 × 480 ) - 381)8.31 ]

T =  1405920 / 1979.442

T =  710.26 K

Therefore, In order to lift off the ground, the air in the balloon must be heated to 710.26 K

The temperature required for the air to be heated is 710.26 K.

Given data:

The mass of a hot air-balloon is, m = 381 kg.

The pressure of air outside the balloon is, [tex]P = 1.01 \times 10^{5} \;\rm Pa[/tex].

The density of air is, [tex]\rho = 1.29 \;\rm kg/m^{3}[/tex].

The volume of heated balloon is, [tex]V = 480 \;\rm m^{3}[/tex].

The condition where the hot air balloon is just about to take off is as follows:

[tex]F-mg - m'g =0[/tex]

Here,

m' is the mass of hot gas inside the balloon and g is the gravitational acceleration and F is the force acting on the balloon in upward direction. And its value is,

[tex]F = V \times \rho \times g[/tex]

Solving as,

[tex](V \times \rho \times g)-mg - m'g =0\\\\ m'=(V \rho )-m[/tex]

Now, apply the ideal gas law as,

PV = nRT

here, R is the universal gas constant and n is the number of moles and its value is,

[tex]n=\dfrac{m'}{M}[/tex]

M is the molecular mass of gas. Solving as,

[tex]PV = \dfrac{m'}{M} \times R \times T\\\\\\T=\dfrac{P \times V\times M}{m'R}\\\\\\T=\dfrac{P \times V\times M}{(V \rho - m)R}[/tex]

Since, the standard value for the molecular mass of air is, [tex]M = 29 \times 10^{-3} \;\rm kg/mol[/tex]. Then solve for the temperature as,

[tex]T=\dfrac{(1.01 \times 10^{5}) \times 480\times 381}{(480 \times (1.29) - 381)8.31}\\\\\\T = 710.26 \;\rm K[/tex]

Thus, we can conclude that the temperature required for the air to be heated is 710.26 K.

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Consulta los dibujos adjuntos

Answer:

Please find the complete question in the attached file.

Explanation:

[tex]V=12\ V\\\\I=1.2\ A\\\\T=5\times 60=300\ second\\\\[/tex]

Calculating the electrical energy dissipated:

[tex]w=p\cdot t=V\cdot I \cdot t\\\\[/tex]

              [tex]=12\times 1.2 \times 300 \ J\\\\=4320\ J[/tex]

[tex]\Delta T=20^{\circ}\ C\\\\W=m\cdot c\cdot \Delta T\\\\4320=m(4186 \times 20)\\\\m=\frac{4320}{4186 \times 20}=51.6 \ grams=0.516 \ kg\\\\[/tex]

Answer:

Explanation:

Let the mass of objects be m₁ and m₂ .

Total kinetic energy = 1/2 m₁ v² + 1/2 m₂ v²= 1/2 ( m₁ + m₂ ) v²

Total kinetic energy after collision= 1/2 ( m₁ + m₂ ) v² / 4  =  1/2 ( m₁ + m₂ ) v² x .25

final KE / initial KE = 1/2 ( m₁ + m₂ ) v² x .25 / 1/2 ( m₁ + m₂ ) v²

= 0.25

b )

Applying law of conservation of momentum to the system . Let m₁ > m₂

m₁ v -  m₂ v = ( m₁ + m₂ ) v / 2

m₁ v -  m₂ v = ( m₁ + m₂ ) v / 2

m₁  -  m₂  = ( m₁ + m₂ )  / 2

2m₁ - 2 m₂ = m₁ + m₂

m₁ = 3m₂

m₁ / m₂ = 3 / 1

Answer:

(a) The ratio is 1 : 4.

(b) The ratio is 1 : 3.

Explanation:

Let the mass of each object is m and m'.

They initially move with velocity v opposite to each other.

Use conservation of momentum

m v - m' v = (m + m') v/2

2 (m - m')  = (m + m')

2 m - 2 m' = m + m'

m = 3 m' .... (1)

(a) Let the initial kinetic energy is K and the final kinetic energy is K'.

[tex]K = 0.5 mv^2 + 0.5 m' v^2 \\\\K = 0.5 (m + m') v^2..... (1)[/tex]

[tex]K' = 0.5 (m + m') \frac{v^2}{4}.... (2)[/tex]

The ratio is

K' : K = 1 : 4

(b) m = 3 m'

So, m : m' = 3 : 1

Answer:

1) ans: The ratio of load to effort in a simple machine is called Mechanical Advantage.

2) ans: Frictuon produced in Simple machine affect the mechanical advantage of a machine.

3) ans: The machine whose efficiency is 100% is called ideal machine.

4) ans: The work done by the machine is called output work.

ans: The work done in the machine is called input work.

5) ans: The turning effect of force is called moment.

Answer:

40 m

Explanation:

Applying,

v² = u²+2as............... Equation 1

Where v = final velocity, u = initial velocity, a = deceleration, s = distance

make s the subject of the equation

s = (v²-u²)/2a............ Equation 2

From the question,

Given: v = 0 m/s (comes to rest), u = 20 m/s, a = -5 m/s² (begins to slow)

Substitute these values into equation 2

s = (0²-20²)/[2(-5)]

s = -400/-10

s = 40 m

Hence the bus was 40 m from the stop

Answer:

the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

Explanation:

Given the data in the question;

wavelength of light; λ = 463 nm = 463 × 10⁻⁹ m

Index of refraction; n = 1.35

Now, the thinnest thickness of the soap film can be determined from the following expression;

[tex]t_{min[/tex] = ( λ / 4n )

so we simply substitute in our given values;

[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 4(1.35)

[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 5.4

[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 4(1.35)

[tex]t_{min[/tex] = 8.574 × 10⁻⁸ m

[tex]t_{min[/tex] = 85.74 × 10⁻⁹ m

[tex]t_{min[/tex] = 85.74 nm

Therefore, the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

It follows from Newton's second law that there is some counteractive force that cancels out the force exerted by the engine - it's most likely drag due to air resistance in combination with static friction between the tires and the road. The car is moving at constant speed past a certain point, so the net force on the car is

∑ F = (force from engine) - (resistive forces) = 0

Answer:

[tex]L=9.76*10^{16}m[/tex]

Explanation:

From the question we are told that:

Time on earth [tex]T_e= 13yrs[/tex]

Time on ship [tex]T_s= 7.9yrs[/tex]

Therefore

[tex]r=\frac{t_e}{t_s}[/tex]

[tex]r=\frac{13}{7.9}[/tex]

[tex]r=1.65[/tex]

Generally the equation for Constant Velocity is mathematically given by

[tex]V=C\sqrt{1-\frac{1}{r^2}}[/tex]

[tex]V=3*10^8\sqrt{1-\frac{1}{1.64^2}}[/tex]

[tex]V=2.38*10^8m/s[/tex]

Therefore

[tex]L=V*t[/tex]

Where

[tex]t=(13*365.25*24*3600)s[/tex]

[tex]t=4.1*10^8[/tex]

[tex]L=2.38*10^8m/s*4.1*10^8[/tex]

[tex]L=9.76*10^{16}m[/tex]

Answer:

[tex]l=12916.5m[/tex]

Explanation:

Distance [tex]d=3600km[/tex]

Since

Density of steel [tex]\rho=7900kg/m^3[/tex]

Stress of steel [tex]\mu= 1*10^9[/tex]

Generally the equation for Stress on Cable is mathematically given by

[tex]S=\frac{F}{A}[/tex]

[tex]S=\frac{\rho Alg}{A}[/tex]

Therefore

[tex]l=\frac{s}{\rhog}[/tex]

[tex]l=\frac{ 1*10^9}{7900kg/m^3*9.8}[/tex]

[tex]l=12916.5m[/tex]

Answer:

Move towards the wall.

Explanation:

When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.

As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.

Answer:

b. 0.20 m/s.

Explanation:

Given;

initial mass, m = 0.2 kg

maximum speed,  v = 0.3 m/s

The total energy of the spring at the given maximum speed is calculated as;

K.E = ¹/₂mv²

K.E = 0.5 x 0.2 x 0.3²

K.E = 0.009 J

If the mass is changed to 0.4 kg

¹/₂mv² = K.E

mv² = 2K.E

[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 0.009}{0.4} } \\\\v = 0.21 \ m/s\\\\v \approx 0.20 \ m/s[/tex]

Therefore, the maximum speed is 0.20 m/s

• Point charge (Q) = 10 μC = 10 × 10⁻⁶ C

• Potential (V) = 1000 V

• Distance (r) = ?

[tex]\implies V = \dfrac{KQ}{r} \\ [/tex]

[tex]\implies r= \dfrac{KQ}{V}[/tex]

[tex]\implies r= \dfrac{9 \times {10}^{9} \times 10 \times {10}^{ - 6} }{1000} \\ [/tex]

[tex]\implies r= \dfrac{90 \times {10}^{3} }{1000} \\ [/tex]

[tex]\implies r= \dfrac{90 \times {10}^{3} }{ {10}^{3} } \\ [/tex]

[tex]\implies r= 90 \times {10}^{3} \times {10}^{ - 3} \\ [/tex]

[tex]\implies\bf r= 90\:m \\ [/tex]

Hence,the option B) 90 m is the correct answer.

Answer:

(a) v = 32 t

h = 16 t^2

g = 32 ft/s^2

(b) 64 ft/s

Explanation:

height, h = 64 feet

g = - 32 ft/s^2

(a) Let the time is t .

Let the velocity after time t is v.

Use first equation of motion

v = u + at

- v = 0 - 32 t

v = 32 t

Let the distance is h from the top.

Use second equation of motion

[tex]h = u t + 0.5 at^2 \\\\- h = 0 - 0.5\times 32 \times t^2\\\\h = 16 t^2[/tex]

The acceleration is constant for entire motion.

(b) Let the velocity is v as it hits the ground. Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 \times 32\times 64\\\\v = 64 feet/s[/tex]

Answer:

Approximately [tex]63\; \rm m[/tex].

Explanation:

Convert the initial speed of this truck to standard units:

[tex]\begin{aligned} v &= 72\; \rm km \cdot h^{-1} \times \frac{1\; \rm h}{3600\; \rm s} \times \frac{1000\; \rm m}{1\; \rm km} \\ &= 20\; \rm m \cdot s^{-1}\end{aligned}[/tex].

Calculate the initial kinetic energy of this truck:

[tex]\begin{aligned}\text{KE} &= \frac{1}{2} \, m \cdot v^{2} \\ &= \frac{1}{2} \times 2600\; \rm kg \times (20\; \rm m \cdot s^{-1})^{2} \\ &= 5.2 \times 10^{5} \; \rm J\end{aligned}[/tex].

The kinetic energy of the truck would be [tex]0[/tex] when the truck is not moving. Hence, the friction on the truck would need to do [tex](-5.2 \times 10^{5}\; \rm J)[/tex] of work on the truck to bring the truck to a stop.

Calculate the displacement required to achieve [tex](-5.2 \times 10^{5}\; \rm J)[/tex] of work at [tex]8200\; \rm N[/tex]:

[tex]\begin{aligned}x &= \frac{W}{F} \\ &= \frac{-5.2 \times 10^{5}\; \rm J}{8200\; \rm N} \approx -63\; \rm m\end{aligned}[/tex].

This result is negative because the direction of friction is opposite to the direction of the motion of the truck.

Hence, the truck would come to a stop after skidding for approximately [tex]63\; \rm m[/tex].

Answer:

In heterogeneous mixture do the physically distinct component parts each have distinct properties.

Answer:

a)  [tex]F=475.7Hz[/tex]

b)  [tex]F'=410.899Hz[/tex]

Explanation:

From the question we are told that:

Velocity of eagle [tex]V_1=35m/s[/tex]

Frequency of eagle [tex]F_1=440Hz[/tex]

Velocity of Black bird [tex]V_2=10m/s[/tex]

Speed of sound [tex]s=343m/s[/tex]

a)

Generally the equation for Frequency is mathematically given by

 [tex]F=f_0(\frac{v-v_2}{v-v_1})[/tex]

 [tex]F=440(\frac{343-10}{343-35})[/tex]

 [tex]F=475.7Hz[/tex]

b)

Generally the equation for Frequency is mathematically given by

 [tex]F'=f_0(\frac{v+v_2}{v+v_1})[/tex]

 [tex]F'=440(\frac{343+10}{343+35})[/tex]

 [tex]F'=410.899Hz[/tex]

Answer:

Len's law

Explanation:

We can explain this exercise using Len's law

when the magnetic flux decreases, a matic flux appears that opposes the decrease, thus maintaining the value of the initial luxury.

Answer:

C is the right answer

Explanation:

fitness logs is a great way to track your progress. You can easily look back and see how you have progressed over time. In addition, it can help you plan and prepare for future workouts, as well as identify patterns of what seems to work well for you and when you have the most success

hope it was useful for you

Answer:

Initial velocity, u = 60 km/h = 16.7 m/s

Final velocity, v = 72 km/h = 20 m/s

time, t = 2 sec

From first equation of motion:

[tex]{ \bf{v = u + at}}[/tex]

Substitute the variables:

[tex]{ \tt{20 = 16.7 + (a \times 2)}} \\ { \tt{2a = 3.3}} \\ { \tt{acceleration = 1.65 \: {ms}^{ - 2} }}[/tex]