How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 36.0°C greater than when they were laid? Their original length is 26.0 m.

Answer:

  P₂ = 2 P₁

we conclude that in the second time the power used is double that in the first rise

Explanation:

In this exercise we are asked the power to climb the stairs, if we assume that we go up with constant speed, we use an energy equal to the potential energy due to the difference in height of the stairs, as this height is constant the potential energy does not change and therefore therefore the energy used by us does not change either.

Now we can analyze the required power,

         P = W / t

From the analysis of the previous paragraph the work is equal to the energy used, according to the work energy theorem,

therefore the first time the power is

           P₁ = E / 10

           P₁ = 0.1 E

for the second time the power is

          P₂ = E / 5

          P₂ = 0.2 E

we see that the power in the second case is

         P₂ = 2 P₁

Therefore, we conclude that in the second time the power used is double that in the first rise.

Answer:

The comoving distance and the proper distance scale

Explanation:

The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster.  Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.

Answer:

Pressure, P = 3724 Pa

Explanation:

Given that,

Depth of water, [tex]h_w=20\ cm =0.2\ m[/tex]

Depth of oil, [tex]h_o=40-20=20\ cm=0.2\ m[/tex]

The density of water, [tex]d_w=1000\ kg/m^3[/tex]

The densinty of oil, [tex]d_o=900\ kg/m^3[/tex]

We need to find the gauge pressure at the bottom of the cylinder. So, total pressure is equal to :

[tex]P=d_wgh_w+d_ogh_o\\\\P=(d_wh_w+d_oh_o)g\\\\P=(1000\times 0.2+900\times 0.2)\times 9.8\\\\P=3724\ Pa[/tex]

So, the gauge pressure at the bottom of the cylinder is 3724 Pa.

Answer:

4.003" (inches )

Explanation:

The maximum distance allowed between the center of hole #2 and datum B can be calculated by adding 4.000" + 0.003" ( perpendicularity of the of hole #2) as seen from the front view of the diagram .

Note :The hole 2 is sited below the workpiece when viewed from the front view while the Datum B is positioned on the left end of the workpiece also note that the diameter is

Answer:

B

Explanation:

Because The Scientific Method is a way to identify a problem or Solve and in the choice above the best way to describe the scientific method will a procedure to achieve something

The procedure is the part of the Scientific Method that describes the steps of the experiment, therefore the correct answer is the option B

Scientific investigation is the process of looking for answers by doing extensive research and finding the answers through experimental results.

The scientific investigation very much relies on true experimental results that can be supported by evidence.

To establish facts or generate information, the scientific method employs a sequence of stages. Although the general procedure is generally known, the particulars of each stage may vary based on what is being examined and who is performing it. Only questions that can be verified or refuted by testing can be addressed using the scientific method.

Thus, The correct response is option B because the procedure is the component of the scientific method that describes the steps of the experiment.

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Answer: force is a push or pull that results in the movement of an object ..

Explanation:

Hope this helps you!!!!

Explanation:

Given that,

Mass of a box is 2.6 kg

(1) We need to find the work done on the box by this force as it is pushed up the 5.00-m ramp to a height of 3.00 m.

It means that the position of the object is 3 m i.e. h = 3 m

Work done = Fd

= mgh

So,

[tex]W=2.6\times 9.8\times 3\\\\W=76.44\ J[/tex]

(2) Now the gravitational potential energy that the box undergoes as it rises to its final height is equal to the work done by the box. So,

Change in potential energy = 76.44 J

Explanation:

Elongation of the wire is:

ΔL = F L₀ / (E A)

where F is the force,

L₀ is the initial length,

E is Young's modulus,

and A is the cross sectional area.

ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)

ΔL = T (1.25×10⁻⁶ m/N)

T = (80,000 N/m) ΔL

Draw a free body diagram of the mass at the bottom of the circle.  There are two forces: tension force T pulling up and weight force mg pulling down.

Sum of forces in the centripetal direction:

∑F = ma

T − mg = mv²/r

T − mg = mω²r

T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)

T − 147 N = (2368.7 N/m) (0.5 m + ΔL)

Substitute:

(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)

(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL

(797631.3 N/m) ΔL = 1331.35 N

ΔL = 0.00167 m

ΔL = 1.67 mm

Answer:

5.448 g/cm³

Explanation:

Density: This is defined as the ratio of the mass of a body to its volume.

The unit of density is kg/m³ other sub units are g/cm³, mg/mm³.

From the question,

D = m/πr²h......................... Equation 1

Where D = density, m = mass, r = radius, h = height.

Given: m = 9.58 g, r = 8/2 mm = 4 mm = 0.4 cm, h = 3.5 cm

Substitute this values into equation 1

D = 9.58/(3.14×0.4²×3.5)

D = 5.448 g/cm³

Hence the density of the metal rod is 5.448 g/cm³

Calculate the mass of the stone.

Kinetic energy = 1/2 × m × v²

We know that,

Kinetic energy = 1/2 × m × v²

⇒ 1892 = 1/2 × m × ( 16 )²

⇒ 1892 = 1/2 × m × 256

⇒ 1892 = 128m

⇒ m = 1892/128

⇒ m = 14.78 kg

Answer:

128 degrees

Explanation:

speed divided by distance travelled

the angle he jumped from the ground is 8.92°

The question above is a projectile motion problem, and we can solve it using the formula of range.

⇒ Formula:

⇒ Where:

From the question,

⇒ Given:

⇒Substitute these values into equation 1

⇒ Solve for ∅

Hence the angle he jumped from the ground is 8.92°

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Answer:

The force on the proton is 7.85 x 10⁻¹² N

Explanation:

Given;

kinetic energy of the proton, K.E = 2MeV = 2 x  10⁶ x 1.602 x 10⁻¹⁹ J

= 3.204 x 10⁻¹³ J

magnitude of the magnetic field, B = 2.5 T

The kinetic energy of the proton is given by;

[tex]K.E = \frac{1}{2} m v^2\\\\v^2 = \frac{2K.E}{m}\\\\v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*3.204*10^{-13}}{1.67 *10^{-27}}} \\\\v = 1.959*10^7 \ m/s[/tex]

The force on the proton moving perpendicular to magnetic field is given by;

F = qvB

F = 1.602  x 10⁻¹⁹ x 1.959 x 10⁷ x 2.5

F = 7.85 x 10⁻¹² N

Therefore, the force on the proton is 7.85 x 10⁻¹² N

Complete question:

Check file uploaded for diagram of the apparatus

Answer:

The pressure of the gas inside the apparatus is 1000 mmHg

Explanation:

Given;

outside pressure, Patm = 750 mmHg

change in height of the two columns, Δh = 25 cm = 250 mm

The pressure of the gas inside the apparatus is given by;

P(gas) = P(atm) + P(Hg)

where;

P(atm) is the outside pressure

P(Hg) is the gauge pressure = height difference of the two columns.

P(gas) = 750 mmHg + 250 mmHg

P(gas) = 1000 mmHg

Therefore, the pressure of the gas inside the apparatus is 1000 mmHg

The pressure of the gas inside the apparatus is 1000 mmHg.

Since

outside pressure, Patm = 750 mmHg

change in height of the two columns, Δh = 25 cm = 250 mm

so we know that

The pressure of the gas inside the apparatus is

P(gas) = P(atm) + P(Hg)

here;

P(atm) is the outside pressure

P(Hg) is the gauge pressure = height difference of the two columns.

So,

P(gas) = 750 mmHg + 250 mmHg

P(gas) = 1000 mmHg

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Answer:

a) 50%

b) 100 J

c) 50 J

Explanation:

The cold temperature of the reservoir = [tex]T_{c}[/tex]

according to the problem, it is stated that the hot reservoir of an engine is twice that of the cold reservoir, therefore,

the hot temperature of the reservoir [tex]T_{h}[/tex] = [tex]2T_{c}[/tex]

The work done by the engine = 50 J

a) The max efficiency obtainable from a heat engine η =  [tex]1 - \frac{T_{c} }{T_{h} }[/tex]

since [tex]T_{h}[/tex] = [tex]2T_{c}[/tex], the equation becomes

η =  [tex]1 - \frac{T_{c} }{2T_{c} }[/tex] =

η =  [tex]1 - \frac{1 }{2 }[/tex] = 0.5 = 50%

b) The heat absorbed per cycle will be gotten from

η =  [tex]\frac{W}{Q}[/tex]

η is the efficiency of the system = 0.5

where W is the work done = 50 J

Q is the heat absorbed = ?

substituting, we have

0.5 =  [tex]\frac{50}{Q}[/tex]

Q = 50/0.5 = 100 J

c) The heat rejected per cycle = 50% of the absorbed heat

==> 0.5 x 100 J = 50 J

Answer:

5 n

Explanation:

25 and 25 cancel each other out and 50-45 is 5

Answer:

Physics is the study of matter and the way living things behave everyday....It is related to maths in how we measure the way objects or people do specific things physics has many branches under it that can be helpful too.....Physics teaches us about the world,about the mechanical things we do about air,space,matter about the solar system and about simple machines and things that we do everyday in life.... What we do everyday is related to physics....

Answer:

R_total = 14.57 Ω ,  V_C = 1.176 V

Explanation:

To solve this circuit we are going to find the equivalent resistance of each branch, let's remember

* Serial resistance  

         [tex]R_{eq}[/tex] = ∑ [tex]R_{i}[/tex]

* For resistance in parallel

        1 / R_{eq} = ∑ 1/R_{i}

We solve the two branches of the wheatstone bridge

Series resistors

Branch B

         R_B = Rb + R4

         R_B = 2 + 18

         R_B = 20 Ω

Branch C

         R_C5 = Rc + R5

         R_C5 = 3 + 12

         R_C5 = 15 Ω

Resistance in parallel R_B and R_C5

         1 / R_BC = 1 / R_B + 1 / R_C5

          1 / R_BC = 1/20 + 1/15 = 0.116666

          R_BC = 8.57 Ω

Now we have a single branch, we solve the series resistance

          R_total = R_A + R_BC

          R_total = 6 + 8.57

          R_total = 14.57 Ω

b) they ask us for the voltage in the resistance R_C

Let's remember that the voltage in a series circuit is the sum of the voltages

           10 = V_a + V_BC

           10 = i R_a + i R_BC = i (R_a + R_BC)

           i = 10 / (R_a + R_BC)

           i = 10 / (14.57)

           i = 0.6863 A

The current in the series circuit is constant

          V_BC = i R_BC

          V_BC = 0.6863 8.57

          V_BC = 5.8819 V

This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.

     Branch C

             V_BC = i R_C5

             i = V_BC / R_C5

             i = 5.8819 / 15

             i = 0.39213 A

In this branch we have two resistors in series, let's remember that the current in a series circuit is constant

             V_C = i R_C

              V_C = 0.39213 3

              V_C = 1.176 V

Answer:

The depth will be equal to 6141.96 m

Explanation:

pressure on the submarine [tex]P_{sea}[/tex] = 62 MPa = 62 x 10^6 Pa

we also know that [tex]P_{sea}[/tex] = ρgh

where

ρ is the density of sea water = 1029 kg/m^3

g is acceleration due to gravity = 9.81 m/s^2

h is the depth below the water that this pressure acts

substituting values, we have

[tex]P_{sea}[/tex] = 1029 x 9.81 x h = 10094.49h

The gauge pressure within the submarine [tex]P_{g}[/tex] = 101 kPa =  101000 Pa

this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.

Equating the pressure [tex]P_{sea}[/tex], we have

62 x 10^6 = 10094.49h

depth h = 6141.96 m

Answer:

1.187 x 10^8 m/s

Explanation:

the potential of the electric field V = 40 kV = 40000 V

the charge on an electron e = 1.6 x 10^-19 C

The energy of an accelerated electron in an electric field is given as

E = eV

E = 1.6 x 10^-19 x 40000 = 6.4 x 10^-15 J

This energy is equal to the kinetic energy with which the electron moves,  according to the conservation of energy.

The kinetic energy = [tex]\frac{1}{2}mv^{2}[/tex]

where

m is the mass of the electron = 9.109 x 10^-31

v is the speed of the electron.

Equating the energy, we have

6.4 x 10^-15 = [tex]\frac{1}{2}*9.109*10^-31*v^{2}[/tex]

6.4 x 10^-15 = 4.55 x 10^-31 [tex]v^{2}[/tex]

[tex]v^{2}[/tex] = 1.41 x 10^16

[tex]x^{2} v = \sqrt{1.41*10^{16}}[/tex] = 1.187 x 10^8 m/s

Answer:

Fluid Flow is a part of fluid mechanics and deals with fluid dynamics. Fluids such as gases and liquids in motion are called fluid flow. It involves the motion of a fluid subjected to unbalanced forces. This motion continues as long as unbalanced forces are applied.

Difference:

Whereas in real fluids velocity varies across the section. But when the size and shape of cross section are constant along the length of channels under consideration, the flow is said to be uniform. A non-uniform flow is one in which velocity is not constant at a given instant.

Answer:

The magnitude of the rate of change of the flux through the coil is 2.82 x 10⁻³ T.m²/s

Explanation:

Given;

radius of coil, r = 8 cm = 0.08 m

number of turns of the coil, N = 110 turns

the induced emf through the coil, E = -310mV

The induced emf through the coil is given by;

[tex]E = - N\frac{d \phi}{dt} \\\\\frac{d \phi}{dt} = \frac{E}{-N}[/tex]

Where;

dФ/dt is the magnitude of the rate of change of the flux through the coil

[tex]\frac{d\phi}{dt} = \frac{-0.31}{110} \\\\\frac{d\phi}{dt} =2.82*10^{-3} \ T.m^2/s[/tex]

Therefore, the magnitude of the rate of change of the flux through the coil is 2.82 x 10⁻³ T.m²/s

Answer:

the distance d traveled by the lighter fragment is 58.1 m.

Explanation:

mass of the lighter fragment = m

the lighter fragment traveled a distance = ?

mass of the heavier fragment = 7m

the distance covered by the heavier fragment = 8.30 m

The two particles will be given the same amount of energy from the explosion. This energy is used to do work by the two fragments.

work done by heavier fragment w = mgd

where m is the mass

g is acceleration due to gravity

d is the distance traveled.

substituting, the work done by the heavier fragment is

w = 7m x g x 8.3 = 58.1mg

The same way, the lighter fragment does work of

w = mgd

equating the two work done since they are given the same amount of energy from the explosion, we have

58.1mg = mgd

mg cancels out, we have

the distance d traveled by the lighter fragment d = 58.1 m

Answer:

Its not A..

Explanation:

I chose A - was incorrect

Answer:

the sum of all force being applied to an object.

Explanation:

Answer:

The magnitude of the acceleration is 5.11 m/s²

Explanation:

Given;

Tension on the rope, T = 164 N

mass of the bucket, m = 11 kg

Weight of the rope is given by;

W = mg = 11 x 9.8 = 107.8 N

According to Newton's second law of motion, the tension on the rope is given by;

T = W - ma

ma = W - T

ma = 107.8N - 164N

ma = -56.2 N

a = -56.2 / m

a = -56.2 / 11

a = -5.11 m/s²

The magnitude of the acceleration is 5.11 m/s²

Answer;

Cos²စ= I/Io

So

A. Cos²စ = 1/2.2

Cosစ= √1/2.2

စ = cos^-1 0.68

= 47.2°

B.

Cosစ = √1/5.2

စ = ,cos^-1 0.4385

= 64°

C.

Cosစ = √1/12

စ = cos^-1 0.2886

= 73.2°

Answer:

Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.

Explanation:

Given;

mass of first marble, m₁ = 50g = 0.05 kg

initial velocity of the first marble, u₁ = 2 m/s

mass of second marble, m₂ = 20 g = 0.02 kg

initial velocity of the second marble, u₂ = 0

final velocity of the first marble, v₁ = 1 m/s

Let the final velocity of the second marble, = v₂

Determine the final velocity of the second marble by applying principle of momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

0.05 x 2 + 0.02 x 0 = 0.05 x 1 + 0.02v₂

0.1 = 0.05 + 0.02v₂

0.02v₂ = 0.1 - 0.05

0.02v₂ = 0.05

v₂ = 0.05 / 0.02

v₂ = 2.5 m/s

During inelastic collision both objects will move at the same velocity after collision.

During elastic collision both objects will move at different velocities after collision.

Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.

Answer:

B) approaches infinity

Explanation:

The capacitive reactance of an AC capacitor is given by;

[tex]X_C = \frac{1}{\omega C } = \frac{1}{2\pi f C}[/tex]

Where;

C is the capacitance

f is the frequency of the ac voltage

[tex]X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi (0) C} \\\\X_C = \frac{1}{0} \\\\X_C = \ infinity[/tex]

Therefore, as the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor approaches infinity.

The correct option is (B) approaches infinity

Answer:

The mass of the chunk of gold is 241.25 g

Explanation:

Since density is defined as mass divided by volume, we can solve for the mass (m) via its equation:

[tex]density=\frac{mass}{volume} \\19.3=\frac{m}{12.5} \\m=19.3\,(12.5)\, grams\\m = 241.25 \,\,g[/tex]

Answer:

(a) When the resistance R is doubled, I = 1 A

(b) When the peak emf εo is doubled, I = 4 A

(c)  When the frequency ω is doubled, I = 2 A

Explanation:

Given;

peak current through the resistor, I = 2.0 A

According to ohms law the peak current through the circuit is given by;

[tex]I = \frac{V}{R}[/tex]

(a) When the resistance R is doubled;

[tex]I = \frac{V_R}{R} \\\\I_1R_1 = I_2R_2\\\\I_2 = \frac{I_1R_1}{R_2} \\\\I_2 = \frac{2*R_1}{2R_1} \\\\I_2 = 1 \ A[/tex]

(b)When the peak emf εo is doubled

[tex]I = \frac{V}{R} = \frac{\epsilon_o}{R} \\\\R = \frac{\epsilon_ o}{I} \\\\\frac{\epsilon_ o_1}{I_1} = \frac{\epsilon_ o_2}{I_2} \\\\I_2 = \frac{\epsilon_ o_2 *I_1}{\epsilon _o_1} \\\\I_2 = \frac{2 \epsilon_ o_1 *2}{\epsilon _o_1} \\\\I_2 = 4 \ A[/tex]

(c) When the frequency ω is doubled

Peak current through resistor is independent of frequency

I₂ = 2.0 A