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3 years ago

An iron railroad rail is 800 ft long when the temperature is 31°C. What is its length (in ft) when the temperature is −17°C?

Physics

1 answer:

natta225 [31]3 years ago

7 0

Answer:

799.54 ft

Explanation:

Linear thermal expansion is:

ΔL = α L₀ ΔT

where ΔL is the change in length,

α is the linear thermal expansion coefficient,

L₀ is the original length,

and ΔT is the change in temperature.

Given:

α = 1.2×10⁻⁵ / °C

L₀ = 800 ft

ΔT = -17°C − 31°C = -48°C

Find: ΔL

ΔL = (1.2×10⁻⁵ / °C) (800 ft) (-48°C)

ΔL = -0.4608

Rounded to two significant figures, the change in length is -0.46 ft.

Therefore, the final length is approximately 800 ft − 0.46 ft = 799.54 ft.

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An engine causes a car to move 10 meters with a force of 100 N. The engine produces 10,000 J of energy. What is the efficiency o

kodGreya [7K]

Answer:

Part A

The efficiency of the engine is 10%

Part B

The change in internal energy is 300 J

Part C

The change in volume is 1 m³ which is one cubic meter

Explanation:

Part A

Efficiency is defined as the ratio of energy output to energy input;

The given parameters are;

The distance the car is moved, d = 10 meters

The force which moves the car, F = 100 N

The amount of energy produced by the engine = 1,000 J

Therefore, we have;

The energy output to the car = The work done on the car = Force applied to the car, F × The distance the car moves, d;

∴ The energy output to the car by the engine = F × d = 100 N × 10 m = 1,000 J

The energy input from the engine = The energy produced by the engine = 10,000 J

The efficiency of the engine = (The energy output)/(The energy input)= 1,000J/10,00J = 0.1

The efficiency in percentage = 0.1 × 100 = 10 %

The efficiency of the engine = 10%

Part B

The amount of heat added to the substance, ΔQ = 1,000J

The amount of work the substance does on the atmosphere, W = 700 J

The change in internal energy, ΔU is given as follows;

ΔQ = W + ΔU

∴ ΔU = ΔQ - W

For the substance, we have;

The change in internal energy, ΔU = 1,000 J - 700 J = 300 J

Part C

The work done by the piston, W = 1,000 J

The pressure, in the piston, P = 1,000 Pa = constant

The work done by the piston in a constant pressure process, W = P × ΔV

Where;

W = The work done

P = The constant pressure applied

ΔV = The change in volume = V₂ - V₁

V₂ = The final volume

V₁ = The initial volume

∴The change in volume ΔV = W/P = 1,000 J/(1,000Pa) = 1 m³

The change in volume ΔV = 1 m³

3 0

3 years ago

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

3 years ago

3, A 4kg block is pushed 2m at an acceleration of 0.2m/s square up a vertical wall by a constant force F applied at an angle of

Andrews [41]

The work done by the applied force on the block against the frictional force is 15.75 J.

<h3>Work done by the applied force</h3>

The work done by the applied force is calculated as follows;

W = Fd

F - Ff = ma

where;

F is applied force
Ff is frictional force

Fcos(37) - μmgsin(37) = ma

Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)

0.799F - 7.077 = 0.8

F = 9.86 N

W = Fdcosθ

W = 9.86 x 2 x cos(37)

W = 15.75 J

Thus, the work done by the applied force on the block against the frictional force is 15.75 J.

Learn more about work done here: brainly.com/question/25573309

#SPJ1

5 0

2 years ago

Calculate the velocity of a 1650kg satellite that is in a circular orbit of 4.2 x 10^6m above the surface of a planet which has

Anastasy [175]

-- We're going to be talking about the satellite's speed.
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.

-- The mass of the satellite makes no difference.

Since the planet's radius is 3.95 x 10⁵m and the satellite is
orbiting 4.2 x 10⁶m above the surface, the radius of the
orbital path itself is

   (3.95 x 10⁵m) + (4.2 x 10⁶m)

   = (3.95 x 10⁵m) + (42 x 10⁵m)

   = 45.95 x 10⁵ m

The circumference of the orbit is (2 π R) = 91.9 π x 10⁵ m.

The bird completes a revolution every 2.0 hours,
so its speed in orbit is

   (91.9 π x 10⁵ m) / 2 hr

= 45.95 π x 10⁵ m/hr x (1 hr / 3,600 sec)

=    0.04 x 10⁵ m/sec

      =    4 x 10³    m/sec

   (4 kilometers per second)

6 0

3 years ago

During a compression at a constant pressure of 290 Pa, the volume of an ideal gas decreases from 0.62 m3 to 0.21 m3. The initial

Aloiza [94]

Answer:

a) -41.1 Joule

b) 108.38 Kelvin

Explanation:

Pressure = P = 290 Pa

Initial volume of gas = V₁ = 0.62 m³

Final volume of gas = V₂ = 0.21 m³

Initial temperature of gas = T₁ = 320 K

Heat loss = Q = -160 J

Work done = PΔV

⇒Work done = 290×(0.21-0.62)

⇒Work done = -118.9 J

a) Change in internal energy = Heat - Work

ΔU = -160 -(-118.9)

⇒ΔU = -41.1 J

∴ Change in internal energy is -41.1 J

b) V₁/V₂ = T₁/T₂

⇒T₂ = T₁V₂/V₁

⇒T₂ = 320×0.21/0.62

⇒T₂ = 108.38 K

∴ Final temperature of the gas is 108.38 Kelvin

5 0

3 years ago