A scientist hypothesizes that the temperature at which an turtle's egg is incubated

A building made with a steel structure is 565 m high on a winter day when the temperature is 0◦F. How much taller is the buildin

anyanavicka [17]

To solve this problem we apply the thermodynamic equations of linear expansion in bodies.

Mathematically the change in the length of a body is subject to the mathematical expression

$\Delta L = L_0 \alpha \Delta T$

Where,

$L_0 =$ Initial Length

$\alpha =$ Thermal expansion coefficient

$\Delta T =$ Change in temperature

Since we have values in different units we proceed to transform the temperature to degrees Celsius so

$0\°F \Rightarrow (0-32)*\frac{5}{9} = -17.77\°C$

$103\°F \Rightarrow (103-32)*(\frac{5}{9})= 39.44\°C$

The coefficient of thermal expansion given is

$\alpha = 1.1*10^{-5}/\°C$

The initial length would be,

$L_0 = 565m$

Replacing we have to,

$\Delta L = L_0 \alpha \Delta T$

$\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))$

$\Delta L = 0.355m$

This means that the building will be 35.5cm taller

3 0

2 years ago

A 2.5m long steel piano wire has a diameter of 0.5cm how great is the tention in the wire if it stretches by 0.45cm when tighten

UkoKoshka [18]

Answer: Their u go i found it their was about 3 pages i did not no what pages u had to do.

Explanation:

Download pdf

7 0

2 years ago

URGENT PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST !!!

kykrilka [37]

Answer:

C would be the answer

Explanation:

Hope this helps! :)

5 0

2 years ago

Read 2 more answers

While playing baseball with your friends your hands begin to sting after you ctach several fast balls.

muminat

Answer:

Explanation:

To stop a ball with high momentum in a small-time imparts a high amount of impact on hands. This is the reason for the stinging of hands.

The momentum of the ball is due to the mass and velocity. To prevent stinging in the hand one needs to lower his hands to increase the time of contact. In this way, the momentum transfer to the hands will be lesser.

7 0

3 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago