Potassium is a crucial element for the healthy operation of the human

The Eiffel tower in Paris is 300 meters tall on a cold day (T = -24 degrees Celsius), what is its height on a hot day when the t

Varvara68 [4.7K]

Answer:

Length of Eiffel tower, when the temperature is 35 degrees = 300.21 m

Explanation:

Thermal expansion is given by the expression

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

Here length of Eiffel tower, L = 300 m

Coefficient of thermal expansion, α = 0.000012 per degree Celsius

Change in temperature, = 35 - (-24) = 59degrees Celsius

Substituting

$\Delta L=L\alpha \Delta T= 300\times 0.000012\times 59=0.2124m \\$

Length of Eiffel tower, when the temperature is 35 degrees = 300 + 0.2124 = 300.21 m

3 0

3 years ago

How is work involved in stopping a car?

vichka [17]

You have to move your foot to stop the car so I guess that would be considered work by moving your foot

5 0

3 years ago

Read 2 more answers

Two moles of helium are initially at a temperature of 21.0 ∘Cand occupy a volume of 3.30×10−2 m3 . The helium first expands at c

Anettt [7]

Answer:

(B) The total internal energy of the helium is 4888.6 Joules

(C) The total work done by the helium is 2959.25 Joules

(D) The final volume of the helium is 0.066 cubic meter

Explanation:

(B) ∆U = P(V2 - V1)

From ideal gas equation, PV = nRT

T1 = 21°C = 294K, V1 = 0.033m^3, n = 2moles, V2 = 2× 0.033=0.066m^3

P = nRT ÷ V = (2×8.314×294) ÷ 0.033 = 148140.4 Pascal

∆U = 148140.4(0.066 - 0.033) = 4888.6 Joules

(C) P2 = P1(V1÷V2)^1.4 =148140.4(0.033÷0.066)^1.4= 148140.4×0.379=56134.7 Pascal

Assuming a closed system

(C) Wc = (P1V1 - P2V2) ÷ 0.4 = (148140.4×0.033 - 56134.7×0.066) ÷ 0.4 = (4888.6 - 3704.9) ÷ 0.4 = 1183.7 ÷ 0.4 = 2959.25 Joules

(C) Final volume = 2×initial volume = 2×0.033= 0.066 cubic meter

6 0

3 years ago

A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes

OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

$y(t) = h +u_y t - \frac{1}{2}gt^2$

where

h = 2.5 km = 2500 m is the initial height

$u_y = 0$ is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

$y(t) = 0$

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

$0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s$

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

$v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s$

So the horizontal distance travelled is

$d=v_x t$

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

$d=(333.3)(22.6)=7533 m = 7.5 km$

7 0

3 years ago

A bag of groceries is on the back seat of your car as you stop for a stop light. The bag does not slide. Apply your analysis to

Romashka [77]

Answer:

This can be the FBD of the bag.

(my bag looks more like a box tho ^^")

7 0

2 years ago