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3 years ago

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The dimensions of a room are 16.40 m long, 4.5 m wide and 3.26 m high. What is the volume of the room in cubic meters? Express y

our answer in scientific notation.

1 answer:

anyanavicka [17]3 years ago

3 0

Answer:

the volume of the room is 240.588 meters3

The picture shows the scientific notation

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Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made o

Colt1911 [192]

Answer:

There is a loss of fluid in the container of 0.475L

Explanation:

To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.

The formula that describes this thermal expansion process is given by:

$\Delta V = \beta V_0 \Delta T$

Where,

$\Delta V =$ Change in volume

$V_0 =$ Initial Volume

$\Delta T =$ Change in temperature

$\beta =$ coefficient of volume expansion (Coefficient of copper and of the liquid for this case)

There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

$\Delta V_T = \Delta V_l - \Delta V_c$

Where,

$\Delta V_l$ = Change in the volume of liquid

$\Delta V_c$ = Change in the volume of copper

Then replacing with the previous equation we have:

$\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T$

$\Delta V = (\beta_l-\beta_c)V_0\Delta T$

Our values are given as,

Thermal expansion coefficient for copper and the liquid to 20°C is

$\beta_c = 51*10^{-6}/\°C$

$\beta_l = 400*10^{-6}/\°C$

$V_0 = 16L$

$\Delta T = (95\°C-10\°C)$

Replacing we have that,

$\Delta V = (\beta_l-\beta_c)V_0\Delta T$

$\Delta V = (400*10^{-6}/\°C-51*10^{-6}/\°C)(16L)(95\°C-10\°C)$

$\Delta V = 0.475L$

Therefore there is a loss of fluid in the container of 0.475L

6 0

3 years ago

The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

(c) Escape velocity on earth is $3.563\times 10^4m/sec$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

(c) Escape velocity is given by

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

5 0

3 years ago

Read 2 more answers

How can I feel pain when I hit a table with my bare hands (Physics)

Sidana [21]

When your hand hits the table the table will vibrate and your hand will be numb for two to three seconds

6 0

3 years ago

A box falls out of a stationary helicopter hovering 135 m above the ground. How long will it take to hit the ground?

Marrrta [24]

We have the equation of motion $s = ut + \frac{1}{2} at^2$ , where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.

Here displacement = 135 m, Initial velocity = 0 m/s, acceleration = 9.81 $m/s^2$

Substituting

$135 = 0*t+\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2 =135\\ \\ t =5.25 seconds$

A box falls out of a stationary helicopter hovering 135 m above the ground will take 5.25 seconds to reach the ground.

4 0

3 years ago

The rotor of an electric motor has an angular velocity of 3600 rpm when the load and power are cut off. the 125-lb rotor, which

velikii [3]

Answer:

θ '= 4975 rev

Explanation:

For this exercise let's use the relationship between work and the change in kinetic energy

W = ΔK

the expression for work is

W = - τ Δθ

where the negative sign indicates that the torque is in the opposite direction to rotation

kinetic energy

K = ½ I w²

we substitute

- τ Δθ = 0 - ½ I w²

θ = $\frac{1}{2} \ \frac{I w^2}{\tau}$

if we approach the rotor to a cylinder with an axis of rotation through its center

I = ½ m r²

we substitute

θ = ½ (½ m r²) $\frac{w^2}{\tau}$

How the measurements are in the English system, the weight

W = m g

m = W / g

let's reduce to the english system

w = 3600 rev / min (2pi rad / 1rev) (1 min / 60s) = 376.99 rad / s

r = 9 in (1 ft / 12in) = 0.75 ft

let's calculate

m = 125/32 = 3.91 slug

θ = ¼ 3.91 0.75² 376.99² / 2.5

θ = 3.126 10⁴ rad

let's reduce to revolutions

θ’= 3.126 10⁴ rad (1rev / 2π rad)

θ’= 4974.9 rev

θ '= 4975 rev

6 0

3 years ago