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2 years ago

Which tool would you use to measure the amount of rainfall?

Physics

2 answers:

alina1380 [7]2 years ago

8 0

Answer:

A. Graduated Cylinder

Explanation:

A thermometer would just tell you the temperature.

A scale would just tell you how much it weighs.

And a stop watch would do nothing but show numbers until it got wet enough and broke.

The only logical answer is A since it is also a measuring flask WITH incriments already on the side of it to tell you how much water you collect.

Hope this helps!!

zheka24 [161]2 years ago

5 0

Answer:

Explanation:

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How much total heat transfer is necessary to lower the temperature of 0.175 kg of steam from 125.5 °C to −19.5 °C, including the

GaryK [48]

Answer:

Explanation:

The heat required to change the temperature of steam from 125.5 °C to 100 °C is:

$Q_1 = ms_{steam} (125.5^0C - 100^0C) \\ \\ Q_1 = 0.175 \ kg ( 1520 \ J/kg.K ) (25.5^0 \ C) \\ \\ Q_1 = 6783 \ J$

The heat required to change the steam at 100°C to water at 100°C is;

$Q_2 = mL_v \\ \\ Q_2 = (0.175 \ kg) (2.25*10^6 \ J/kg ) \\ \\ Q_2 = 393750 \ J$

The heat required to change the temperature from 100°C to 0°C is

$Q_3 = ms_{water} (100^) \ C) \\ \\ Q_3 = (0.175\ kg)(4186 \ J/kg.K) (100 ^0c ) \\ \\ Q_3 = 73255 \ J$

The heat required to change the water at 0°C to ice at 0°C is:

$Q_4 = mL_f \\ \\ Q_4 = (0.175 \ kg)(3.34*10^5 \ J/kg) \\\\ Q_4 = 58450 \ J$

The heat required to change the temperature of ice from 0°Cto -19.5°C is:

$Q_5 = ms _{ice} (100^0 C) \\ \\ Q_5 = (0.175 \ kg)(2090 \ J/kg.K)(19.5^0C) \\ \\ Q_5 = 7132.125 \ J$

The total heat required to change the steam into ice is:

$Q = Q_1 + Q_2 + Q_3 + Q_4 +Q_5 \\ \\Q = (6788+393750+73255+58450+7132.125)J \\ \\ Q = 539325.125 \ J \\ \\ Q = 5.39*10^5 \ J$

The time taken to convert steam from 125 °C to 100°C is:

$t_1 = \frac{Q_1}{P} = \frac{6738 \ J}{835 \ W} = 8.12 \ s$

The time taken to convert steam at 100°C to water at 100°C is:

$t_2 = \frac{Q_2}{P} =\frac{393750}{834} =471.56 \ s$

The time taken to convert water to 100° C to 0° C is:

$t_3 = \frac{Q_3}{P} =\frac{73255}{834} = 87.73 \ s$

The time taken to convert water at 0° to ice at 0° C is :

$t_4 = \frac{Q_4}{P} =\frac{58450}{834} = 70.08 \ s$

The time taken to convert ice from 0° C to -19.5° C is:

$t_5 = \frac{Q_5}{P} =\frac{7132.125}{834} = 8.55 \ s$

5 0

3 years ago

Find the mass of a sample that has a dersity of 2 g/mL and a volume of 10 mL

In-s [12.5K]

Answer:

mass = volume x density.

Explanation:

2 x 10 = 20g

4 0

2 years ago

A circular wooden loop of mass m and radius R rest on a flat horizontal friction less surface. A bullet, also of mass m, and mov

bearhunter [10]

Answer:

w = vR/3

Explanation:

The centre of mass of the loop to bullet system is given by D / 4 from centre of loop, which is equivalent to R / 2 from its centre.

From the principle of conservation of linear momentum , we have

m*v = 2*m* Vcm

Where v = velocity of bullet, Vcm = velocity of wood

Hence, we have

Vcm = v2

Also, from the conservation of angular momentum about the centre of mass.

M*V*(R/2) = Ic*w - equation (I)

where Ic = moment of inertia and w = angular velocity

Ic for a ring is given by $mr^2 + m(r/2)^2$

Ic of a bullet is given by $m(r/2)^2$

Hence, the moment of inertia of the system is given by the summation of the two moments of inertia Ic(ring) + Ic(bullet) which gives

Ic(system) = $3*m*R^2/2$

Substituting back into equation (I), we have

$m*v*R^2=3*m*R^2*w/2$

Hence, we obtain w =vR/3

w=v3R

4 0

3 years ago

A swan on a lake gets airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.40

Veseljchak [2.6K]

Answer:

53.895 m.

Explanation:

Using the equation of motion,

v² = u² + 2as .............. Equation 1

Where v = final velocity of the swan, u = initial velocity of the swan, a = acceleration of the swan, s = distance covered by the swan.

make s the subject of the equation,

s = (v² - u²)/2a----------- Equation 2

Given: v = 6.4 m/s, u = 0 m/s ( from rest) a = 0.380 m/s².

Substitute into equation 2

s = (6.4²-0²)/(2×0.380)

s = 40.96/0.76

s = 53.895 m.

Hence the swan will travel 53.895 m before becoming airborne.

6 0

3 years ago

A gas cloud surrounds a dying star against a dark background. The star heats the gases in the cloud. What type of spectrum would

8090 [49]

There are several different types of spectrums that you could expect to find from
the gas cloud, but the best option from the list would be "<span>high-frequency spectrum".</span>

3 0

3 years ago