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Otrada [13]
3 years ago
14

A brass washer has an outside diameter of 4.50 cm with a hole of diameter 1.25 cm and is 1.50 mm thick. The density of brass is

8599 kg/m3. If you put this washer on a laboratory balance, what will it "weigh" in grams?
Physics
1 answer:
-Dominant- [34]3 years ago
6 0
We will find the mass from 
mass = density x volume 
We are told the density and must find the volume from the dimensions given 
the volume of the washer will be the area x thickness (remembering to convert all measurements to meters) 
if the washer had no hole, its area would be pi (0.0225m)^2 (remember to convert to meters and to use radius) 
the area of the hole is pi(0.00625m)^2 
so the area of the washer is pi[(0.0225m)^2 - (0.00625m)^2] = 1.5x10^-3 m 
the volume of the washer is 1.5x10^-3 m x 1.5x10^-3 m = 2.25x10^-6 m^3 (the thickness of the washer is 1.5 mm = 1.5x10^-3m) 
thus, the mass of the washer = 8598kg/m^3 x 2.25x10^-6m^3 = 0.0189kg = 18.9 grams
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A projectile is launched vertically from the surface of the Moon with an initial speed of 1360 m/s. At what altitude is the proj
LenaWriter [7]

Answer:

485520 m

Explanation:

v_{o} = initial velocity of the projectile = 1360 m/s

v_{f} = final velocity of the projectile = \left ( \frac{2}{5} \right )v_{_{o}} = \left ( \frac{2}{5} \right )(1360) = 544 m/s

a = acceleraton due to gravity on moon = - 1.6 m/s²

h = Altitude of the projectile

Using the kinematics equation

v_{f}^{2} = v_{o}^{2} + 2 a h

Inserting the values

544^{2} = 1360^{2} + 2 (-1.6) h

h = 485520 m

3 0
3 years ago
15 to who helps me out!
s2008m [1.1K]

Answer:

a. same

b. less

c. same

d. same?

Explanation:

the mass will always be the same no matter where it is. the weight however depends on the gravity.

7 0
3 years ago
A battery is replaced with one of lower emf. State and explain how the resistance of the lamps would have to change in order to
kow [346]

The brightness of the lamp is proportional to the current flowing through the lamp: the larger the current, the brighter the lamp.

The current flowing through the lamp is given by Ohm's law:

I=\frac{V}{R}

where

V is the potential difference across the lamp, which is equal to the emf of the battery, and R is the resistance of the lamp.

The problem says that the battery is replaced with one with lower emf. Looking at the formula, this means that V decreases: if we want to keep the same brightness, we need to keep I constant, therefore we need to decrease R, the resistance of the lamp.

3 0
3 years ago
Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum
cestrela7 [59]

Answer:

T_{2}=278.80 K

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

Let's recall that the term nR is a constant. That is why we can match these equations.  

We can find a relation between the volumes of the initial and the final state.

\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}

Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40

Finally, T2 will be:

T_{2}=278.80 K

6 0
3 years ago
In terms of saturation (unsaturated, saturated, super-saturated). How would you classify the following?
miskamm [114]

<u>Answer:</u>

<em>1. A NaCl solution with a concentration of 50g/100mL of water at 40°C:</em> The NaCl solution with a given concentration is saturated at this temperature .As the temperature increases the solution will more dissolves.

<em>2. A sugar solution with a concentration of 200g/100mL of water at 40°C: </em>The sugar solution with a given concentration is saturated at this temperature. As the temperature increases the solution will more dissolves.

<em>3. A sugar solution with a concentration of 240g/100mL of water at 40°C:</em> The sugar solution with a given concentration is saturated at given temperature.

7 0
3 years ago
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