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Crazy boy [7]
2 years ago
5

Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o

f air is 1.29 kg/m3kg/m3, and the density of aluminum is 2.7×103kg/m32.7×103kg/m3.
Physics
1 answer:
spin [16.1K]2 years ago
6 0

Answer:

The true weight of the aluminium is m_{alu} = 4.5021 kg

Explanation:

Given data

m_{app} = 4.5 kg

\rho_{air} = 1.29 \frac{kg}{m^{3} }

\rho_{al} = 2.7× 10^{3} \frac{kg}{m^{3} }

The true mass of the aluminium is given by

m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }

Put all the values in above equation we get

m_{alu} = \frac{(2700)(4.5)}{2700-1.29}

m_{alu} = 4.5021 kg

Therefore the true weight of the aluminium is m_{alu} = 4.5021 kg

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Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:45 A.M., bu
Veronika [31]

Answer:

Jari

Explanation:

The question requires to know who is traveling faster. This is done by comparing the gradients. The steeper the slope (high gradient), the faster the speed and vice versa.

From Jari's line, the starting point is (0, 0) and another point is (6, 7)

The gradient being change in y to change in x

Change in y=7-0=7

Change in x=6-0=6

Slope is 7/6

For Jade, first point is (0, 10) then another point is (6, 16)

Change in y=16-10=6

Change in x=6-0=6

Slope is 6/6=1

Clearly, 7/6 is greater than 6/6 or 1 hence Jari is faster than Jade

3 0
2 years ago
At room temperature, a typical person loses energy to the surroundings at the rate of 62 W. If this energy loss has to be made u
Alex_Xolod [135]

To solve this problem it is necessary to use the given proportions of power and energy, as well as the energy conversion factor in Jules to Calories.

The power is defined as the amount of energy lost per second and whose unit is Watt. Therefore the energy loss rate given in seconds was

P = \frac{E}{t} \rightarrow E= Energy, t = time

P = 62W = 62 \frac{J}{s}

The rate of energy loss per day would then be,

P = 62\frac{J}{s} (\frac{86400s}{1day})

P = 5356800 \frac{J}{day}

That is to say that Energy in Jules per lost day is 5356800J

By definition we know that 1KCal = 4.184*10^{6}J

In this way the energy in Cal is,

E = 5356800J \frac{1KCal}{4.184*10^{6}J}

E = 1279.694 KCal

The number of kilocalories (food calories) must be 1279.694 KCal

4 0
2 years ago
What landforms form at convergent boundaries where two oceanic plates collide?
Bad White [126]
It forms mountains and sometimes islands.
4 0
2 years ago
Read 2 more answers
a crane lifts a 35000 N steel girder a distance of 25 m in 45 s. How much power did the crane require to lift the girder
Angelina_Jolie [31]
Hello.

The formula for Power is Work divided by Time; however, we do not have our value for Work - yet.
To find for the Work inputted, we need to use its formula: Force * Distance.

Let's multiply our Force by our Distance. Remember that our Force is always  measured in Newtons (N), and our Distance is measured by Meters (M).
35,000 * 25 = 875,000 J (Unit for Work is "J" or "Joules")

Now that we have the value for Work, let's apply it to our Power formula.
P = 875,000 / 45; 19,444.44~

The Power required to lift the girder is 1944.44~ W (Unit for Power is "W" or "Watts").

I hope this helps!
7 0
2 years ago
jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en
d1i1m1o1n [39]

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

\theta = \frac{2\pi}{6}

\theta = \frac{\pi}{3}

so the angular speed of the train is given as

\omega = \frac{\theta}{t}

\omega = \frac{\pi}{30} rad/s

now we have time period of the train given as

T = \frac{2\pi}{\omega}

T = \frac{2\pi}{\frac{\pi}{30}}

T = 60 s

5 0
2 years ago
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