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Aleonysh [2.5K]

3 years ago

11

A steel tape (a=1.300 E-5 per Celsius degrees) measures 200.0 m at 15.00 degrees celsius. It’s length at 55.00 degrees celsius i

s

1 answer:

Verizon [17]3 years ago

8 0

Answer:

200.1 m is the answer

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If the diameter of the space station is 730 m , how many revolutions per minute are needed for the "artificial gravity" accelera

nignag [31]

For circular motion we know that

<span>F=ma=v^2/r </span>

<span>Therefore: </span>

<span>v = sqrt (rma) </span>

<span>Also, for cicular motion: </span>

<span>rev/min. = 60v/(2r*pi) </span>

<span>So your equation is: </span>

<span>rev./min = 60sqrt(rma)/(2pi*r) </span>

<span>For the mass (m) we can just use 1 kg.
</span>
rev./min = 60sqrt(730*1*9.8)/(2pi*730) =60sqrt(7154)/(4584.4)
rev./min = 60sqrt(7154)/(4584.4) =1.11 rev/min
<span>
the answer is </span>1.11 rev/min<span>

</span>

4 0

3 years ago

Help me with this question please

vichka [17]

Answer:

Its true i'm pretty sure

Explanation:

7 0

3 years ago

Read 2 more answers

A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr

const2013 [10]

Answer:

$388 cm^3$

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

$pV=const.$

which can also be rewritten as

$p_1 V_1 = p_2 V_2$

In our case, we have:

$p_1 = 75.9 kPa$ is the initial pressure

$V_1 = 639 cm^3$ is the initial volume

$p_2 = 125 kPa$ is the final pressure

Solving for V2, we find the final volume:

$v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3$

7 0

3 years ago

An aeroplane flew at 600mph for 7 hours. How far did it travel?

____ [38]

Answer:

The aeroplane flew 4200 miles for 7 hours.

600 times 7 is 4200.

6 0

2 years ago

A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find

FrozenT [24]

Answer:

The time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

Explanation:

Given :

Current $I = 0.106$ A

Area of plate $A = 36 \times 10^{-4}$ $m^{2}$

Plate separation $d = 4 \times 10^{-3}$ m

(A)

First find the capacitance of capacitor,

$C = \frac{\epsilon _{o} A }{d}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

$C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4} }{4 \times 10^{-3} }$

$C = 7.9 \times 10^{-12}$ F

But $C = \frac{Q}{V}$

Where $Q = It$

$C = \frac{It}{V}$

$V = \frac{It}{C}$

Now differentiate above equation wrt. time,

$\frac{dV}{dt} = \frac{I}{C}$

$= \frac{0.106}{7.9 \times 10^{-12} }$

$= 1.34 \times 10^{10}$ $\frac{V}{s}$

Therefore, the time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

8 0

3 years ago