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2 years ago

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A balloon of hydrogen is put into to pressure chamber. The initial pressure and volume of hydrogen is 1 atm and .5 cm3. The pres

sure of the chamber is increased to 2 atm. What is the new volume? Use the formula for Boyle's Law: P1V1 = P2V2

1 answer:

Sedbober [7]2 years ago

3 0

Answer:

2.5 cm³

Explanation:

Use Boyle's Law

1×5=2×V

5=2V

V=5/2=2.5 cm³

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A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to

jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

b) Current at the center of the circular coil $=2.48\times 10^{5}\ A$

c) Current near the center of a solenoid $236.8\ A$

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

Conversion $1\times 10^6 A = 1\ MA$ ,and $2cm=\frac{2}{100}=0.02\ m$

$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

8 0

2 years ago

If the distance d (in meters) traveled by an object in time t (in seconds) is given by the formula d=a+bt2, the si units of a an

ss7ja [257]

Distance , d = a+b $t^2$

The unit of d is in meter and t is in seconds.

So the unit of a a must be meter.

Now we have unit of b $t^2$ is meter.

So unit of b* $second^2$ = meter

Unit of b = meter/ $second^2$

So unit of a = m and unit of b = m/ $s^2$ .

8 0

2 years ago

You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.

valentinak56 [21]

Answer:

$F=6\times 10^{-7}\ N$

Explanation:

Given:

quantity of point charge, $q=-4\times 10^{-9}\ C$

radial distance from the linear charge, $r=0.09\ m$

linear charge density, $\lambda=3\times 10^{-9}\ C.m^{-1}$

<u>We know that the electric field by the linear charge is given as:</u>

$E=\frac{\lambda}{2\pi.\epsilon_0.r}$

$E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}$

$E=150\ N.C^{-1}$

<u>Now the force on the given charge can be given as:</u>

$F=E.q$

$F=150\times 4\times 10^{-9}$

$F=6\times 10^{-7}\ N$

3 0

3 years ago

(I will give brainliest whoever helps me !!)

lorasvet [3.4K]

C. Forces have mass and take up space

3 0

2 years ago

Nataliya [291]

Answer: b

Explanation: the two pieces will repel as both have obtained a static charge.

5 0

3 years ago