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alexgriva [62]
4 years ago
5

. If 540.0 mL of nitrogen at 0.00 °C is heated to a temperature of 100.0 °C what will be the new volume of the gas?​

Physics
2 answers:
Afina-wow [57]3 years ago
8 0

Explanation:

jsjsjsshshsgeyyeyey373772ii2owpqlqlammzbxdh

nevsk [136]3 years ago
4 0

Answer:737.66mL.

Upon hearing, the volume will increase. You first need to convert the degrees to Kelvin then solve the final volume

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konstantin123 [22]
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As the distance between two objects decreases, what happens to the force of gravity between the two objects
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The force of gravity between them increases.
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4 years ago
A 16.5-kg crate starts at rest at the top of a 60.0° incline. The coefficients of friction are μs = 0.400 and μk = 0.300. The cr
irga5000 [103]

Answer:

t = 1.62 s

Explanation:

given,

mass of the block m₁ = 16.5 Kg

m₂ = 8 Kg

angle of inclination = 60°

μs = 0.400 and μk = 0.300

time to slide 2 m = ?

a) let a is the acceleration of the block m₁ downward.

Net force acting on m₂,

F₂ = T - m₂ g

m₂a = T - m₂ g

a = \dfrac{T}{m_2} - g.......(1)

net force acting on m₁

F₁ = m₁g sin(60°) - μ_k m₁g cos (60°) - T

m₁ a = m₁g sin(60°) - μ_k m₁g cos (60°) - T

a = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}.........(2)

from equations 1 and 2

\dfrac{T}{m_2} - g = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}

\dfrac{T}{m_2} +\dfrac{T}{m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)

 T\dfrac{m_1+m_2}{m_2\times m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)

 T = \dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{\dfrac{m_1+m_2}{m_2\times m_1}}

 T = {m_2\times m_1}\dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{{m_1+m_2}}  

T = {16.5\times 8}\dfrac{9.8 + 9.8 sin(60^0) - 0.3\times 9.8 cos (60^0)}{{16.5+8}}

T = 90.61 N

from equation (1)

a = \dfrac{90.61}{8} - 9.8.......(1)

a = 1.52 m/s²

let t is the time taken

Apply,

d = ut + 0.5 a t²

2 = 0 + 0.5 x 1.52 x t²

t = \sqrt{2.63}

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5 0
3 years ago
A deuteron particle consists of one proton and one neutron and has a mass of 3.34x10^-27 kg. A deuteron particle moving horizont
nikklg [1K]

_dThe radius of curvature of a subatomic particle under a magnetic field is given by the following formula:

r=\frac{mv}{qB}

Where:

\begin{gathered} r=\text{ radius} \\ v=\text{ velocity} \\ q=\text{ charge} \\ B=\text{ magnetic field} \end{gathered}

We can determine the quotient between the velocity and the charge of the deuteron particle from the formula. First, we divide both sides by the mass:

\frac{r_d}{m_d}=\frac{v}{q_B_}

Now, we multiply both sides by the magnetic field "B":

\frac{Br_d}{m_d}=\frac{v}{q}

Since the charge of the deuterion is the same as the charge of the proton and the velocity we are considering are the same this means that the quotient between velocity and charge is the same for both particles. Therefore, we can apply the formula for the radius again, this time for the proton:

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And substitute the quotient between velocity and charge:

r_p=\frac{m_p}{B}(\frac{Br_d}{m_d})

Now, we cancel out the magnetic field:

r_p=\frac{m_pr_d}{m_d}

Now, we substitute the values:

r_p=\frac{(1.67\times10^{-27}kg)(0.385m)}{(3.34\times10^{-27}kg)}

Solving the operations:

r_p=0.193m=19.3cm

Therefore, the radius is 19.3 cm.

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Answer:

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