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2 years ago

Calculate the mass of a gas with a density of 0.0065 g/cm3 and volume of 260 cm3

Physics

1 answer:

Anuta_ua [19.1K]2 years ago

6 0

$\sf Density = \frac{mass}{volume }\\\\0.0065 = \frac{mass}{260} \\\\{\boxed {\sf {mass =1.69 ~grams}}$

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Un coche inicia un viaje de 450 km a las ocho de la mañana con una velocidad media de 90 km/h. ¿A qué hora llegará a su destino?

Artyom0805 [142]

Answer:

Llegara a su destino a la 1:00 pm

Explanation:

Si el coche va a 90 km/h buscamos un numero q al multiplicarlo por 90 nos de 450. Entonces 90×5 = 450, si hacemos la cuenta desde las ocho de la mañana mas las 5 horas del viaje terminaria llegando a su destino a la 1:00 pm.

5 0

2 years ago

The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m

Julli [10]

Answer:

$v=(6ti+6k)\ m/s$

Explanation:

Given that,

The position of a particle is given by :

$r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m$

Let us assume we need to find its velocity.

We know that,

$v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s$

So, the velocity of the particle is $(6ti+6k)\ m/s$ .

5 0

2 years ago

Two charged particles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged par

Murrr4er [49]

Answer:

X₃₁ = 0.58 m and X₃₂ = -1.38 m

Explanation:

For this exercise we use Newton's second law where the force is the Coulomb force

F₁₃ - F₂₃ = 0

F₁₃ = F₂₃

Since all charges are of the same sign, forces are repulsive

F₁₃ = k q₁ q₃ / r₁₃²

F₂₃ = k q₂ q₃ / r₂₃²

Let's find the distances

r₁₃ = x₃- 0

r₂₃ = 2 –x₃

We substitute

k q q / x₃² = k 4q q / (2-x₃)²

q² (2 - x₃)² = 4 q² x₃²

4- 4x₃ + x₃² = 4 x₃²

5x₃² + 4 x₃ - 4 = 0

We solve the quadratic equation

x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5

x₃ = [-4 ± 9.80] 10

X₃₁ = 0.58 m

X₃₂ = -1.38 m

For this two distance it is given that the two forces are equal

7 0

2 years ago

If the intensity of a loud car horn is 0.005 W/m^2 when you are 2 meters away from the source. Calculate the sound intensity lev

nekit [7.7K]

Answer:

Explanation:

We have given the intensity of the loud car horn $I=0.005w/m^2$

We know that $I_O=10^{-12}w/m^2$

Now the sound intensity level is given by $\beta =10log\frac{I}{I_0}=10log\frac{0.005}{10^{-12}}=96.98dB$ , which is nearly equal to 97

So the sound intensity level will be 97 dB

So option (c) will be the correct option

4 0

2 years ago

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

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The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

2 years ago