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Len [333]
4 years ago
7

From measurements on a uniformly sized material from a dryer, it was found that the surface area of the material is 1200 m2. If

the density of the material is 1450 kg m-3 and the total weight is 360 kg calculate the equivalent diameter of the particles.
Physics
1 answer:
amm18124 years ago
3 0

Answer:

D = 2.07\times 10^{-4} m

Explanation:

We know that the volume of the material

= mass/density

mass= 360 Kg

density = 1450 kg/m^3

Volume= 360/1450= 0.248 m^3

and Volume = Area× equivalent Diameter

Area= 1200 m^2

⇒Equivalent Diameter= Volume/Area= 0.248/1200=

D = 2.07\times 10^{-4} m

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Sergio039 [100]
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4 0
3 years ago
Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star.
Vsevolod [243]

Answer:

The angular speed of the neutron star is 3130.5 rad/s.

Explanation:

Given that,

Initial radius r_{1}=7\times10^{5}\ km

Final radius r_{2}=18 km

Density of a neutron \rho= 10^{14}

Equal masses of two stars m_{1}=m_{2}

Suppose, If the original star rotated once in 35 days, find the angular speed of the neutron star

Time period of original star T =  35 days = 3024000 s

We need to calculate the initial angular speed of original star

Using formula of angular star

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

Put the value into the formula

\omega_{1}=\dfrac{2\pi}{3024000}

\omega_{1}=0.00207\times10^{-3}\ rad/s

Let the initial moment of inertia of the star is

I_{1}=m_{1}r_{1}^2

Final moment of inertia of the star is

I_{2}=m_{2}r_{2}^2

From the conservation of angular momentum

I_{1}\omega_{1}=I_{2}\omega_{2}

\omega_{2}=\dfrac{I_{1}\omega_{1}}{I_{2}}

\omega_{2}=\dfrac{m_{1}r_{1}^2\omega_{1}}{m_{2}r_{2}^2}

Put the value into the formula

\omega_{2}=\dfrac{(7.0\times10^{5})^2\times0.00207\times10^{-3}}{18^2}

\omega_{2}=3130.5\ rad/s

Hence, The angular speed of the neutron star is 3130.5 rad/s.

8 0
3 years ago
A hockey goalie is standing on ice. Another player fires a puck (m = 0.170 kg) at the goalie with a velocity of +41.0 m/s. (a) I
viktelen [127]

Answer:

2991.42 N

Explanation:

For this problem, we'll use the equations: momentum= mass x velocity and impulse = change in momentum, and impulse=force x time.

initial momentum; p1 = 0.17 x 41 = 6.97 kg.m/s

final momentum; p2 = 0, because final velocity is 0 m/s

Thus,

impulse = p1 - p2= 6.97 - 0 = 6.97 kg.m/s

Finally, impulse= Force x time,

Thus, Force = Impulse/time

Force= 6.97/ (2.33 x 10^(-3)) = 2991.42 N

4 0
4 years ago
A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x
DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is \lambda

Total charge, Q = 7 nC = 7\times 10^{- 9} C

At x = 2L,

Electric field, \vec{E_{2L}} = 500N/C

Coulomb constant, K = 8.99\times 10^{9} N.m^{2}/C^{2}

Now, we know that:

\vec{E} = K\frac{Q}{x^{2}}

Also the line charge density:

\lambda = \frac{Q}{L}

Thus

Q = \lambda L

Now, for small element:

d\vec{E} = K\frac{dq}{x^{2}}

d\vec{E} = K\frac{\lambda }{x^{2}}dx

Integrating both the sides from x = L to x = 2L

\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx

\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]

\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}

Similarly,

For the field in between the range 2L< x < 3L:

\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx

\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]

\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}

Now,

If at x = 2L,

\vec{E_{2L}} = 500 N/C

Then at x = 3L:

\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C

4 0
4 years ago
Why do storm go to Galveston?​
notsponge [240]

Hurricanes and tropical storms can also spawn tornadoes and microbursts, create storm surges ... If you are unable to evacuate, go to your wind-safe room.

8 0
4 years ago
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