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ASHA 777 [7]
3 years ago
8

A cube made Of an unknown material has a height of 9 symeters the mass of this cube is 3645 g. calculate the density of this cub

e given this information
Physics
1 answer:
kherson [118]3 years ago
8 0

Answer: 5 \frac{g}{cm^{3}}

Explanation:

The density \rho of a material is given by:

\rho=\frac{m}{V} (1)

Where:

m=3645 g is the mass of the cube

V is the volume of the cube

Now, the volume of a cube is equal to the length L of its edge to the power of 3:

V=L^{3} (2)

If we know L=9 cm, the volume of this cube is:

V=(9 cm)^{3}=729 cm^{3} (3)

Substituting (3) in (1):

\rho=\frac{3645 g}{729 cm^{3}} (4)

\rho=5 \frac{g}{cm^{3}} This is the density of the cube

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What makes a ship float​
stellarik [79]

Answer:

Ships can float because a ship is less dense than that of the water that it floats on.

Explanation:

Hope this helps!

7 0
3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
3 years ago
The velocity of a ball changes from ‹ 9, −6, 0 › m/s to ‹ 8.96, −6.12, 0 › m/s in 0.02 s, due to the gravitational attraction of
Alenkasestr [34]

Answer:

a) a=(-2,-6,0)m/s^2, with a magnitude of 6.3m/s^2

b) \frac{\Delta p}{\Delta t}=(-0.24,-0.72,0)Kgm/s^2, with a magnitude of 0.76Kgm/s^2

c) F=(-0.24,-0.72,0)N, with a magnitude of 0.76N

Explanation:

We have:

v_{ix}=9m/s, v_{iy}=-6m/s, v_{iz}=0m/s\\v_{fx}=8.96m/s, v_{fy}=-6.12m/s, v_{fz}=0m/s\\t=0.02s, m=0.12Kg

We can calculate each component of the acceleration using its definition a=\frac{\Delta v}{\Delta t}

a_x=\frac{v_{fx}-v_{ix}}{t} = \frac{(8.96m/s)-(9m/s)}{0.02s} =-2m/s^2\\a_y=\frac{v_{fy}-v_{iy}}{t} = \frac{(-6.12m/s)-(-6m/s)}{0.02s} =-6m/s^2\\a_y=\frac{v_{fz}-v_{iz}}{t} = \frac{(0m/s)-(0m/s)}{0.02s} =0m/s^2\\

The rate of change of momentum of the ball is \frac{\Delta p}{\Delta t} = \frac{\Delta mv}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma

So for each coordinate:

\frac{\Delta p_x}{\Delta t}=-0.24Kgm/s^2\\\frac{\Delta p_y}{\Delta t}=-0.72Kgm/s^2\\\frac{\Delta p_z}{\Delta t}=0Kgm/s^2

And these are equal to the components of the net force since F=ma.

If magnitudes is what is asked:

a=\sqrt{a_x+a_y+a_z} =6.3m/s^2\\F=ma=\frac{\Delta p}{\Delta t}=0.76N

<em>(N and </em>Kgm/s^2<em> are the same unit).</em>

3 0
2 years ago
Weight of 1 kg becomes 1/6 on moon. If radius of moon is 1.76×10^6 m, then the mass of moon will be​
igomit [66]

Answer:

7.65 x 10^22kg

sorry if im wrong!

5 0
2 years ago
According to the Big Bang theory, when the universe first formed, the temperature was extremely hot. As it began to cool, proton
alex41 [277]

hydrogen would be the first Element

5 0
3 years ago
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