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Mariana [72]
3 years ago
12

Use the equation d=m/v. If a rock has a density of 2g/cm^3 and a volume of 8cm^3, what is the mass?

Physics
1 answer:
Ray Of Light [21]3 years ago
8 0

Answer:

The mass is 16 [g]

Explanation:

We have to remember the formula that defines density, which mentions that the density is equal to mass divided by the volume.

Density = mass / volume\\where:\\density = 2[\frac{g}{cm^{3} } ]\\Volume = 8 [cm^{3}]

Now we clear the mass:

mass= density*volume\\mass = 2[\frac{g}{cm^{3} }]*8[cm^{3} ]\\mass = 16 [g]

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Light incident on a glass sheet is partly reflected and partly refracted. How is the reflected ray different from the refracted
kkurt [141]

Answer:

E. The refracted ray is vertically polarized whereas the reflected ray is horizontally polarized.

Explanation:

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3 years ago
Which activity is a method of habitat restoration?
irina1246 [14]

A. cleaning up after environmental disasters

7 0
3 years ago
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Calculate earth's mass given the acceleration due to gravity at the north pole is 9.830 m/s2 and the mean radius of the earth at
valentinak56 [21]

Answer: M = 5.98\times 10^{24} kg

Explanation:

We know that force acting on an object due to Earth's gravity on the surface is given by:

mg = G\frac{Mm}{r^2}\\ \Rightarrow g = \frac{GM}{r^2}

where g is the acceleration due to gravity, r would be radius of Earth, M is the mass of Earth and G is the gravitational constant.

It is given that at pole, g = 9.830 m/s² and r = 6371 km = 6371 × 10³ m

\Rightarrow M = \frac{g\times r^2}{G}

M = \frac{9.830 m/s^2 \times (6371 \times 10^3 m )^2}{6.67 \times 10^{-11} m^3 kg^{-1} s^{-2}}

\Rightarrow M = 5.98\times 10^{24} kg

Hence, Earth's mass is 5.98\times 10^{24} kg

5 0
3 years ago
At t=0, a train approaching a station begins decelerating from a speed of 80 mi/hr according to the acceleration function a(t)=−
vredina [299]

Answer:

a)  Δx = 49.23 mi , b)  Δx = 5.77 mi

Explanation:

As we have an acceleration function we must use the definition of kinematics

     a = dv / dt

     ∫dv = ∫ a dt

we integrate and evaluators

      v - vo = ∫ (-1280 (1 + 8t)⁻³ dt = -1280 ∫ (1+ 8t)⁻³ dt

We change variables

       1+ 8t = u

       8 dt = du

       v - v₀ = -1280 ∫ u⁻³ du / 8

       v -v₀ = -1280 / 8 (-u⁻²/2)

       v - v₀ = 80 (1+ 8t)⁻²

We evaluate between the initial t = 0 v₀ = 80 and the final instant t and v

      v- 80 = 80 [(1 + 8t)⁻² - 1]

      v = 80 (1+ 8t)⁻²

We repeat the process for defining speed is

     v = dx / dt

    dx = vdt

    x-x₀ = 80 ∫ (1-8t)⁻² dt

    x-x₀ = 80 ∫ u⁻² dt / 8

    x-x₀ = 80 (-1 / u)

    x-x₀ = -80 (1 / (1 + 8t))

We evaluate for t = 0 and x₀ and the upper point t and x

   x -x₀ = -80 [1 / (1 + 8t) - 1]

We already have the function of time displacement

a) let's calculate the position at the two points and be

t = 0 h

     x = x₀

t = 0.2 h

    x-x₀ = -80 [1 / (1 +8 02) -1]

    x-x₀ = 49.23

displacement is

  Δx = x (0.2) - x (0)

   Δx = 49.23 mi

b) in the interval t = 0.2 h at t = 0.4 h

t = 0.4h

     x- x₀ = -80 [1 / (1+ 8 0.4) -1]

     x-x₀ = 55 mi

    Δx = x (0.4) - x (0.2)

     Δx = 55 - 49.23

     Δx = 5.77 mi

3 0
3 years ago
The current theory of the structure of the
IRISSAK [1]

1) The mass of the continent is 3.3\cdot 10^{21} kg

2) The kinetic energy of the continent is 624 J

3) The speed of the jogger must be 4 m/s

Explanation:

1)

We start by finding the volume of the continent. We have:

L = 5850 km = 5.85\cdot 10^6 m is the side

t = 35 km = 3.5\cdot 10^4 m is the depth

So the volume is

V=L^2 t = (5.85\cdot 10^6)^2 (3.5\cdot 10^4)=1.20\cdot 10^{18} m^3

We also know that its density is

d=2750 kg/m^3

Therefore, we can find the mass by multiplying volume by density:

m=dV=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg

2)

The kinetic energy of the continent is given by:

K=\frac{1}{2}mv^2

where

m=3.3\cdot 10^{21} kg is its mass

v = 3.2 cm/year is the speed

We have to convert the speed into m/s. We have:

3.2 cm = 0.032 m

1 year = 1(365)(24)(60)(60)=3.15\cdot 10^7 s

So, the speed is:

v=\frac{0.032 m}{3.15 \cdot 10^7 s}=1.02\cdot 10^{-9} m/s

So, we can now find the kinetic energy:

K=\frac{1}{2}(1.20\cdot 10^{21})(1.02\cdot 10^{-9})^2=624 J

3)

Here we have a jogger of mass

m = 78 kg

And the jogger has the same kinetic energy of the continent, so

K = 624 J

The kinetic energy of the jogger is given by

K=\frac{1}{2}mv^2

where v is the speed of the jogger.

Solving for v, we find the speed that the jogger must have:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(624)}{78}}=4 m/s

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0
3 years ago
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