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xxMikexx [17]
3 years ago
14

The mass of 2 cm3 of gold is 38.6 grams. find the density of the gold. Could you please help me??

Physics
1 answer:
Alex Ar [27]3 years ago
6 0

Density = mass/ volume (so here this is how you would solve the problem)

<span>D = 38.6 g/ 2 cm3 (first step)</span>

<span>D= 19.3 g/cm3  ( Do math and then you would get this)</span>

<span>
</span>

<span>Hope this helps!! :) </span>


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the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp
Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

Explanation:

From the question given above, the following data were obtained:

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

Cross multiply

1.0006 (1 + 100α) = (1 + 159α)

1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

α = 1.02×10¯⁵ C¯¹

Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

6 0
3 years ago
3. An object of mass 90 kg travels down a slide.
Gwar [14]

Answer:

3) Ep = 13243.5[J]

4) v = 17.15 [m/s]

Explanation:

3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.

m = mass = 90 [kg]

h = elevation = 15 [m]

Potential energy is defined as the product of mass by gravity by height.

E_{p}=m*g*h\\E_{p}=90*9.81*15\\E_{p}=13243.5[J]

This energy will be transformed into kinetic energy.

Ek = 13243.5 [J]

4) The velocity can be determined by defining the kinetic energy, as shown below.

E_{k}=\frac{1}{2} *m*v^{2}  \\v = \sqrt{\frac{2*E_{k} }{m} }\\ v= \sqrt{\frac{2*13243.5 }{90} }\\v=17.15[m/s]

4 0
3 years ago
A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.
Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

\Delta K = \Delta E_{p} = q\Delta V

<u>Where</u>:

K: is the kinetic energy

E_{p}: is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV                                    

\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J

Now, the kinetic energy of the particle as it moves through point B is:

\Delta K = K_{f} - K_{i}

K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!      

8 0
3 years ago
How is temperature related to the physical change of a substance?
WINSTONCH [101]
When you heat something of cool it down you don't change the substance you might change the why is looks, but it is still the same substance. For example you cool water to 0 degrees Celsius it turns into ice but it still is two parts hydrogen and one part oxygen H2O. Physical changes will change state and/or form but it will still be what it originally was on the molecular level. Hope that helped. 
8 0
3 years ago
Read 2 more answers
A -kilogram car travels at a constant speed of 20. meters per second around a horizontal circular track. The diameter of the tra
Tatiana [17]

Answer:

The centripetal acceleration of the car is 8\ m/s^2.

Explanation:

Let the mass of the car, m=10^3\ kg

Diameter of the circular path, d = 100 m

Speed of car, v = 20 m/s

Radius, r = 50 m

When an object moves in a circular path, the centripetal acceleration acts on it. It is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(20\ m/s)^2}{50\ m}

a=8\ m/s^2

So, the centripetal acceleration of the car is 8\ m/s^2. Hence, this is the required solution.

4 0
3 years ago
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