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katrin2010 [14]
4 years ago
9

How is the density of a substance calculated?

Physics
1 answer:
den301095 [7]4 years ago
3 0
1).  Take a sample of the substance.  The sample should be the largest
possible that will allow it to be be easily handled and the following steps
to be performed with it. 
(The density doesn't depend on the size of the sample, and every sample
of the same substance has the same density.  But using a larger sample
can improve the accuracy of the measurements you make, and therefore
improve the accuracy of the density you derive for the substance.)

2). Ask or measure the mass of the sample.

3). Ask or measure the volume of the sample.

4). Divide the mass by the volume. Their quotient is the density
of the substance.
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HEY CAN ANYONE ANSWER DIS QUESTION PLS!!!!!!!!
Valentin [98]
It’s very big and very small numbers
6 0
4 years ago
A cannon is mounted on a cart which sits on the ground, supported by frictionless wheels. The mass of the cannon and cart is 4.6
Kitty [74]

Answer:

The velocity of the launcher after the projectile is launched is  -5.011 m/s

Explanation:

Here we have the mass of the cannon and cart, m₁ =  4.65 kg

Velocity of cannon and cart, v₁ = 2.00 m/s

Mass of projectile, m₂ = 50.0 g = 0.05 kg

Velocity of projectile, v₂ = 647 m/s

Velocity of the launcher, v₃ = Required

Mass of cannon and cart, launcher after launching projectile m₃ = 4.65-0.05

= 4.6 kg

Therefore, from the principle of the conservation of linear momentum, we have

Total initial momentum = Total final momentum

m₁ × v₁ = m₂ × v₂ + m₃ × v₃

Substituting gives

4.65 kg × 2.00 m/s = 0.05 kg × 647 m/s + 4.6 kg × v₃

4.65 kg × 2.00 m/s - 0.05 kg × 647 m/s = 4.6 kg × v₃

-23.05 kg·m/s = 4.6 kg × v₃

v_3 = \frac{-23.05 \, kg\cdot m/s}{4.6 \, kg} =  \frac{-461}{92} m/s

v₃ = -5.011 m/s.

3 0
3 years ago
Read 2 more answers
If I push a box at a constant rate is there friction force acting on it?
yanalaym [24]
Yes, the friction is acting in the opposite direction you are pushing.
3 0
3 years ago
A rubber band has a spring constant of 75 N/m. How much force is required to stretch the rubber band 0.02 m past its natural len
ale4655 [162]

Answer:

(75 x 0.02) = 1.5 N

Explanation:

7 0
3 years ago
A parallel-plate, air-gap capacitor has a capacitance of 0.14 mu F. The plates are 0.5 mm apart, What is the area of each plate?
Marysya12 [62]

Answer:

7.9060 m²

8.57 Volts

5.142×10⁻⁶ Joule

1.2×10⁻⁶ Coulomb

Explanation:

C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F

d = Distance between plates = 0.5 mm = 0.5×10⁻³ m

Q = Charge = 1.2 μC = 1.2×10⁻⁶ C

ε₀ = Permittivity = 8.854×10⁻¹² F/m

Capacitance

C=\frac{\epsilon_{0}A}{d}\\\Rightarrow A=\frac{Cd}{\epsilon_{0}}\\\Rightarrow A=\frac{0.14\times 10^{-6}\times 0.5\times 10^{-3}}{8.854\times 10^{-12}}\\\Rightarrow A=7.9060\ m^2

∴ Area of each plate is 7.9060 m²

Voltage

V=\frac{Q}{C}\\\Rightarrow V=\frac{1.2\times 10^{-6}}{0.14\times 10^{-6}}\\\Rightarrow V=8.57\ Volts

∴ Potential difference between the plates if the capacitor is charged to 1.2 μC  is 8.57 Volts.

Energy stored

E=0.5CV²

⇒E = 0.5×0.14×10⁻⁶×8.57²

⇒E = 5.142×10⁻⁶ Joule

∴ Stored energy is 5.142×10⁻⁶ Joule

Charge

Q = CV

⇒Q = 0.14×10⁻⁶×8.57

⇒Q = 1.2×10⁻⁶ C

∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb

6 0
3 years ago
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