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3 years ago

What are the steps to find the mass of 30 mL of water ...?

1 answer:

8 0

The steps in order to calculate the mass of a given volume of water is to use density. Density is a physical property of a substance that represents the mass of that substance per unit volume. It is a property that can be used to describe a substance. Density slightly varies with changes in temperature so that it is important to determine the temperature.

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The position of a 50 g oscillating mass is given by x(t)=(2.0cm)cos(10t−π/4), where t is in s. If necessary, round your answers

Leona [35]

For the answer to the question above, this is the maximum displacement, the spring has only elastic potential energy.
spring is constant @ 5 N/m
maximum displacement = 2 cm = 0.02 m
elastic potential energy = 1/2 kx²
= 0.5 x 5 x 0.02²
So the answer would be
= 0.001 Joules

5 0

3 years ago

A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to

Nostrana [21]

Answer:

a) r=4.24cm d=1 cm

b) $Q=5x10^{-10} C$

Explanation:

The capacitance depends only of the geometry of the capacitor so to design in this case knowing the Voltage and the electric field

$V=1.00x10^{2}v\\E=1.00x10^{4} \frac{N}{C}$

$V=E*d\\d=\frac{V}{E}\\d=\frac{1.0x10^{2}}{1.0x10^{4}}\\d=0.01m$

The distance must be the separation the r distance can be find also using

$C=\frac{Q}{V_{ab}}$

But now don't know the charge these plates can hold yet so

a).

d=0.01m

$C=E_{o}*\frac{A}{d}\\A=\frac{C*d}{E_{o}}$

$A=\frac{5pF*0.01m}{8.85x10^{-12}\frac{F}{m}}\\A=5.69x10^{-3}m^{2}$

$A=\pi *r^{2}\\r=\sqrt{\frac{A}{r}}\\r=\sqrt{\frac{5.64x10^{-3}m^{2} }{\pi } } \\r=42.55x^{-3}m$

b).

$C=\frac{Q}{V_{ab}}$

$Q=C*V\\Q=5x10^{-12} F*1x10^{2}\\Q=5x10^{-10}C$

5 0

3 years ago

Resistor 1 has a resistance of 10 N and

Arlecino [84]

Answer:

Resistance 2.

Explanation:

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

$V = IR$

Where;

V represents voltage measured in voltage.
I represents current measured in amperes.
R represents resistance measured in ohms.

Note: Voltage is the same as potential difference.

Given the following data;

Resistance 1 = 10 Ohms

Current 1 = 3 Amperes

Potential difference 1 = current * resistance

Potential difference 1 = 3 * 10

Potential difference 1 = 30 V

To find the potential difference in 2;

Resistance 2 = 5 Ohms

Current 2 = 10 Amperes

Potential difference 2 = 10 * 5

Potential difference 2 = 50 V

Therefore, Resistance 2 has greater potential difference.

5 0

2 years ago

What is the displacement of a NASCAR stock car after he completes the Indianapolis 500?

Mnenie [13.5K]

I'm such a devoted NASCAR fan that I don't even know if the starting line and finish line at Indy are the same line ... or maybe they're in different places, like the 100-meter sprint in track.

I have no idea, but I'm going to guess that the start and finish at Indy are the SAME line.

If that's true, then the displacement of a car that runs the whole 500 miles is very close to ZERO.

Displacement is the distance and direction between the place where the object starts out and the place where it ends up. The route it follows to get from the start to the finish is completely ignored, and doesn't matter.

(Do they the Indianapolis 500 in "stock" cars ? ?)

3 0

3 years ago

For a bronze alloy, the stress at which plastic deformation begins is 267 MPa and the modulus of elasticity is 115 GPa. (a) What

choli [55]

Answer

given,

Stress for plastic deformation = 267 MPa

modulus of elasticity = 115 GPa

cross sectional area = 377 mm²

a) maximum load (in N) that may be applied to a specimen

= σ x A

= 267 x 10⁶ x 377 x 10⁻⁶

= 100659 N

b) modulus of elasticity = stress/strain

115 x 10⁹ = $\dfrac{267 \times 10^6}{\dfrac{\Delta l}{L}}$

L = 127 mm

115 x 10⁹ = $\dfrac{267 \times 10^6}{\dfrac{\Delta l}{127}}$

$\dfrac{\Delta l}{127}=\dfrac{267 \times 10^6}{115\times 10^9}$

Δ l = 0.295 mm

maximum length after the stretched = 127 mm + 0.295 mm

= 127.295 mm

4 0

3 years ago