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Maksim231197
1 month ago
6

What is the mass of a cylinder of lead that is 1.80 in in diameter, and 4.12 in long. the density of lead is 11.4 g/ml?

Physics
2 answers:
Maru [3.3K]1 month ago
7 0

Response: The lead cylinder weighs 1,995.898 g.

Clarification:

The radius of the cylinder, r =\frac{Diameter}{2}=\frac{1.80 inches}{2}=0.9 inches

Height of the cylinder, h = 4.12 inches

Cylinder volume: \pi r^2h=3.14\times (0.9 inches)^2\times 4.12 inches=10.47 inches^3

inches^3=16.3871 ml

Cylinder volume in ml= 10.47\times 16.3871 ml=171.57 ml

Density of the lead in the cylinder = 11.4 g/ml

density=\frac{Mass}{Volume}

Mass=11.4 g/ml\times 171.57 ml=1,955.898 g

Thus, the lead cylinder's mass is 1,995.898 g.

Ostrovityanka [3.2K]1 month ago
5 0
1950 g This is the result of lead being spread out in kilograms
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Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2
Ostrovityanka [3204]

The complete question is;

Block 1 sits on the floor with block 2 resting atop it. Block 3, which is stationary on a frictionless table, is attached to block 2 via a string that passes over a pulley depicted in the illustration below. Both the string and pulley have negligible mass.

Once block 1 is taken away without impacting block 2.

Derive an equation for the acceleration of block 3 considering arbitrary values for m3 and m2. Express your answer in terms of m3, m2, and relevant physical constants as needed.

Answer:

a = (m2)g/(m3 + m2)

Explanation:

Examining the attached illustration, by analyzing the free body diagram for block 3 and utilizing Newton's first law of motion, we reach the following formula;

T = (m3)a - - - (eq 1)

where;

T is the tension in the string

a is acceleration

m3 is the mass of block 3

Simultaneously, doing the same for Block 2, the free body diagram yields the equation; (m2)g - T = (m2)a

Rearranging for T results in;

T = (m2)g - (m2)a - - - (eq 2)

where;

g represents acceleration due to gravity

T is the tension in the string

a is acceleration

m2 is the mass of block 2

To deduce the acceleration, we will substitute (m3)a in place of T in eq 2.

Thus;

(m3)a = (m2)g - (m2)a

(m3)a + (m2)a = (m2)g

a(m3 + m2) = (m2)g

a = (m2)g/(m3 + m2)

3 0
28 days ago
The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that
serg [3582]

Response:

Reasoning:

We will utilize a Gaussian surface that resembles the curved wall of a cylinder, with a radius of 3mm and a length of 1 unit directed parallel to the wire axis.

The charge within this cylinder amounts to 250 x 10⁻⁹ C.

Let E denote the electric field at the curved surface, perpendicular to it.

The total electric flux leaving the curved surface

is calculated as 2π r x 1 x E

or 2 x 3.14 x 3 x 10⁻³ E

According to Gauss's law, the total flux is given by the charge within divided by ε (the charge inside the cylinder being 250 x 10⁻⁹C)

equals 250 x 10⁻⁹ / 2.5 x 8.85 x 10⁻¹²   (where ε = 2.5 ε₀ = 2.5 x 8.85 x 10⁻¹²)

resulting in 11.3 x 10³ weber.

Thus,

2 x 3.14 x 3 x 10⁻³ E = 11.3 x 10³

E =  11.3 x 10³ /  2 x 3.14 x 3 x 10⁻³

=.599 x 10⁶ N /C.

4 0
1 month ago
Two of the types of ultraviolet light, uva and uvb, are both components of sunlight. their wavelengths range from 320 to 400 nm
Sav [3153]

In terms of light energy, a higher frequency corresponds to increased energy within the light.

We establish that frequency is essentially the inverse of wavelength:

frequency = 1 / wavelength

Calculating frequencies:

f UVA = 1/320 to 1/400

f UVA = 0.0031 to 0.0025

 

f UVB = 1/290 to 1/320

f UVB = 0.0034 to 0.0031

Since UVB occupies a higher frequency range, it consequently possesses greater energy than UVA.

7 0
1 month ago
Read 2 more answers
The acceleration of segment D is m/s2. Rank segments A, B , C from least accelerations to greatest acceleration. Least
Ostrovityanka [3204]

Answer:

D, C, B, A

Explanation:

The derivative from a velocity-time graph provides the acceleration value.

Segment A

\frac{dy}{dx} = \frac{15m/s}{1s} = 15m/s^2

Segment B

\frac{dy}{dx} = \frac{5m/s}{1s} = 5m/s^2

Segment C

\frac{dy}{dx} = \frac{0m/s}{2s} = 0m/s^2

Segment D

\frac{dy}{dx} = \frac{-20m/s}{1s} = -20m/s^2

Sorted from the lowest to the highest acceleration:

D, C, B, A

8 0
1 month ago
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10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????
serg [3582]

Answer: Tension = 47.8N, Δx = 11.5×10^{-6} m.

              Tension = 95.6N, Δx = 15.4×10^{-5} m

Explanation: The speed of a wave on a string under tension can be determined using the following:

|v| = \sqrt{\frac{F_{T}}{\mu} }

F_{T} denotes tension (N)

μ refers to linear density (kg/m)

Calculating the velocity:

|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }

|v| = \sqrt{0.00874 }

|v| = 0.0935 m/s

Distance a pulse traveled in 1.23ms:

\Delta x = |v|.t

\Delta x = 9.35.10^{-2}*1.23.10^{-3}

Δx = 11.5×10^{-6}

With a tension of 47.8N, the distance a pulse will cover is Δx = 11.5×10^{-6}  m.

When tension is doubled:

|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }

|v| = \sqrt{2.0.00874 }

|v| = \sqrt{0.01568}

|v| = 0.1252 m/s

Distance in the same time:

\Delta x = |v|.t

\Delta x = 12.52.10^{-2}*1.23.10^{-3}

\Delta x = 15.4×10^{-5}

With the increased tension, it moves \Delta x = 15.4×10^{-5} m

4 0
2 months ago
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