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3 years ago

A mass m = 4.6 kg hangs on the end of a massless rope L = 2.16 m long. The pendulum is held horizontal and released from rest.

1 answer:

4 0

ind the velocity my using the conservation of energy mgh=1/2mv^2. At the bottom, the mass is turning in a circle with tension pointing up towards the center of the circle and weight pointing down. Use the centripetal force requirement mv^2/r=T-W and solve for T. Hope this helps.

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Number of conducting plates of a multiplate capacitor is 5. The no. Of capacitors is

Ivahew [28]

Answer:

4 capacitors

Explanation:

Given

$n = 5$ --- conducting plates

Required

The number of capacitor (c)

This is calculated as:

$c = n - 1$

So, we have:

$c = 5 - 1$

$c = 4$

8 0

2 years ago

g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located

Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is: P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P /RT ⋯ ⋯( I )

Here

m is the mass of the air,

v is the volume of the air,

P is the atmospheric pressure,

T is the standard temperature at the atmospheric pressure and

R is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101 kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here A is the area of the windmill

Expression to calculate the A is

A = π /4 d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94 k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94 k g / s ( (6m/s)²/2 )

P w = 160144.92 W × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0

3 years ago

Why might a major volcanic eruption lead to cooler global temperatures? Explain your answer.

irina [24]

Answer:

Volcanic eruptions cool down the planet

Explanation:

Volcanic eruptions actually cool the planet because the particles ejected from volcanoes shade incoming solar radiation. ... The small ash and aerosol particles decrease the amount of sunlight reaching the surface of the Earth and lower average global temperatures.

Hope this helps!!! :D

4 0

3 years ago

A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across

NARA [144]

Answer:

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:

$A=\frac{\pi }{4}d^{2} \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\$

a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper </em>

L= 2.00 m

From Table Copper Resistivity $p$ = 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

$R=\frac{pL}{A}$

=0.0165Ω

The Potential difference across the copper wire is:

V=IR

=2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:

$R=\frac{pL}{A}$

=0.014Ω

The Potential difference across the Silver wire is:

V=IR

=1.75*10^-4

4 0

3 years ago

Which statements describe using genetic factors to influence the growth of organisms? Select the three (3) that apply.

Damm [24]

Answer:

- increasing use of hybrid crops

- altering genes in DNA to create new plants

- developing disease or pest resistant crops

Explanation:

The use of genetic factors to influence the growth of a plant encompasses manipulating the genetic constituent (gene) of such plant.

For example,

- Increasing use of hybrid crops entails mating two pure bred plants based on a gene of interest responsible for a particular trait, to form a hybrid.

- Altering genes in DNA to create new plants is also a genetic factor as it has to with gene modification.

- developing disease or pest resistant crops means that the genetic make up of such plant has been modified to be resistant to pest/disease.

6 0

2 years ago