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3 years ago

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1. In an urban area, what can you expect

2 answers:

pantera1 [17]3 years ago

8 0

Answer: Most inhabitants of urban areas have nonagricultural jobs meaning not of, relating to, or used in farming and agriculture nonagricultural products/land. Urban areas are very developed, meaning there is a density of human structures such as houses, commercial buildings, roads, bridges, and railways.

Explanation:

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zysi [14]3 years ago

8 0

A huge population and a great number of buildings. Urban is not where you find the country, it is the city, where you find many different places, people, buildings, and it is very developed, or is developed.

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A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and release

iragen [17]

Answer:

The frequency of the oscillations in terms of fo will be f2=fo/3

E xplanation:

T= $2pie\frac sqrt {m}{k}$

$\frac {{f2}/times {fo}}$ =1:3

⇒f2=fo\3

Here frequency f is inversely poportional to square root of mass m.

so the value of remainder of frequency f2 and fo is equal to 1:3.

⇒ $\frac{f2} {f1}$ = $\frac sqrt{m1}[m2}$

⇒ $\frac{f2}{fo}$ = 1:3

⇒f2= $\frac{fo} {3}$

6 0

2 years ago

A wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same m

Nina [5.8K]

Answer: $0.05\ mm$

Explanation:

Given

Cross-sectional area of wire $A_1=4\ mm^2$

Extension of wire $\delta l=0.1\ mm$

Extension in a wire is given by

$\Rightarrow \delta l=\dfrac{FL}{AE}$

where, $E=\text{Youngs modulus}$

$\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)$

for same force, length and material

$\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)$

Divide (i) and (ii)

$\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm$

5 0

2 years ago

What term is defined as the amount of charge stored per volt?

Shtirlitz [24]

Capacitance is a measure of charge stored per volt.

3 0

3 years ago

Read 2 more answers

Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is

dimaraw [331]

Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

$W_b = m_bg = \\ \\ W_b = (0.730 \ kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N$

The mass of the water accumulated in the bucket after 3.20s is:

$m_w= (0.20 \ L/s) ( 3.20)s$

$m _w=0.64 \ kg$

To determine the weight of the water accumulated in the bucket, we have:

$W_w = m_w g$

$W_w = ( 0.64 \ kg )(9.80\ m \ /s^2)$

$W_w = 6.272 \ N$

For the speed of the water before hitting the bucket; we have:

$v = \sqrt{2gh}$

$v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}$

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

$F = v ( \dfrac {dm}{dt} )$

$F = (8.4 \ m/s)( 0.200 \ L/s)$

F = 1.68 N

Finally, the reading scale is:

$F_{scale$ = 7.154 N + 6.272 N + 1.68 N

= 15.106 N

6 0

2 years ago

Read 2 more answers

Analogy: atom is to element as ____ is to compound

Masja [62]

Molecule is the answer to what you are looking for

5 0

3 years ago