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3 years ago

7

Which properties make a metal a good material to use for electrical wires? malleability and reactivity conductivity and ductilit

y ductility and malleability reactivity and conductivity

2 answers:

konstantin123 [22]3 years ago

9 0

Conductivity ... So it can carry electric current without energy loss.

Ductility ... So it can be stretched out to make wire without breaking.

vesna_86 [32]3 years ago

8 1

Answer:

d

Explanation:

i did the quiz

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Hey can someone help me with my physics exam

ivanzaharov [21]

Answer:

by answering your question

4 0

2 years ago

Read 2 more answers

A backyard swimming pool with a circular base of diameter 6.00m is filled to depth 1.50m. (b) Two persons with combined mass 150

Mashutka [201]

The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.

<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

$\rho \mathrm{Vg}=2 \mathrm{mg} \\$

$V=\frac{2 m}{\rho} \\$

$V=\frac{2 \times 150}{1000} \\$

$\mathrm{~V}=0.3 \mathrm{~m}^3$

Area of pool $=\frac{\pi}{4} d^2$

$&=\frac{\pi}{4} \times 6^2 \\$

$&=28.27 \mathrm{~m}^2$

Height of the water rise

$h &=\frac{V}{A} \\$

$h &=\frac{0.3}{28.27} \\$

$& \mathrm{~h}=0.0106 \mathrm{~m}$

Pressure increased

P = ρ g h

P = 1000 x 10 x 0.0106

P = 106.103 Pa

To learn more about absolute pressure refer to:

brainly.com/question/17200230

#SPJ4

4 0

1 year ago

A mass of 3.6 kg oscillate on a horizontal spring with a spring constant of 160 N/m.

Darya [45]

Answer:

48.7 J

Explanation:

For a mass-spring system, there is a continuous conversion of energy between elastic potential energy and kinetic energy.

In particular:

- The elastic potential energy is maximum when the system is at its maximum displacement

- The kinetic energy is maximum when the system passes through the equilibrium position

Therefore, the maximum kinetic energy of the system is given by:

$KE=\frac{1}{2}mv^2$

where

m is the mass

v is the speed at equilibrium position

In this problem:

m = 3.6 kg

v = 5.2 m/s

Therefore, the maximum kinetic energy is:

$KE=\frac{1}{2}(3.6)(5.2)^2=48.7 J$

6 0

2 years ago

What is the beat frequency of two keys on a phone, one with a frequency of 210 Hz and the other one with a frequency of 280 Hz?

vodka [1.7K]

The best frequency should be 70 Hz you simply find the difference between 280 Hz and 210 Hz

6 0

2 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0

2 years ago