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3 years ago

Which properties are characteristics of non metals

Physics

1 answer:

Natasha2012 [34]3 years ago

4 0

Non - Metals:

Non- metals are bad conductor of electricity ( except Graphite ).

Non- metals are also bad conductor of heats. ( except diamond )

Non - metal are neither malleable nor ductile. They are brittle.

Non-metal are not strong.

Non-metal are not sonorous.

Some terms:

★ Malleable: This means that metals can be beaten into thin sheets with hammer.

★ Ductile: This means that metals can be drawn into thin wires.

★ Brittle: This means that non-metal break into pieces on hammering.

★ Sonorous: This means capable of producing a ringing sound.

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What four regions of the electromagnetic spectrum

MArishka [77]

There are a lot.
Some of them are

Gamma radiation
X-ray radiation
Ultraviolet radiation
Visible radiation
Infrared radiation
Microwave radiation

6 0

3 years ago

a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe

Elden [556K]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

$Mg(r - L/2) = mg(L/2 - 8)$

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

$400,000(r - 10) = 19,600 (10 - 8)$

$r - 10 = 0.098$

$r = 10.098 m$

so pillar is at distance 10.098 m from one end of the slab

3 0

3 years ago

Weddell seals make holes in sea ice so that they can swim down to forage on the ocean floor below. Measurements for one seal sho

Mariulka [41]

Answer:

B. 1 m/s

Explanation:

Metric unit conversions:

0.3 km = 300m

5 minutes = 5*60 = 300 seconds

So if a seal can reach a depth of 300m in a time of 300 seconds, its diving speed is the distance divided by time duration

v = s/t = 300/300 = 1m/s

So B is the correct answer

3 0

3 years ago

Air enters a turbine operating at steady state at 8 bar, 1400 K and expands to 0.8 bar. The turbine is well insulated, and kinet

vladimir2022 [97]

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

Where

$T_1 =$ Temperature at inlet of turbine

$T_2 =$ Temperature at exit of turbine

$P_1 =$ Pressure at exit of turbine

$P_2 =$ Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

$m_i = m_0 = m$

$m(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W$

Where,

m = mass

$m_i$ = mass at inlet

$m_0$ = Mass at outlet

$h_i$ = Enthalpy at inlet

$h_0$ = Enthalpy at outlet

W = Work done

Q = Heat transferred

$V_i$ = Velocity at inlet

$V_0$ = Velocity at outlet

$Z_i$ = Height at inlet

$Z_0$ = Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

$h_i = h_0 + W$

$W = h_i -h_0$

Using the relation T-P we can find the final temperature:

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

$\frac{T_2}{1400K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}$

$T_2 = 725.126K$

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

So:

$W = h_i -h_0$

$W = C_p (T_1-T_2)$

$W = 1.005(1400-725.126)$

$W = 678.248kJ/Kg$

Therefore the maximum theoretical work that could be developed by the turbine is 678.248kJ/kg

5 0

4 years ago

A golf club rotates 215 degrees and has a length (radius) equal to 29 inches. The time it took to swing the club was 0.8 seconds

vichka [17]

Answer:

The average linear velocity (inches/second) of the golf club is 136.01 inches/second

Explanation:

Given;

length of the club, L = 29 inches

rotation angle, θ = 215⁰

time of motion, t = 0.8 s

The angular speed of the club is calculated as follows;

$\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega = (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s$

The average linear velocity (inches/second) of the golf club is calculated as;

v = ωr

v = 4.69 rad/s x 29 inches

v = 136.01 inches/second

Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

8 0

3 years ago