What does properties mean

What is the difference between a constellation and an “asterism”? Name at least one commonly known asterism.

marta [7]

A "constellation" is the total, full, complete thing in the night sky ...
ALL of the stars that go together to make up the complete dog or
the queen or the fish or the bear, or whatever else the story of
that particular part of the sky involves.

An "asterism" is a small, easily recognized group of stars, that
can be spotted right away, although they're only a small part
of a constellation.

The Big Dipper, the Little Dipper, Orion's Belt, and the Great Square
are asterisms. They're parts of the constellations of the Big Bear,
the Little Bear, Orion the Great and Mighty Hunter, and Pegasus,
respectively.

7 0

3 years ago

On isolated ground receptacles, the metal yoke ____ integrally bonded to the equipment grounding terminal of the receptacle.

Ede4ka [16]

On isolated ground receptacles, the metal yoke is not allowed to be integrally bonded to the equipment grounding terminal of the receptacle.

Any device with two distinct switches or receptacles is a duplex device. It can be shaped to fit a Decora opening or a typical duplex plate opening. It should be noted that they can be combination devices with a switch/outlet, switch/pilot light, etc.

Because of grounding connection removal and receptacle, it is utterly undesirable to connect the two bare equipment grounding conductors together directly.

The equipment grounding conductor associated with those circuits must be connected to the box when circuit conductors are terminated on equipment inside a metal box to prevent unneeded current discharge.

Learn more about grounding conductors here brainly.com/question/14886979

#SPJ4.

4 0

2 years ago

Fifteen identical particles have various speeds: one has a speed of 2.00 m/s, two have speeds of 3.00 m/s, three have speeds of

Evgesh-ka [11]

Answer:

a) $V=7.5m/s$

b) $rms=8.4m/s$

c) Generally the most probable speed is 8m/s as it the most posses by particles being the average

Explanation:

From the question we are told that:

Sample size N=15

$Speed 1 v_1=2m/s\\\\Speed 2 v_2=3m/s\\\\Speed 3 v_2=5m/s\\\\Speed 4 v_4=8m/s\\\\Speed 3 v_5=9m/s\\\\Speed 2 v_6=15m/s\\\\$

Generally the equation for Average speed is mathematically given by

$V_{avg}=\frac{\sum(nv)}{N}$

Therefore

$V_{avg}=\frac{(2+2(3)+3(5)+4(8)+3(9)+2(15))}{15}$

$V=7.5m/s$

b)

Generally the equation for RMS speed of the particle is mathematically given by

$rms=\sqrt{\frac{\sum(nv^2)}{N}}$

$rms=\sqrt{\frac{2^2+2(3)^2+3(5)^2+4(8)^2+3(9)^2+2(15)^2}{15}}$

$rms=\sqrt{69.73}$

$rms=8.4m/s$

c

Generally the most probable speed is 8m/s as it the most posses by particles being the average

4 0

3 years ago

A 1200 kg car traveling at 60 mph quickly brakes to a halt. The kinetic energy of the car is converted to thermal energy of the

ololo11 [35]

Answer:

ΔT = 59.9 ° C

Explanation:

For this exercise the brake energy is totally converted into heat

Let's calculate the vehicle energy

K = ½ m v²

Let's reduce the units to the SI system

v = 30 mph (1609.34 m / mile) (1h / 3600s) = 13.41 m / s

Em = K = ½ 1200 13.41²

K = 1.079 105 J

All this energy is transformed into heat

Em = Q

The expression for heat is

Q = m $c_{e}$ ΔT

ΔT = Q / m $c_{e}$

The specific heat of iron is $c_{e}$ = 450 J / Kg ºC

ΔT = 1,079 105 / (4.0 450)

ΔT = 59.9 ° C

8 0

3 years ago

Read 2 more answers

When an object such as a plastic comb is charged by rubbing it with a cloth, the net charge is typically a few microcoulombs. If

masya89 [10]

The concept required to solve this problem is quantization of charge.

First the number of electrons will be calculated and then the total mass of the charge.

With these data it will be possible to calculate the percentage of load in the mass.

$Q= ne$

Here Q is the charge, n is the number of electrons and e is the charge on the electron

$n = \frac{Q}{e}$

Replacing,

$n = \frac{4*10^{-6}C}{1.6*10^{-19}}$

$n = 2.5 * 10^{13}77$

According to the quantization of charge the charge is defined as product of the number of electron and the charge on the electron

The total mass of the charge is

$m= nm_e$

Here,

m = Mass of the charge

n = Number of electrons

$m_e$ = Mass of the electron

$\text{Percentage change} = \frac{nm_e}{M}*100$

Replacing we have

$\text{Percentage change} = \frac{(2.5*10^13)(9.1*10^{-28})}{33}*100$

$\text{Percentage change} = 6.9*10^{-14} \%$

6 0

4 years ago