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Digiron [165]
3 years ago
5

What type of weather is associated with a stationary front

Physics
1 answer:
Annette [7]3 years ago
4 0

Answer:

A warm front brings gentle rain or light snow, followed by warmer, milder weather. Stationary front Forms when warm and cold air meet and neither air mass has the force to move the other. They remain stationary, or “standing still.” Where the warm and cold air meet, clouds and fog form, and it may rain or snow.

Hope this helps :)

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Which of the following is not an indication of a chemical reaction? a substance dissolves formation of a precipitate color chang
SVETLANKA909090 [29]

a substance dissolves.  


like adding a soluble salt to water, it just dissolves, i.e dissociates homogeneously as water is able to dissociate salts (ionic compounds) into its ions. (it can also dissociate other non-ionic compounds like HCL)  


the salt still remains chemically as a salt and is unchanged chemically thus it is not an indication of a chemical reaction as no chemical reaction has taken place.  


the formation of a precipitate is a chemical reaction because a new substance (i.e new chemical) is formed. For example adding aqueous sodium hydroxide into an aqueous solution with CU2+ cations will form a blue precipitate (that is copper (II) hydroxide which is insoluble, hence it precipitates). Since a new chemical is formed, a chemical reaction has taken place and thus indicates a chemical reaction.  


color change... im not sure but usually a color change will only occur when a new substance is formed. Like iron corrodes (i.e rust) slowly in moist air to form hydrated iron (III) oxide that is rust. (brown color).  


usually adding a mixture to a mixture has little energy change, i.e little heat taken in by the reaction mixture or little heat given out by the reaction mixture. Whereas when a new substance is formed, there is usually noticeable energy change like the container gets colder or hotter (without heat being supplied of course). For example dissolving basic oxides into water releases energy ( more energy released than gained = exothermic reaction).  


i think that should be the answer... hope it helped :D

4 0
3 years ago
Read 2 more answers
One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi
Masteriza [31]

Answer:

a)S_1=-9.65}\ J/K

b)S_2=71.18\ J/K

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

a)

The entropy change of reservoir 745 K

S_1=-\dfrac{7190}{745}\ J/K

Negative sign because heat is leaving.

S_1=-9.65}\ J/K

b)

The entropy change of reservoir 101 K

S_2=\dfrac{7190}{101}\ J/K

S_2=71.18\ J/K

c)

The entropy change of the rod will be zero.

d)

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

3 0
3 years ago
Three out of four crashes happen at a speed of __________ miles per hour or less.
Nitella [24]
<span>The answer is 45 miles per hour or even less. Most crashes occur at a speed of 45 miles per hour or even less, and most of these accidents occur close to our homes. These crashes may also be caused by different factors, such as being drunk or sudden occurrences that are not controllable, which is why it is best to precede with caution when driving at crowded areas.</span>
5 0
3 years ago
Why doesn’t a machine that increases force break the law of conservation of energy?
USPshnik [31]

Answer:

A machine in which work input equals work output. energy can be used to do work, work can be used to transfer energy. The change in the kinetic energy of an object is equal to the net work done on the object.

hope this helps

8 0
3 years ago
A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it
Degger [83]

Answer:

a)\omega_1=8.168\,rad.s^{-1}

b)n_1=7.735 \,rev

c)\alpha_1 =0.6864\,rad.s^{-2}

d)\alpha_2=4.1454\,rad.s^{-2}

e)t_2=1.061\,s

Explanation:

Given that:

  • initial speed of turntable, N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}
  • full speed of rotation, N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}
  • time taken to reach full speed from rest, t_1=11.9\,s
  • final speed after the change,  N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}
  • no. of revolutions made to reach the new final speed,  n_2=11\,rev

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

\omega=\frac{2\pi.N}{60}\, rad.s^{-1}

where N = angular speed in rpm.

putting the respective values from case 1 we've

\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}

\omega_1=8.168\,rad.s^{-1}

(c)

using the equation of motion:

\omega_1=\omega_0+\alpha . t_1

here α is the angular acceleration

78=0+\alpha_1\times 11.9

\alpha_1 = \frac{8.168 }{11.9}

\alpha_1 =0.6864\,rad.s^{-2}

(b)

using the equation of motion:

\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1

8.168^2=0^2+2\times 0.6864\times n_1

n_1=48.6003\,rad

n_1=\frac{48.6003}{2\pi}

n_1=7.735\, rev

(d)

using equation of motion:

\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2

12.5664^2=8.168^2+2\alpha_2\times 11

\alpha_2=4.1454\,rad.s^{-2}

(e)

using the equation of motion:

\omega_2=\omega_1+\alpha_2 . t_2

12.5664=8.168+4.1454\times t_2

t_2=1.061\,s

4 0
3 years ago
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