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mezya [45]
3 years ago
13

A An ____ occurs when a very low resistance circuit is formed and can easily start a fire.

Physics
1 answer:
Andrews [41]3 years ago
6 0

Answer:

1- Short circuit

2- ammeter

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Identify the solutions (homogeneous mixtures) in the list below. check all that apply. lead, an alloy of tin and lead ammonia, a
Goryan [66]

The solutions and homogeneous mixtures in the given list are:

A. Lead solder, an alloy of tin and lead.

D. Window cleaner, a mixture of ammonia and coloring dissolved in water.

E. Gasoline, a mixture of organic liquids with a fixed composition throughout.

<h3>What is a solution?</h3>

A solution can be defined as a special type of homogeneous mixture that comprises a solute and a solvent.

A homogeneous mixture can be defined as any solid, liquid, or gaseous mixture which has an identical (uniform) composition and properties throughout any given sample of the mixture.

In Science (Physics), all solutions are considered to be a homogeneous mixture because their constituents are uniformly (evenly) distributed.

In conclusion, the solutions and homogeneous mixtures in the given list are:

  • Lead solder, an alloy of tin and lead.
  • Window cleaner, a mixture of ammonia and coloring dissolved in water.
  • Gasoline, a mixture of organic liquids with a fixed composition throughout.

To know more about solution follow

brainly.com/question/22070951

4 0
2 years ago
Read 2 more answers
A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to
irinina [24]

When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

<h3>Frictional force between the block and the horizontal surface</h3>

The frictional force between the block and the horizontal surface is determined by applying Newton's law;

∑F = ma

F - Ff = ma

Ff = F - ma

Ff = 4 - 2(1.2)

Ff = 4 - 2.4

Ff = 1.6 N

When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;

F - Ff = ma

5 - 1.6 = 2a

3.4 = 2a

a = 3.4/2

a = 1.7 m/s²

Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

Learn more about frictional force here: brainly.com/question/4618599

8 0
2 years ago
Can someone please help me with these physics problems? I just don’t even know where to start.
KIM [24]

#1

for the block of mass 5 kg normal force is given as

F_n = mg

F_n = 5*9.8 = 49 N

friction force is given as

F_f = \mu F_n

F_f = 0.1*49 = 4.9 N

Net force is given as

F_{net} = ma

F_{net} = 5*2 = 10 N

now we know that

F_{net} = F_{app} - F_f

10 = F_{app} - 4.9

F_{app} = 14.9 N

#2

Normal force is given as

F_n = mg

F_n = 6*9.8

F_n = 58.8 N

now we know that

F_{net} = F_{app} - F_f

F_{net} = 0

as object moves with constant velocity

F_{app} = F_f = 15 N

now for coefficient of friction we can use

F_f = \mu F_n

15 = \mu * 58.8

\mu = 0.255

#3

net force upwards is given as

F = 1.2 * 10^{-4} N

mass is given as

m = 7 * 10^{-5} kg

now as per newton's law we can say

F = ma

1.2 * 10^{-4} = 7 * 10^{-5} * a

a = 1.71 m/s^2

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

F_{net} = F_{app} - F_f

ma = F_{app} - F_f

5*6 = 40 - F_{f}

F_f = 40 - 30 = 10 N

part b)

for coefficient of friction we can use

F_f = \mu F_n

10 = \mu * F_n

here normal force is given as

F_n = mg = 5*9.8 = 49 N

now we have

\mu = \frac{10}{49} = 0.204

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

d = v_i*t + \frac{1}{2}at^2

20 = 0 + \frac{1}{2}a(5^2)

a = 1.6 m/s^2

now net force is given as

F_{net} = ma

F_{net} = 10*1.6 = 16 N

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

a = \frac{v_f - v_i}{t}

a = \frac{0 - 25}{1.5} = -16.67 m/s^2

now the force is gievn as

F = ma

F = 5*16.67 = 83.3 N

3 0
4 years ago
When air pressure increases the liquid in a mercury barometer?
MrMuchimi
It forces the mercury to rise, being pushed up the tube by pressing down on the dish.

I hope I could help :)
7 0
4 years ago
Read 2 more answers
HELP!!
QveST [7]

Answer:

Total displacement 155 m

Explanation:

From the question, the formula to apply will be that of cosine rule where:

c=\sqrt{a^2+b^2-2ab*cosC}

where

a= 43 m , b=130m and c =? , <C=28° + 90° =118°

Substituting values in the equation as;

c=\sqrt{43^2+130^2-2*43*130*cos118}

c=\sqrt{1849+16900-11180*-0.46947} \\\\c=\sqrt{1849+16900+5249} \\\\c=\sqrt{23998} \\\\\\c=155m

6 0
3 years ago
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