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Stells [14]
3 years ago
5

Which of the following is true about inertia

Physics
1 answer:
Tamiku [17]3 years ago
3 0

Answer:

The correct answers are It is the resistance of an object to changes in its motion, and It is a force

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This is the si unit of current
skelet666 [1.2K]

Answer:

ampere

and it is denoted by A

3 0
3 years ago
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1) Do liquids and gases also exert pressure?​
Marta_Voda [28]

Explanation:

Liquids also exert pressure in all directions on the walls of the container they are stored in. We see water coming out from leaking pipes and taps. ... Gases (Air) also exert pressure in all directions

6 0
3 years ago
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HELP PLEASE 20 POINTS SHOW WORK, ALL EQUATIONS
nataly862011 [7]

Answer:

s = 3 m

Explanation:

Let t be the time the accelerating car starts.

Let's assume the vehicles are point masses so that "passing" takes no time.

the position of the constant velocity and accelerating vehicles are

s = vt = 40(t + 2)  cm

s = ½at² = ½(20)(t)² cm

they pass when their distance is the same

½(20)(t)² = 40(t + 2)

10t² = 40t + 80

0 = 10t² - 40t - 80

0 = t² - 4t - 8

t = (4±√(4² - 4(1)(-8))) / 2(1)

t = (4± 6.928) / 2  ignore the negative time as it has not occurred yet.

t = 5.464 s

s = 40(5.464 + 2) = 298.564 cm

300 cm when rounded to the single significant digit of the question numerals.

7 0
3 years ago
A refrigerator removes 55.0 kcal of heat from the freezer and releases 73.5 kcal through the condenser on the back.How much work
sammy [17]

Here refrigerator removes 55 kcal heat from freezer

Refrigerator releases 73.5 kcal heat to surrounding

So here we can use energy conservation principle by II Law of thermodynamics

the law says that

Q_1 = Q_2 + W

here we know that

Q_1 = heat released to the surrounding

Q_2 = heat absorbed from freezer

W = work done by the compressor

now using above equation we can write

73.5 = 55 + W

W = 73.5 - 55

W = 18.5 kcal

So here compressor has to do 18.5 k cal work on it

5 0
3 years ago
a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column
scoray [572]

Answer:

The answer is "1155\ \frac{kg}{m^3}"

Explanation:

Please find the complete question in the attached file.

p = p_0 + ?gh

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

h_1 = 11 \ cm\\\\h_2= 3 \ cm

The pressures would be proportional to the quantity 11-3 = 8 cm from below the surface at the interface between both the oil and the liquid.

\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\

       = \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}

8 0
3 years ago
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