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3 years ago

If temperature and the number of particles remain constant, and increase in pressure will cause volume to

1 answer:

5 0

The correct answer is remain the same because if the particles and temperature remain constant, when the pressure increases the temperature is not increasing so it would just stay the same

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7) Which statement below best describes the motion of the cart under the conditions shown in the image below?

arlik [135]

The cart is going left is your answer

4 0

2 years ago

A horizontal pipe of inner diameter 2.2 cm carries water with a density of 1000.0 kg/m3 flowing at a rate of 1.5 kg/s. If the pi

EleoNora [17]

The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

The initial diameter is, $\rm d_1 = 2.2 \ cm$

initial radius,

$r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm$

The initial crossection area;

$\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2$

The final crossection area;

$\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2$

The initial flow rate is;

R = density ×velocity ×area

$\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1 = 3.947 \ m/sec$

The speed of the water in the wider part will be;

From the continuity equation;

$\rm A_1 V_1 = A_2V_2 \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec$

Hence, the speed of the water in the wider part will be 1.194 m/sec.

To learn more about the speed, refer to the link;

brainly.com/question/7359669

#SPJ1

7 0

1 year ago

you throw a rock with a horizontal velocity at 36.6 m/s out a window that is 29.5 m above the ground. What was it’s total time i

ANEK [815]

29.5-4.9x^2= your answer

4 0

2 years ago

Question 1

svetlana [45]

300

Explanation:

100 x 3 =300 simple and easy

7 0

1 year ago

After a model rocket reached its maximum height, it then took 5.0 seconds to return to the launch site. what is the approximate

kifflom [539]

Air resistance is ignored.
g = 9.8 m/s².
At maximum height, the vertical velocity is zero.

Let h = the maximum height reached.
Let u = the vertical launch velocity.

Because ot takes 5.0 seconds to reach maximum height, therefore
(u m/s) - (9.8 m/s²)*(5 s) = 0
u = 49 m/s

The maximum height reached is
h = (49 m/s)*(5 s) - (1/2)*(9.8 m/s²)*(5 s)²
= 122.5 m

Answer: 122.5 m

3 0

2 years ago