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3 years ago

PHYSICAL SCIENCE: What are some methods used to control the types and sources of pollution?

1 answer:

3 0

Gaseous emissions are industrial products such as sulfur dioxide, carbon monoxide, and oxides of nitrogen also released during various manufacturing operations. Particulate control. Methods for particulate control tend to operate on a common principle.

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Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.001

Vlad1618 [11]

Answer:

F = 1.128 10⁸ Pa

Explanation:

Pressure is defined by

P = F / A

If the gas is ideal for equal force eds on all the walls, so on the piston area we have

F = P A

We reduce the pressure to the SI system

P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa

we calculate

F = 150 10³ / 0.00133

F = 1.128 10⁸ Pa

3 0

3 years ago

Pls help i begg youuuuu

Anna [14]

Pitch is the sensation of certain frequencies to the ear. High frequency = high pitch, low frequency = low pitch.

f = c(speed of the wave) / λ (wavelength)

1. 343m/s / 0.77955m = 439.99 Hz
This corresponds to pitch A

2. 343m/s / 0.52028m = 659.26 Hz
 This corresponds to pitch E

3. 343m/s / 0.65552m = 523.349 Hz
 This corresponds to pitch C

4. using f = c / λ
λ = c / f
= 343m/s / 587.33 = 0.583999 m = 0.584 m

3 0

3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

5 0

3 years ago

A cyclist turns a corner with a radius of 50m at a speed of 10m/s. What is the cyclist's acceleration?

garri49 [273]

Answer:

2 m/s^2

Explanation:

a = v^2/r

a = (10m/s)^2 / 50m

a = 2 m/s^2

Leave a like and mark brainliest if this helped

5 0

3 years ago

A 21.0 kg shopping cart is moving with a velocity of 6.0 m/s. It strikes a 11.0 kg box that is initially at rest. They stick tog

DedPeter [7]

Answer:

a) 126 kgm/s

b) 0 kgm/s

c) 3.9 m/s

Explanation:

To solve this question, we will use the law of conservation of momentum.

Momentum before collision = momentum after collision

m1v1 + m2v2 = (m1 + m2)v, where

m1 = mass of the shopping cart, 21 kg

m2 = mass of the box, 11 kg

v1 = initial velocity of the shopping cart, 6 m/s

v2 = initial velocity of the box, 0 m/s

v = final velocity of the box+cart

Momentum of the shopping cart before collision = P

P = mv

P = 21 * 6

P = 126 kgm/s = c

Momentum of the box before collision

Like in question a above, the momentum of the box is P

P = mv

P = 11 * 0

P = 0 kgm/s = b

Velocity of the combined shopping cart wreckage after collision is

m1v1 + m2v2 = (m1 + m2)v

(21 * 6) + (11 * 0) = (21 + 11)v

126 + 0 = 32v

32v = 126

v = 126/32

v = 3.9375 m/s, on approximating to 1 decimal place, we have 3.9 m/s and option b as the answer.

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3 years ago