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3 years ago

15

Particles q1, q2, and q3 are in a straight line. Particles q1 = -28.1 μC, q2 = +25.5 μC, and q3 = -47.9 μC. Particles q1 and q2

are separated by 0.300 m. Particles q2 and q3 are separated by 0.300 m. What is the net force on q3?

2 answers:

Arturiano [62]3 years ago

3 0

Answer:-88.4

Explanation:

Digiron [165]3 years ago

3 0

Answer:

-88.4

Explanation:

hope it helps

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A backyard swimming pool with a circular base of diameter 6.00m is filled to depth 1.50m. (b) Two persons with combined mass 150

Mashutka [201]

The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.

<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

$\rho \mathrm{Vg}=2 \mathrm{mg} \\$

$V=\frac{2 m}{\rho} \\$

$V=\frac{2 \times 150}{1000} \\$

$\mathrm{~V}=0.3 \mathrm{~m}^3$

Area of pool $=\frac{\pi}{4} d^2$

$&=\frac{\pi}{4} \times 6^2 \\$

$&=28.27 \mathrm{~m}^2$

Height of the water rise

$h &=\frac{V}{A} \\$

$h &=\frac{0.3}{28.27} \\$

$& \mathrm{~h}=0.0106 \mathrm{~m}$

Pressure increased

P = ρ g h

P = 1000 x 10 x 0.0106

P = 106.103 Pa

To learn more about absolute pressure refer to:

brainly.com/question/17200230

#SPJ4

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2 years ago

Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the s

Nadusha1986 [10]

Answer:

Explanation:

Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

velocity of approach = 1.5 - 0 = 1.5

velocity of separation = v₁ + v₂

coefficient of restitution = velocity of separation / velocity of approach

.8 = v₁ + v₂ / 1.5

v₁ + v₂ = 1.2

applying law of conservation of momentum

m x 1.5 + 0 = mv₂ - mv₁

1.5 = v₂ - v₁

adding two equation

2 v ₂= 2.7

v₂ = 1.35 m /s

v₁ = - .15 m / s

During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

For second collision ,

coefficient of restitution = velocity of separation / velocity of approach

.5 = v₃ + v₄ / 1.35

v₃ + v₄ = .675

applying law of conservation of momentum

m x 1.35 + 0 = mv₄ - mv₃

1.35 = v₄ - v₃

adding two equation

2 v ₄= 2.025

v₄ = 1.0125 m /s

v₃ = - 0 .3375 m / s

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4 years ago

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Answer:

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3 years ago

Read 2 more answers