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4 years ago

Which of the following is NOT NECESSARILY a property of an air mass?

Physics

1 answer:

lilavasa [31]4 years ago

4 0

Answer:

Explanation:

The property of air mass include

1) it must be large.

2) it must have relatively uniform properties.

3)it must travel as a recognizable entity.

It must have a warm front at its leading edge, is not necessarily a property of an air mass because Not all air masses have a warm front at their leading edge. There are five types of air masses and different types of the front can be developed.

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Review. From a large distance away, a particle of mass 2.00 g and charge 15.0σC is fired at 21.0 i^ m/s straight toward a second

MissTica

(a)

Determine the system's initial configuration at ri = infinite particle separation and the system's final configuration at the point of closest approach.

Since the two-particle system is not being affected by any outside forces, we may treat it as an isolated system for momentum and use the momentum conservation law.

m1v1 + m1v2 = (m1+m2)v

The second particle's starting velocity is zero, so:

m1v1 = (m1+m2)v

After substituting the values we get,

v = 6i m/s

(b)

Since the two particle system is also energy-isolated, we may use the energy-conservation principle.

dK + dU = 0

Ki +Ui = Kf + Uf

Substituting the values,

1/2m1v1^2i + 1/2 m2v2^2i + 0 = 1/2m1v1^2f + 1/2m2v2^2f +ke q1q2/rf

The second particle's initial speed is 0 (v2 = 0). Additionally, both the first and second particle's final velocity have the same value, v. Put these values in place of the preceding expression:

1/2m1v1^2i = 1/2m1v1^2 + 1/2m2v2^2 +ke q1q2/rf

After solving we get,

rf = 2ke q1q2 / m1v1^2 - (m1+m2)v^2

Substituting the values we get,

rf = 3.64m

(c)

v1f = (m1-m2 / m1 + m2) v1i

v1f = -9i m/s

(d)

v2f = (2m1/ m1 +m2) v1i

After substituting the values,

v2f = 12i m/ s

Question :

Review. From a large distance away, a particle of mass 2.00 g and charge 15.0 \muμC is fired at 21.0 m/s straight toward a second particle, originally stationary but free to move, with mass 5.00 g and charge 8.50 \muμC. Both particles are constrained to move only along the x axis. (a) At the instant of closest approach, both particles will be moving at the same velocity. Find this velocity. (b) Find the distance of closest approach. After the interaction, the particles will move far apart again. At this time, find the velocity of (c) the 2.00-g particle and (d) the 5.00-g particle. \hat{i}

To learn more about momentum conservation law click on the link below:

brainly.com/question/7538238

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5 0

2 years ago

A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One

Andre45 [30]

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

$\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\$

To get the required work done, we will divide the mass by the speed of one knot to have:

$w=\frac{7230}{0.51}\\w= 14,176.47Joules$

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

Learn more here: brainly.com/question/25573786

8 0

3 years ago

Please help on this one?

kupik [55]

Answer: A)30V. First find the current of the circuit. I=V/R(total resistance). So I=60/120=0.5. Now to find voltage drop in R3 use ohms law as given. V(of 3)=(0.5)(60)=30V

7 0

3 years ago

A 3.3 kg ball sits on the ground and is kicked with a FAPP of 36N

jeyben [28]

a) 32.3 N

The force of gravity (also called weight) on an object is given by

W = mg

where

m is the mass of the object

g is the acceleration of gravity

For the ball in the problem,

m = 3.3 kg

g = 9.8 m/s^2

Substituting, we find the force of gravity on the ball:

$W=(3.3)(9.8)=32.3 N$

b) 48.3 N

The force applied

$F_{app} = 36 N$

The ball is kicked with this force, so we can assume that the kick is horizontal.

This means that the applied force and the weight are perpendicular to each other. Therefore, we can find the net force by using Pythagorean's theorem:

$F=\sqrt{W^2+F_{app}^2}$

And substituting

W = 32.3 N

Fapp = 36 N

We find

$F=\sqrt{32.3^2+36^2}=48.3 N$

c) $14.6 m/s^2$

The ball's acceleration can be found by using Newton's second law, which states that

F = ma

where

F is the net force on an object

m is its mass

a is its acceleration

For the ball in this problem,

m = 3.3 kg

F = 48.3 N

Solving the equation for a, we find

$a=\frac{F}{m}=\frac{48.3}{3.3}=14.6 m/s^2$

8 0

4 years ago

In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue

agasfer [191]

Answer:

a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s

Explanation:

a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

a) the initial impulse is

p₀ = m v₁₀ + 0

p₀ = 0.6 2

p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

we write the final moment for each axis

X axis

p₀ₓ = 1.2 kg m / s

p_{fx} = m v1f cos 20 + m v2f cos θ

p₀ = p_f

1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

1.2482 = v_{2f} cos θ

Y axis

p_{oy} = 0

p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

0.2736 = v_{2f} sin θ

we write our system of equations

0.2736 = v_{2f} sin θ

1.2482 = v_{2f} cos θ

divide to solve

0.219 = tan θ

θ = tan⁻¹ 0.21919

θ = 12.36

let's look for speed

0.2736 = v_{2f} sin θ

v_{2f} = 0.2736 / sin 12.36

v_{2f} = 1.278 m / s

7 0

2 years ago