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3 years ago

Explain why atmospheric pressure decreases towards higher altitudes

Physics

1 answer:

denpristay [2]3 years ago

3 0

Answer:

At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels. ... Since most of the atmosphere's molecules are held close to the earth's surface by the force of gravity, air pressure decreases rapidly at first, then more slowly at higher levels.

Explanation:

happy to help:)

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Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v

wlad13 [49]

1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we call $v_f$ the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
$m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f$ (1)
where
$m_A=7 kg$ is the mass of ball A
$m_B=2 kg$ is the mass of ball B
$v_{Ai}=6 m/s$ is the initial velocity of ball A
$v_{Bi}=-12 m/s$ is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find $v_f$ , we find that the final velocity of the balls is
$v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s$
and the positive sign means the two balls are going to the right.

2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
$m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}$
$\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2$
where
$v_{fA}$ is the final velocity of ball A
$v_{fB}$ is the final velocity of ball B

If we solve simultaneously the two equations, we find:
$v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s$
$v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s$
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
$p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s$
and the total momentum after the collision is:
$p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s$

3 0

3 years ago

After Gru shrinks the moon are the side effects realistic to what would happen if the moon actually shrunk? Explain.

Elena-2011 [213]

Answer: No, water in the ocean wouldn't have tides wouldn't be as strong anymore.

Explanation:

7 0

2 years ago

You toss a 0.40-kg ball at 9.0 ms/ to a 14-kg dog standing on an iced-over pond. The dog catches the ball and begins to slide on

just olya [345]

Answer:

a) $v_{f}$ = 0.25 m / s b) u = 0.25 m / s

Explanation:

a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved

We will write the data

m₁ = 0.40 kg

v₁₀ = 9.0 m / s

m₂ = 14 kg

v₂₀ = 0

Initial

po = m₁ v₁₀

Final

$p_{f}$ = (m₁ + m₂) vf

po = pf

m₁ v₁₀ = (m₁ + m₂) $v_{f}$

$v_{f}$ = v₁₀ m₁ / (m₁ + m₂)

$v_{f}$ = 9.0 (0.40 / (0.40 +14)

$v_{f}$ = 0.25 m / s

b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass

u = 0.25 m / s

In the direction of movement of the ball

c) Let's calculate the kinetic energy in both moments

Initial

K₀ = ½ m₁ v₁₀² +0

K₀ = ½ 0.40 9 2

K₀ = 16.2 J

Final

$K_{f}$ = ½ (m₁ + m₂) $v_{f}$ 2

$K_{f}$ = ½ (0.4 +14) 0.25 2

$K_{f}$ = 0.45 J

ΔK = K₀ - $K_{f}$

ΔK = 16.2-0.445

ΔK = 1575 J

These will transform internal system energy

d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.

v₁₀’= v₁₀ -u

v₁₀’= 9 -.025

v₁₀‘= 8.75 m / s

v₂₀ ‘= v₂₀ -u

v₂₀‘= - 0.25 m / s

$v_{f}$ ‘= $v_{f}$ - u

$v_{f}$ = 0

Initial

K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²

Ko = ½ 0.4 8.75² + ½ 14.0 0.25²

Ko = 15.31 + 0.4375

K o = 15.75 J

Final

$k_{f}$ = ½ (m₁ + m₂) vf’²

$k_{f}$ = 0

All initial kinetic energy is transformed into internal energy in this reference system

3 0

3 years ago

A 1.0 m string with a 5 g stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of th

Andrew [12]

Answer:

Explanation:

A )

At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy

1/2 m V² = mg x 2r + 1/2 mv²

m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top

1/2 V² = g x 2r + 1/2 v²

V² = g x 4r + v²

V² = 9.8 x 4 + 8²

V² = 103.2

V = 10.16 m/s

B )

If T be the tension at the top

Net downward force

= mg + T . This force provides centripetal force for the circular motion

mg +T = mv² / r

T = mv²/r -mg

= m ( v²/r - g )

= .005 ( 8²/1 -g )

= .005 x 54.2

= .27 N .

C ) At the bottom

Net force = T - mg , T is tension at the bottom , V is velocity at bottom

T-mg = mV²/r

T = m ( V²/r +g )

= .005 ( 10.16²/1 +9.8)

= .005 x 113

= .56 N .

3 0

3 years ago

The solid rod shown has a radius of 0.75 in. if it is subjected to the force the 500lb determine the max normal stress developed

Kipish [7]

The normal stress follows the formula written below:

σ = F/A

There are two types of stress, axial and tangential. Since we are only given with the dimension of the radius (and not the length), the possible stress is axial. So, the area is,

A = πr² = π(0.75 in)² = 1.767 in²

So,
σ = F/A = 500 lb/1.767 in² = <em>282.94 psi</em>

7 0

3 years ago