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3 years ago

What type of pressure is associated with more dense, sinking air?

Physics

2 answers:

NISA [10]3 years ago

6 0

B. low
it’s more dense

Verizon [17]3 years ago

3 0

B I hope it helps you

You might be interested in

a light bulb has a resistance of 360 . what is the current in the bulb when it has a potential difference of 120 v across it? 0.

xenn [34]

Presume we are looking for the current:

V = IR

120 = I*360

120/360 = I

1/3 = I

I = 1/3 = 0.333..

Current ≈ 0.33 Ampere.

7 0

3 years ago

Read 2 more answers

Two cars, one of mass 1100 kg, and the second of mass 2500 kg, are moving at right angles to each other when they collide and st

Pie

Answer:

$v_{f}=17.47 m/s$

Explanation:

Let's use the conservation of momentum to solve it.

$p_{initial}= p_{final}$ (1)

The total initial momentum will be: $m_{1}v_{1i}+m_{2}v_{2i}$
The total final momentum will be: $m_{1}v_{1f}+m_{2}v_{2f}$ , but as they stick together after the collision, v1f = v2f = vf.

So we can rewrite (1), using the above information:

$m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{f}+m_{2}v_{f}$

$m_{1}v_{1i}+m_{2}v_{2i}=v_{f}(m_{1}+m_{2})$

$v_{f}=\frac{m_{1}v_{1i}+m_{2}v_{2i}}{m_{1}+m_{2}}$

$v_{f}=\frac{1100\cdot 14+2500\cdot 19}{1100+2500}$

Finally, the magnitude of the velocity of the wreckage of the two cars immediately after the collision is:

$v_{f}=17.47 m/s$

I hope it helps you!

5 0

3 years ago

How do i calculate this?

Lesechka [4]

CAR 1

Momentum = Mass/Velocity

M = 2100/20

M = 105 m/s^2

CAR 2

Momentum = Mass/Velocity

M = 2100/30

M = 70 m/s^2

8 0

3 years ago

A satellite with mass 500 kg is placed in a circular orbit about Earth (Mass= 5.98 x 10^24 kg), radius = (6.4 x 10^6), a distanc

gizmo_the_mogwai [7]

Explanation:

a) F = GmM / r²

F = (6.67×10⁻¹¹) (500) (5.98×10²⁴) / (6.4×10⁶ + 1.5×10⁶)²

F = 3200 N

b) F = ma

3200 = 500a

a = 6.4 m/s²

c) a = v² / r

640 = v² / (6.4×10⁶ + 1.5×10⁶)

v = 7100 m/s

4 0

3 years ago

A pulley of radius 8.0 cm is connected to a motor that rotates at a rate 7000 rad s-1 and then decelerate uniformly at a rate of

zlopas [31]

Answer:

(a) α = - 1000 rad/s²

Negative sign represents deceleration.

(b) θ = 3581 rotations

(d) a = - 80 m/s²

Explanation:

(a)

using First equation of motion for angular motion:

ωf = ωi + αt

where,

ωf = Final Angular Speed = 2000 rad/s

ωi = Initial Angular Speed = 7000 rad/s

α = Angular Acceleration = ?

t = time = 5 s

Therefore,

2000 rad/s = 7000 rad/s + α(5s)

α = (2000 rad/s - 7000 rad/s)/5 s

α = - 1000 rad/s²

Negative sign represents deceleration.

(b)

Using second equation of motion:

θ = ωi t + (1/2)αt²

where,

θ = No. of Rotations = ?

Therefore,

θ = (7000 rad/s)(5 s) + (1/2)(- 1000 rad/s²)(5 s)²

θ = 35000 rad - 12500 rad

θ = (22500 rad)(1 rotation/2π rad)

θ = 3581 rotations

(c)

Length of String = L = (Circumference of Pulley)(θ)

L = [2π(0.08 m)][3581 rotations]

L = 1800 m

(d)

Tangential Acceleration = a = rα

a = (0.08 m)(-1000 rad/s²)

a = - 80 m/s²

4 0

3 years ago