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3 years ago

12

"The Mount Pinatubo Particle Model shows that the initial atmospheric flow of erupted fine ash and sulfur oxide was from _______

_. Subsequently, the prevailing westerlies dispersed the material to higher latitudes."

1 answer:

Leya [2.2K]3 years ago

6 0

Answer:

what

Explanation:

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(4 points) A mother with mass m1 is skating at velocity v1 behind her daughter whose mass is m2, who is skating at v2 . Instead

STatiana [176]

Answer:

B) collision is inelastic because they stick together after collision and share a common final velocity Vf

C) M1V1 + M2V2 = (M1 + M2)Vf

D) Vf = 6.33m/s

E) force = 3040N

Explanation:

Detailed explanation and calculation is shown in the image below

3 0

3 years ago

Someone help me haha;(

Nikitich [7]

Answer: vf = 51 m/s

d = 112 m

Explanation: Solution attached:

To find vf we use acceleration equation:

a = vf - vi / t

Derive to find vf

vf = at + vi

Substitute the values

vf = 3.5 m/s² ( 8.0 s) + 23 m/s

= 51 m/s

To solve for distance we use

d = (∆v)² / 2a

= (51 m/s - 23 m/s )² / 2 ( 3.5 m/s²)

= (28 m/s)² / 7 m/s²

= 784 m/s / 7 m/s²

= 112 m

6 0

3 years ago

A comet with a mass of 7.85 × 1011 kg is moving with a velocity of 25,000 m/s. calculate its kinetic energy.

Nadusha1986 [10]

The kinetic energy of an object is given by:
$K= \frac{1}{2}mv^2$
where
K is the kinetic energy
m is the mass of the object
v is its velocity.

The comet in our problem has a mass of $m=7.85 \cdot 10^{11} kg$ and a velocity of $v=25000 m/s$ , so its kinetic energy is:
$K= \frac{1}{2}(7.85 \cdot 10^{11} kg)(25000 m/s)^2=2.45 \cdot 10^{20}J$

6 0

3 years ago

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

Gnesinka [82]

Answer:

a). $a_p=-2.39x10^{-12} rad/s^2$

b). $t=1016298.8 years$

c). $T_i=80.58x10^{-3}s$

Explanation:

a).

The acceleration for definition is the derive of the velocity so:

$a_p=\frac{dw}{dt}$

$w=\frac{2\pi}{t}$

$a_p=\frac{dw}{dt}=-\frac{2\pi}{t^2}*\frac{dT}{dt}$

$dT=0.0808s$

$dt=1 year*\frac{365d}{1year} \frac{24hr}{1d} \frac{60minute}{1hr} \frac{60s}{1minute}=31.536x10^{6}s$

Replacing

$a_p=-\frac{2\pi}{0.082s^2}*\frac{9.84x10^{-7}}{31.536x10^{6}s}= -2.39x10^{-12} rad/s^2$

b).

If the pulsar will continue to decelerate at this rate, it will stop rotating at time:

$t=\frac{w}{a_p}$

$w=\frac{2\pi }{t}=\frac{2\pi }{0.0820s}=76.62 rad/s$

$t=\frac{76.62 rad/s}{2.39x10^{-12}rad/s^2}= 3.2058x10^{13}s$

$t=1016298.8 years$

c).

582 years ago to 2019

1437

$T_i=0.0820-9.84x10^{-7}*1437)=80.58x10^{-3}s$

5 0

3 years ago

Who wrote the universal gravitation

ira [324]

I think it was Isaac Newton

3 0

3 years ago