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xxTIMURxx [149]
3 years ago
15

A galaxy whose stars are arranged in a disk with arms that surround a central bulge is a _____ galaxy. irregular elliptical clus

ter spiral
Physics
2 answers:
maks197457 [2]3 years ago
5 0
A galaxy whose stars are arranged in a disk with arms that surround a central bulge is a <span>spiral galaxy. the rest of the choices do not answer the question above</span>
qaws [65]3 years ago
3 0

Answer;

Spiral galaxy

Explanation;

In a spiral galaxy, the stars, gas and dust are gathered in spiral arms that spread outward from the galaxy's center. Spiral galaxys are divided into three main types depending on how tightly wound their spiral arms are: Sa, Sb and Sc.

All spiral galaxies can broadly be described visually as having a central bulge of old stars surrounded by a flattened disk of young stars, gas and dust.

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For the following elementary reaction 2br• -&gt; br2-. The rate of consumption of the reaction and the rate of formation of prod
Scorpion4ik [409]

Answer: -\frac{1}{2}\times \frac{d[Br^.]}{dt}=+\frac{d[Br_2]}{dt}

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

2Br^.\rightarrow Br_2

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate=-\frac{d[Br^.]}{2dt}

or Rate=+\frac{d[Br_2]}{dt}

Thus -\frac{d[Br^.]}{2dt}=+\frac{d[Br_2]}{dt}

4 0
3 years ago
he coefficent of static friction between the floor of atruck and a box resting on it is 0.30. The truck is travelingat 80.0 km/h
Vedmedyk [2.9K]

Answer:

83.97 m

Explanation:

Without any other external force causing motion in the horizontal direction, it is the frictional force that brings the box to rest. The frictional force has to match the force due to deceleration experienced by the truck to not move.

So,

Frictional force = ma

Then, we calculate frictional force now,

In the vertical direction, the force balance has the weight and the normal reaction equal

N = W = mg

And frictional force = μN = μmg

where μ is the coefficient of friction = 0.3

ma = μmg

a = μg = 0.3 × 9.8 = 2.94 m/s²

Then, using the equations of motion, we obtain the distance over which this deceleration stops the truck

u = initial velocity of the truck = 80 km/h = 22.22 m/s

v = final velocity of the truck = 0 m/s, since the truck comes to rest

a = - 2.94 m/s² (negative sign because it's a deceleration)

x = distance covered during this deceleration = ?

v² = u² + 2ax

0² = 22.22² + 2(-2.94)(x)

5.88x = 493.73

x = 83.97 m

4 0
3 years ago
Read 2 more answers
A car is safely negotiating an unbanked circular turn at a speed of 25 m/s. The road is dry, and the maximum static frictional f
Brums [2.3K]

Answer:

The velocity must be reduced to one third to stay on the road

Explanation:

The sideways force that friction must resist comes from the centrifugal acceleration due to the turn.

fc=mv2Rfc=mv2R

the frictional force is given by

ff=μmgff=μmg where μμ is the static friction coefficient

if the car is not to skid

fc≤fffc≤ff so    

mv2R≤μmgmv2R≤μmg

v≤μgR−−−−√v≤μgR

thus vv varies as the square root of μμ

so if μμ is reduced by 9, vv must be reduced by 9–√=39=3

and thus the speed must be reduced to<u> 26</u> m/s

                                                                    3

8 0
3 years ago
The 8.00-cm long second hand on a watch rotates smoothly.
dolphi86 [110]
The watch hand covers an angular displacement of 2π radians in 60 seconds.

ω = 2π/60
ω = 0.1 rad/s

v = ωr
v = 0.1 x 0.08
v = 8 x 10⁻³ m/s
4 0
3 years ago
Please help/ show work !!! Also don’t answer by putting a link!!
Zepler [3.9K]
I hope this helps you out!

5 0
3 years ago
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