Hey can yall help me out in dis

How many stars are contained in the milking way galaxy A) 0 b)1 c) billions d)infinite

skelet666 [1.2K]

Answer:

c) billions

Explanation:

there are more than 0 stars, because our sun is a star

and there are more than one because the milky way has a lot of suns

there aren't infinite because the milky way is finite it has some volume and even you made the smallest star possible you wouldn't have infinite stars

so the only option is billions

7 0

3 years ago

2) [25 pts] The bob of mass m=5 kg shown in the figure is being held by a force F. If the angle is 0⃗

Colt1911 [192]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The tension is $T = 48.255 \ N$

b

The time taken is $t = 0.226 \ s$

c

The position for maximum velocity is

S = 0

d

The maximum velocity is $V_{max} =0.384 \ m/s$

Explanation:

The free body for this question is shown on the second uploaded image

From the question we are told that

The mass of the bob is $m_b = 5 \ kg$

The angle is $\theta = 10^o$

The length of the string is $L = 0.5 \ m$

The tension on the string is mathematically represented as

$T = mg cos \theta$

substituting values

$T = 5 * 9.8 cos(10)$

$T = 48.255 \ N$

The motion of the bob is mathematically represented as

$S = A sin (w t + \frac{\pi }{2} )$

=> $S = A sin (wt)$

Where $w$ is the angular speed

and $\frac{\pi}{2}$ is the phase change

At initial position S = 0

So $wt = cos^{-1} (0)$

$wt = 1$

Generally $w$ can be mathematically represented as

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

Where T is the period of oscillation which i mathematically represented as

$T = 2 \pi \sqrt{\frac{L}{g} }$

So

$t = \frac{1}{w}$

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

$t = \sqrt{\frac{L}{g} }$

substituting values

$t = \sqrt{\frac{0.5}{9.8} }$

$t = 0.226 \ s$

Looking at the equation

$wt = 1$

We see that maximum velocity of the bob will be at S = 0

i. e $w = \frac{1}{t}$

The maximum velocity is mathematically represented as

$V_{max} = w A$

Where A is the amplitude which is mathematically represented as

$A = L sin \theta$

So

$V_{max} = \frac{2 \pi }{T } L sin \theta$

$V_{max} = \frac{2 \pi }{2 \pi } \sqrt{\frac{g}{L} } L sin \theta$ Recall $T = 2 \pi \sqrt{\frac{L}{g} }$

$V_{max} = \sqrt{gL} sin \theta$

substituting values

$V_{max} = \sqrt{9.8 * 0.5 } sin (10)$

$V_{max} =0.384 \ m/s$

6 0

2 years ago

A train accelerates from 23m/s to 190m/s in 54 seconds. What was its acceleration?

Ierofanga [76]

Answer:

Use the method on the image and solve it.

3 0

3 years ago

Porque crees que el agua no podia subir mas de 10 m de altura con ese tipo de bomba

Vinvika [58]

I don’t speak this language

5 0

3 years ago

problems like this A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet fi

Ymorist [56]

This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

Answer:

(a) Xs=0.459m

(b) t=0.984 s

(c) Vc=6.65 m/s

Explanation:

(a) To reach maximum distance

$g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\ x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m$

(b) For Time

To find t we must find t1 and t2

as

t=t1+t2

For T1

$t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s$

For T2

$x_{l}=Vbt+(1/2)gt_{2}^{2}\\ as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s$

For Total Time

t=t1+t2

t=0.306+0.6789

t=0.984s

(c) To find Vc

Vc=Vb+gt2

Vc=(0)+(9.8)(0.6789)

Vc=6.65 m/s

7 0

2 years ago