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2 years ago

14

The Milky Way Galaxy is (a) another name for our solar system; (b) a small group of stars visible in our night sky; (c) a collec

tion of more than 100 billion stars, of which our Sun is one

1 answer:

gtnhenbr [62]2 years ago

3 0

Answer:

C

Explanation:

please give brainliest

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The scale is 1:30 and their are 10 boulders. How many boulders are their in the real pond

Alenkasestr [34]

Answer:

the correct answer is C

Explanation:

When we express that the scale is 1:30 we mean that the objects of the realization are reduced by a factor of 30 in the graph, for example a distance of 30 cm in the graph is represented by a distance of 1 cm.

Therefore something that in the graph has n value to bring it to real size must be multiplied by the scale.

Applying this to our case if there is

10 boulder on the chart

in reality there are #_boulder = 10 30

#_boulder = 300 boulder

so the correct answer is C

7 0

3 years ago

A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w

kramer

Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = ?

$v_{1i}$ = initial velocity of the first body before collision = $v$

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

$v_{1f}$ = final velocity of the first body after collision =

using conservation of momentum equation

$m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}$

Using conservation of kinetic energy

$m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35$

b)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = 1.35 kg

$v_{1i}$ = initial velocity of the first body before collision = 4 ms⁻¹

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

$v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67$ ms⁻¹

8 0

3 years ago

How many layers does the world have?

Lemur [1.5K]

The world has a total of 7 layers

6 0

3 years ago

Read 2 more answers

As you slide a heavy box across the floor, friction applies a force of -100 N

nirvana33 [79]

Answer:

A -500J

Explanation:

because W=Fs

100 × 5 = 500

4 0

3 years ago

Read 2 more answers

Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge of Object 2 is tripled, then the new

Leya [2.2K]

Answer:

The force will be 54.0 units

Explanation:

The magnitude of the electrostatic force between two charged objects is given by Coulomb's Law:

$F=\frac{kq_1 q_2}{r^2}$

where

k is Coulomb's constant

q1, q2 are the magnitude of the two charges

r is the separation between the two charges

From the equation, we see that the magnitude of the force is directly proportional to the charge of object 2:

$F\propto q_2$

In this problem, the initial force between the two objects is

F = 18.0 N

And so, when the charge on object 2 is tripled,

$q_2'=3q_2$

The new electrostatic force will be

$F'\propto q_2' = (3q_2) = 3F$

So, the force will also triple: since the original force was 18.0 units, the new force will be

$F'=3F=3(18.0)=54.0$

5 0

3 years ago