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AfilCa [17]
3 years ago
5

A mass of 5 kg is moving at 5 m/s when it collides with another mass of 2 kg moving in the same direction at 2 m/s. After the co

llision they become bonded together. What is the final velocity of the two masses together? 2.22 m/s
6.23 m/s

3.56 m/s

4.14 m/s
Physics
1 answer:
Paul [167]3 years ago
6 0

Answer:

4.14 m/s

Explanation:

m¹u¹+m²u²=(m¹+m²)v

5(5)+2(2)=(5+2)v

25+4=7v

29=7v

29/7=v

v=4.14 m/s

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The magnetic field at the center of a wire loop of radius , which carries current , is 1 mT in the direction (arrows along the w
Citrus2011 [14]

Complete Question

 The complete question is shown on the first uploaded image

Answer:

The magnetic field is B_{net} = \frac{1}{4}  * mT

And the direction is  -\r k

Explanation:

      From the question we are told that

                 The magnetic field at the center is B = 1mT

Generally magnetic field is mathematically represented as

              B = \frac{\mu_o I}{2R}

We are told that it is equal to 1mT

So

                B = \frac{\mu_o I}{2R} = 1mT

From the first diagram we see that the effect of the current flowing in the circular loop is  (i.e the magnetic field generated)

                         \frac{\mu_o I}{2R} = 1mT

 This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)

                   B_1 = \frac{1}{2} \frac{\mu_o I}{2R}

and  for the larger semi-circular loop  is

                 B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}

Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction

    So the net magnetic field would be

                   B_{net} = B_1 - B_2

                        =  \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}

                        =\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}

                        =\frac{\mu_o I}{8R}

                        = \frac{1}{4} \frac{\mu_o I}{2R}

Recall  \frac{\mu_o I}{2R} = 1mT

    So  

             B_{net} = \frac{1}{4}  * mT

Using the Right-hand rule we see that the direction is into the page which is -k

3 0
3 years ago
Choose the word that BEST completes the following sentence. Saturn’s rings have been assigned a letter, based on their location.
puteri [66]

Answer:

The correct answer would be Saturn's Cassini Division.

Explanation:

Read about it here.

https://caps.gsfc.nasa.gov/simpson/kingswood/rings/

Hope this helps! :)

5 0
2 years ago
ILL ADD BRAINLIEST
Marina86 [1]

This is kinda confusing. I wish u just to a screenshot of the problem but here goes...

Forest at highest latitudes- Hardwood trees/deer, squirrel, foxes

Praries/temperate climate- Mostly small mammals/scrubs/steppes

High humidity/rainfall near equator- Abundant thick vegatation/manny species

No trees/ polar bears/ mosses- 25cm rain/few animals

6 0
3 years ago
A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900
Whitepunk [10]
It would be 17 m/s

If we use

V2 = V1 + a*t
Sub in 5 for v1
2m/s*2 for a
And
6 for t
That should give you the answer.
5 0
2 years ago
Read 2 more answers
Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the
stealth61 [152]

Answer:

\frac{50}{\pi }Hz

Explanation:

In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;

V(t) = V sin (ωt + Ф)            -----------------(i)

Where;

V = amplitude value of the voltage

ω = angular frequency = 2 π f        [f = cyclic frequency or simply, frequency]

Ф = phase difference between voltage and current.

<u><em>Now,</em></u>

From the question,

V(t) = 230 sin (100t)              ---------------(ii)

<em><u>By comparing equations (i) and (ii) the following holds;</u></em>

V = 230

ω = 100

Ф = 0

<em><u>But;</u></em>

ω = 2 π f = 100

2 π f = 100             [divide both sides by 2]

π f = 50

f = \frac{50}{\pi }Hz

Therefore, the frequency of the voltage is \frac{50}{\pi }Hz

7 0
2 years ago
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