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vekshin1
2 years ago
6

What are three elements that are in the same group

Physics
1 answer:
cluponka [151]2 years ago
3 0
Iodine, chloride, and bromide
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A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m
kifflom [539]

<u>Answer</u>:

(a) magnitude of F = 797 N

(b)the total work done  W = 0

(c)work done by the gravitational force =  -1.55 kJ

(d)the work done by the pull  = 0

(e) work your force F does on the crate = 1.55 kJ

<u>Explanation</u>:

<u>Given</u>:

Mass of the crate, m =  220 kg

Length of the rope, L = 14.0m

Distance, d =  4.00m

<u>(a) What is the magnitude of F when the crate is in this final position</u>

Let us first determine vertical angle as follows

=>Sin \theta = \frac{d }{L}

=> \theta = Sin^{-1} \frac{d}{L} =

Now substituting thje values

=> \theta = Sin^{-1} \frac{4}{12} =

=> \theta = Sin^{-1} \frac{1}{3}

=> \theta = Sin^{-1}(0.333)

=> \theta = 19.5^{\circ}

Now the tension in the string resolve into components

The vertical component supports the weight

=>Tcos\theta = mg

=>T = \frac{mg}{cos\theta}

=>T = \frac{230 \times 9.8 }{cos(19.5)}

=>T = \frac{2254 }{cos(19.5)}

=>T = \frac{2254 }{0.9426}

=>T =2391N

Therefore the horizontal force

F = TSin(19.5)

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

<u>c) The work done by the gravitational force on the crate</u>

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values                                                            

= -230 \times 9.8\times 12 ( 1 - cos(19.5) )

= -230 \times 9.8\times 12 ( 1 - 0.9426) )

= -230 \times 9.8\times 12 (0.0574)

= -230 \times 9.8\times 0.6888

=  -230 \times 6.750

= -1552.55 J

The work done by gravity = -1.55 kJ

<u>d) the work done by the pull on the crate from the rope</u>

Since the pull  is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg  

WF = -(-1.55)

WF = 1.55 kJ

<u>(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)</u>

Here the work done by force is not equal to F*d  

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)      

7 0
3 years ago
HELP ME ASAP!!!! PIC IS DOWN BELOW!
vagabundo [1.1K]
I think it is B Because it sounds like the best choice
7 0
2 years ago
Read 2 more answers
PLS HELP
Ahat [919]
<h3>B. True</h3>

"This was the idea that non-living objects can give rise to living organisms."

7 0
3 years ago
A 100 mH inductor whose windings have a resistance of 6.0 Ω is connected across a 12 Vbattery having an internal resistance of 3
Alexus [3.1K]

Answer:

The store energy in the inductor is 0.088 J

Explanation:

Given that,

Inductor = 100 mH

Resistance = 6.0 Ω

Voltage = 12 V

Internal resistance = 3.0 Ω

We need to calculate the current

Using ohm's law

V = IR

I=\dfrac{V}{R+r}

Put the value into the formula

I=\dfrac{12}{6.0+3.0}

I=1.33\ A

We need to calculate the store energy in the inductor

U=\dfrac{1}{2}LI^2

U=\dfrac{1}{2}\times100\times10^{-3}\times(1.33)^2

U=0.088\ J

Hence, The store energy in the inductor is 0.088 J

7 0
2 years ago
A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. Suppos
brilliants [131]

Answer:

v₂=- 34 .85 m/s

v₁=0.14 m/s

Explanation:

Given that

m₁=70 kg ,u₁=0 m/s

m₂=0.15 kg ,u₂=35 m/s

Given that collision is elastic .We know that for elastic  collision

Lets take their final speed is v₁ and v₂

From momentum conservation

m₁u₁+m₂u₂=m₁v₁+m₂v₂

70 x 0+ 0.15 x 35 = 70 x v₁ + 0.15 x v₂

70 x v₁ + 0.15 x v₂=5.25                   --------1

v₂-v₁=u₁-u₂            ( e= 1)

v₂-v₁ = -35        --------2

By solving above equations

v₂=- 34 .85 m/s

v₁=0.14 m/s

4 0
3 years ago
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