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2 years ago

Que la tarcer ley de newton

Physics

1 answer:

slava [35]2 years ago

3 0

Explanation:

If you write it in English so I can help u if you need it

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If you add air to a flat tire through a single small entry hole, why does the air spread out to fill the tire

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The tire fills up just like any thing else that holds air when u pump a ball or tire up it fills all the way up cause it is a small confined space and after filling it with air the atoms of the air fill the tire up

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2 years ago

Read 2 more answers

While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, cau

eduard

Answer:

The speed is $v =8.17 m/s$

Explanation:

From the question we are told that

The angle of slant is $\theta = 37.0^o$

The weight of the toolbox is $W_t = 92.0N$

The mass of the toolbox is $m = \frac{92}{9.8} = 9.286kg$

The start point is $d = 4.25m$ from lower edge of roof

The kinetic frictional force is $F_f = 22.0N$

Generally the net work done on this tool box can be mathematically represented as

$Net \ work done = Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction$

The workdone due to weigh is = $mgsin \theta * d$

The workdone due to friction is = $F_f \ cos\theta * d$

Substituting this into the equation for net workdone

$W_{net} = mgsin\theta * d + F_f \ cos \theta *d$

Substituting values

$W_{net} = 92 * sin (37) * 4.25 + 22 cos (37) * 4.25$

$= 309.98 J$

According to work energy theorem

$W_{net} = \Delta Kinetic \ Energy$

$W_{net} = \frac{1}{2} m (v - u)^2$

From the question we are told that it started from rest so u = 0 m/s

$W_{net} = \frac{1}{2} * m v^2$

Making v the subject

$v = \sqrt{\frac{2 W_{net}}{m} }$

Substituting value

$v = \sqrt{\frac{2 * 309.98}{9.286} }$

$v =8.17 m/s$

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2 years ago

A 6 kg object falls 10 m. The object is attached mechanically to a paddle-wheel which rotates as the object falls. The paddle-wh

maksim [4K]

Answer: $16.234^{\circ}C$

Explanation:

Given

Mass of object (m)=6 kg

falling height(h)=10 m

mass of water( $m_w$ )=600 gm

temperature of water =15

specific heat of water $=4.186 j/g-^{\circ}C$

Let T be the Final Temperature of water

Here Object Potential Energy is converted into Heat energy which will be absorbed by water

Potential Energy(P.E.) $=mgh=6\times 9.81\times 10=588.6 J$

Heat supplied $=m_wc(\Delta T)$

$H.E.=600\times 4.186\times (T-16)$

$588.6=2511.6\times (T-16)$

T-16=0.234

$T=16.234^{\circ}C$

This is not an efficient way of heating water as there is only $0.234^{\circ}C$ increase in temperature.

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If i try to fail and succeed which one did I do

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You've failed because you failing becomes a statement rather than it becoming fact or what actually happened.

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Time Warner Cable's leadership development program that spanned over 30 days and included weekly videos, practice exercises, and

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Answer: A.

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Explanation:

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