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3 years ago

Calculate the energy extracted from 50g water vapour 100 ℃ to transform it into water at 80 Celsius.

Physics

1 answer:

myrzilka [38]3 years ago

4 0

Answer:lefmprkfniou4gjkfjrnwerkjdkcheouvwe

Explanation:fefefefrgff

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The plates on a capacitor have a radius of 1.5 mm and are separated by a distance of 0.43 mm. The space between the plates is fi

zhenek [66]

So, C = kE°A/d

putting the values,

C
= 3.8 × 8.85×10^(-12) × 3.14×1.5×1.5 × 10^(-6)/0.43 × 10^(-3)

so, 1.02 × 10^(-13)

so the most appropriate answer is 2 ...that is
1.4 × 10^(-13) ....answer !!

3 0

3 years ago

Please help ASAP....

Solnce55 [7]

Explanation:

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2 years ago

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

3 0

3 years ago

A machine carries a 100kg cargo to a boat at a rate of 10m/s2. The distance between the ground to the boat is 50ft. If the machi

klasskru [66]

Answer:

50.8 watt

Explanation:

we know that P=W÷t

W=F.S S-->distance=50 ft= 15.24 m

F=ma

=100×10=1000 N

SO W= 1000×15.24

=15240 J

NOW

P=W÷t t=5 mints = 5×60=300 sec

P=15240÷300

P=50.8 watt

8 0

3 years ago

Sediment laid down by glacial meltwater is called

Mrac [35]

<span>Your answer isStratified drift</span>

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3 years ago

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