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mote1985 [20]
3 years ago
11

The heat vaporization for methyl alcohol is 1100 kj/kg. It is 2257 KJ/Kg for water. Thus means that______________.

Physics
1 answer:
dusya [7]3 years ago
7 0

Answer: B) Methyl alcohol requires less than half as much energy per kg to evaporate than water does

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 kilo gram of liquid into its vapor state without change in its temperature.

Heat of vaporization is more for water than for methyl alcohol which means more heat is required to convert from liquid to vapour form.

As the Heat of vaporization for methyl alcohol  (1100) is almost half as that of Heat of vaporization for water (2257) , it means  Methyl alcohol requires less than half as much energy per kg to evaporate than water does.

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<h3>Answer : </h3><h3 /><h3>A ) The larger gear can be moved by applying a relatively small force on the smaller gear.</h3>

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D ) the system would continue to move without any further, after and initial force has set in motion.

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3 years ago
Light intensity is?
Fittoniya [83]
The correct answer is:
<span>The rate at which a waves energy flows through a given unit of area

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</span>I= \frac{P}{A}
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3 years ago
Read 2 more answers
A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about
Olin [163]

Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity  Vₓ is 0

The accelerationis given by =1000 m/s²

the stopping distance = x-x₀=?

So we can wind the stopping distance by following formula

2 (-1000)(x-x₀)=[0^{2} -11^{2}]

x-x₀=6.05*10^{-2} m

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3 0
3 years ago
One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing
Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

m = m_{1}=m_{2}

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Along X- axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}

v_{1}=u_{1}+u_{2}

Put the value into the formula

v_{1}=2.2i-0.80i

v_{1}=1.4i\ m/s

Along Y-axis

0=m_{1}v_{1}+m_{2}v_{2}

m_{1}v_{1}=-m_{2}v_{2}

v_{1}=-v_{2}

Put the value into the formula

v_{1}=-1.36j\ m/s

Then the final speed of the first ball

v_{1}=\sqrt{(1.4)^2+(1.36)^2}

v_{1}=1.95\ m/s

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

\tan\theta=\dfrac{v_{2}}{v_{1}}

\tan\theta=\dfrac{-1.36}{1.4}

\theta=\tan^{-1}\dfrac{-1.36}{1.4}

\theta=-44.16^{\circ}

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

7 0
3 years ago
falling objects drop with an average acceleration of 9.8m/sec/sec. if an object falls from a tall building how long will it take
sveta [45]

Answer:

5 seconds

Explanation:

<em>Acceleration = (final velocity - initial velocity) ÷ time</em>

<em>a =  \frac{v - u}{t}</em>

<em>9.8 =  \frac{49 - 0}{t}</em>

<em>9.8 =  \frac{49}{t}</em>

<em>9.8t = 49</em>

<em>t =  \frac{49}{9.8}</em>

<em>t = 5</em>

8 0
3 years ago
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