Graphing Motion

The latent heat of fusion of alcohol is 25 kcal/kg and its melting point is -114 o C. It has a specific heat of 0.60 in its liqu

UNO [17]

Answer:

=170kcal

Explanation:

We first calculate the amount of energy required to melt the alcohol using the formula: MLf, where Lf is the latent heat of fussion

We then calculate amount of heat required to raise the temperature of liquid alcohol to -14° C using MC∅. We then add the two.

Thus ΔH=MLf+MC∅

ΔH=2kg×25kcal/kg+ 2kg×(0.6kcal/kg.K×(-14-⁻114)

=50kcal+120kcal

=170kcal

3 0

2 years ago

I hope you are able to read this question?? Help ASAP this question is on the quiz tommorow

Masteriza [31]

You would be correct.

Because you have only JUST released the arrow, and how close he is to the target, it would have the same amount of energy when it strikes the target. Yes, the kinetic energy would be destroyed when you hit the target but not right away. And yes, the potential energy would also be destroyed once you release the arrow, but it goes straight back once it stops moving, aka when it hits the target, although it has only just stopped moving.

Hope this helps!

8 0

2 years ago

Is the frictional force the same as the applied force when the net force equals zero?

DerKrebs [107]

Answer:

Since the net force is to the right (in the direction of the applied force), then the applied force must be greater than the friction force. The friction force can be determined using an understanding of net force as the vector sum of all the forces.

Explanation:

6 0

2 years ago

You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel wi

Elenna [48]

Answer:

$\frac{g_{2}}{g_{1}} = \frac{1}{4}$

Explanation:

The period of the simple pendulum is:

$T = 2\pi\cdot \sqrt{\frac{l}{g} }$

Where:

$l$ - Cord length, in m.

$g$ - Gravity constant, in $\frac{m}{s^{2}}$ .

Given that the same pendulum is test on each planet, the following relation is formed:

$T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}$

The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:

$\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}$

$\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}$

$\frac{g_{2}}{g_{1}} = \frac{1}{4}$

5 0

2 years ago

An automobile engine delivers 47.4 hp. how much time will it take for the engine to do 6.82 × 105 j of work? one horsepower is e

kogti [31]

1 horsepower is equal to 746 W, so the power of the engine is
$P=47.4 hp \cdot 746 \frac{W}{hp}=35360 W$
The power is also defined as the energy E per unit of time t:
$P= \frac{E}{t}$
Where the energy corresponds to the work done by the engine, which is $E=6.82 \cdot 10^5 J$ . Re-arranging the formula, we can calculate the time t needed to do this amount of work:
$t = \frac{E}{P}= \frac{6.82 \cdot 10^5 J}{35360 W}=19.3 s$

8 0

2 years ago