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3 years ago

What is it that lives if it is fed, and dies if you give it a drink?

Physics

1 answer:

IceJOKER [234]3 years ago

6 0

A fire because you feed it wood, it lives longer but if you give it a drink, water, then it dies.

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All objects, regardless of their mass, fall with the same rate of acceleration on

tamaranim1 [39]

Answer:

A -TRUE

Explanation:

The mass, size, and shape of the object are not a factor in describing the motion of the object. So all objects, regardless of size or shape or weight, free fall with the same acceleration.

4 0

3 years ago

Read 2 more answers

Answer either one or both questions! Must explain and show work to receive brainliest!

MariettaO [177]

<span>F x L = W x X whereW=weight is total load = 80, L is length from fulcrum which is the unknown and what we are solving for. x= length we know. and F equals 50 force we know. So (W*X)/F=LL equals 64</span>

3 0

3 years ago

We are going to make an imaginary engine using water. We are going to heat 100 grams of water to 120 C from its initial temperat

Svetach [21]

Answer:

The work done by this engine is 800 cal

Explanation:

Given:

100 g of water

120°C final temperature

22°C initial temperature

30°C is the temperature of condensed steam

Cw = specific heat of water = 1 cal/g °C

Cg = specific heat of steam = 0.48 cal/g °C

Lw = latent heat of vaporization = 540 cal/g

Question: How much work can be done using this engine, W = ?

First, you need to calculate the heat that it is necessary to change water to steam:

$Q_{1} =m_{w} C_{w} (100-22)+m_{w}L_{w}+m_{w}C_{g}(120-100)$

Here, mw is the mass of water

$Q_{1} =(100*1*78)+(100*540)+(100*0.48*20)=62760cal$

Now, you need to calculate the heat released by the steam:

$Q_{2} =m_{w}C_{g}(120-100)+m_{w}L_{w}+m_{w}C_{w}(100-30)=(100*0.48*20)+(100*540)+(100*1*70)=61960cal$

The work done by this engine is the difference between both heats:

$W=Q_{1}-Q_{2}=62760-61960=800cal$

8 0

3 years ago

(b) Fig. 1.1 shows two airports A and C. north SE с sea land क WE not to scale Fig. 1.1 An aircraft flies due north from A for a

GalinKa [24]

The addition of vectors and the uniform motion allows to find the answers for the questions about distance and time are:

The distance to go between airports A and C is 373.6 10³ m

The time to go from airport A to B is 2117 s

Vectors are quantities that have modulus and direction, so their addition must be done using vector algebra.

In this case the plane flies towards the North a distance of y = 360 10³ m at an average speed of v = 170 m / s, when arriving at airport B it turns towards the East and travels from x = 100 10³ m, until' it the distance reaches the airport C

Let's use the Pythagoras theorem to find the distance traveled

R = Ra x² + y²

R = 10³

R = 373.6 10³ m

They indicate the average speed for which we can use the uniform motion ratio

v = $\frac{\Delta y }{t}$

t = $\frac{\Delta y}{v}$

They ask for the time in in from airport A to B, we calculate

t = 360 10 ^ 3/170

t = 2.117 10³ s

In conclusion we use the addition of vectors the uniform motion we can find the answer for the question of distance and time are:

The distance to go between airports A and C B is 373.6 10³ m

The time to go from airport A to B is 2117 s

Learn more here: brainly.com/question/15074838

5 0

3 years ago

When an element ejects an alpha particle, the mass number of that element

Katyanochek1 [597]

Answer:

A.reduce by 4.

Explanation:

In Chemistry, an alpha particle which is typically a helium nucleus has an atomic mass (number of nucleons) of four (4). Therefore, the ejection of an alpha particle by any chemical element simply means that, the mass number of that particular chemical element would be reduced by four (4).

Basically, this process is referred to as an alpha decay and can be defined as a radioactive disintegration of a chemical element that causes it to emit an alpha particle and as a result reducing its mass number by four (4).

Hence, when an element ejects an alpha particle, the mass number of that element reduces by 4.

8 0

3 years ago