I have a question

A new company is making security cameras. These security cameras will be different than other security cameras because instead o

babunello [35]

Answer:

b

Explanation:

4 0

2 years ago

An engine extracts 441.3kJ of heat from the burning of fuel each cycle, but rejects 259.8 kJ of heat (exhaust, friction,etc) dur

Lilit [14]

Answer:

The thermal efficiency is 0.4113.

Explanation:

We know that the thermal efficiency is the ratio of work done by the engine over the heat taken

$\eta = \frac{W_{made}}{Q_{in}}$

Now, how much work the engine do in a cycle?

We know that the work done in a cycle must be equal to the heat taken minus the heat rejected

$W{made} = Q_{in} - Q_{rejected}$

So, the thermal efficiency will be:

$\eta = \frac{Q_{in} - Q_{rejected}}{Q_{in}}$

$\eta = \frac{Q_{in}}{Q_{in}} - \frac{Q_{rejected}}{Q_{in}}$

$\eta = 1 - \frac{Q_{rejected}}{Q_{in}}$

Putting the values of the problem

$\eta = 1 - \frac{259.8 kJ }{441.3kJ}$

$\eta = 0.4113$

7 0

4 years ago

What is the specific heat of the masses in this experiment? Infer the substance the masses are made of and explain your inferenc

Liono4ka [1.6K]

The metal whose specific heat capacity is close to the obtained value is aluminum.

<h3>What is specific heat capacity</h3>

The specific heat capacity of an object is the heat required to raise a unit mass of the substance by 1 kelvin.

$Q= mc\Delta \theta$

where;

c is the specific heat capacity
Δθ is change in temperature

Let the mass of the water = 50 g

mass of the metal for this first trial = 50 g

The heat gained by the water is calculated as follows

$Q = 50 \times 4.184 \times 8.4\\\\Q = 1757.28 \ J$

Specific heat capacity of the metal for the first trial is calculated as follows;

Heat gained by water = Heat lost by metal

$C = \frac{Q}{m\Delta T} = \frac{1757.28}{50\times 8.4} = 4.184 \ J/g^oC$

Specific heat capacity of the metal for the second trial;

mass of metal = 200 - mass of water = 150 g

$C_2 = \frac{1757.28}{150 \times 15.2} = 0.77 \ J/g^oC$

Specific heat capacity of the metal for the third trial;

$C_3 = \frac{1757.28}{250 \times 20.8} = 0.34\ J/g^oC$

Specific heat capacity of the metal for the fourth trial;

$C_4 = \frac{1757.28}{350 \times 25.4} = 0.19\ J/g^oC$

Specific heat capacity of the metal for the fifth trial;

$C_5 = \frac{1757.28}{450 \times 29.6} = 0.13\ J/g^oC$

Average specific heat capacity

$C = \frac{4.184 + 0.77 + 0.34+ 0.19 + 0.13 }{5} = 1.12 \ J/g^oC = 1120 J/kg^oC$

The metal whose specific heat capacity is close to the above value is aluminum.

Learn more about specific heat capacity here: brainly.com/question/16559442

8 0

2 years ago

Even through there is equal and opposite reaction,usually the two forces are not seen balanced.Why?

Tomtit [17]

Answer:

This may refer to a situation like:

"one person pushes a box, if there is equal and opposite reaction why the box moves and the person does not?"

Remember the second Newton's law:

F = m*a

suppose that the mass of the person is 3 times the mass of the box.

So, if the box has a mass M, the person will have a mass 3*M

Then the Newton's equation for the box when the person pushes with a force F is:

F = M*a

solving for the acceleration, we get:

F/M = a

While the person is also pushed by the box with a force with the same magnitude, then the equation for the person is:

F = (3*M)*a'

Solving for the acceleration, we get:

F/(3M) = a'

Now we can compare the acceleration of the box (F/M) with the acceleration of the person (F/3M).

Is easy to see that the acceleration of the box is 3 times the acceleration of the person.

So regardless of the fact that both the box and the person experience a force with the same magnitude, the box will move more due to this force.

This is why in situations like this, the forces do not seem balanced.

5 0

3 years ago

Two objects, X and Y, move toward one another and eventually collide. Object X has a mass of 2M and is moving at a speed of 2vo

Nady [450]

Answer:

The force exerted by X on Y is F to the right, and the force exerted by Yon X is F to the left.

Explanation:

There are two objects X and Y. Mass of object X is 2M, that is moving with a speed of $2v_o$ and that of object Y is M that is moving with a speed of $v_o$ .

When both of the objects collide, the magnitude of the forces F the objects exert on each other is equal to :

The force exerted by X on Y is F to the right, and the force exerted by Yon X is F to the left using Newton's third law of motion. Hence, the correct option is (c).

3 0

3 years ago