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3 years ago

Which best describes how combustion works?

Physics

2 answers:

miv72 [106K]3 years ago

6 0

Answer: A) It breaks and forms bonds of fuel to release thermal energy.

Explanation:

Chemical energy is the energy stored in the bonds of molecules.

Thermal energy is the energy possessed by a body by virtue of its temperature.

Combustion is a chemical reaction in which a fuel is burnt in the presence of oxygen to form carbon dioxide, water and heat.

$C_xH_y+x+\frac{y}{4}O_2\rightarrow xCO_2+\frac{y}{2}H_2O+Energy$

Thus the chemical energy stored in the bonds of reactants is converted to thermal energy and heat is released in the combustion reaction.

jekas [21]3 years ago

3 0

The answer is A, it breaks down and releases thermal energy.

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4.-Una atleta patea un balon (masa de 650gr y angulo de 45 grados) arriba de un edificio con una velcidad 21 m/s, el edificio mi

marta [7]

Answer:

v = 25 m/s

Explanation:

½mv₀² + mgh = ½mv₁²

v₁² = v₀² + 2gh

v₁ = √(21² + 2(9.8)(10))

v₁ = 25.238858..

3 0

3 years ago

A car traveling at 30 m/s drives off a cliff that is 50 meters high? How far away does it land?

Semenov [28]

Answer:

The maximum range $R_{max}= 132. 72$ m

Explanation:

Given,

The initial velocity of the car, u = 30 m/s

The height of the cliff, h = 50 m

Let the car drives off the cliff with a horizontal velocity of 30 m/s.

The formula for a projectile that is projected from a height h from the ground is given by the relation

$R_{max}= \frac{u}{g}\sqrt{u^{2} + 2gh }$ m

Where,

g - acceleration due to gravity

Substituting the values in the above equation

$R_{max}= \frac{30}{9.8}\sqrt{30^{2} + 2X9.8X50 }$

= 132.72 m

Hence, the car lands at a distance, $R_{max}= 132. 72$ m

3 0

3 years ago

A block is held at rest against a wall by a force of magnitude F exerted at an angle theta from the horizontal, as shown in the

wel

Answers:

B.) $F cos\theta=F_{n}$

C.) $F sin\theta=F_{g} \pm F_{f}$

Explanation:

The image attached shows the way the force $F$ is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:

Net Force in X:

$-F_{n}+F cos\theta=0$ (1)

Where:

$F_{n}$ is the Normal force

$F$ is the magnitude of the force exerted on the block

$\theta$ is the angle

Net Force in Y:

$F sin\theta \pm F_{f}-F_{g}=0$ (2)

Where:

$F_{f}$ is the Friction force (it is expresed with the $\pm$ sign because this force may be up or down, we cannot know because the block is at rest)

$F_{g}$ is the gravity force

Rewrittin (1):

$F cos\theta=F_{n}$ (3) This is according to option B

Rewritting (2):

$F sin\theta=F_{g}\pm F_{f}$ (3) This is according to option C

3 0

3 years ago

A cannon is mounted on a cart which sits on the ground, supported by frictionless wheels. The mass of the cannon and cart is 4.6

Kitty [74]

Answer:

The velocity of the launcher after the projectile is launched is -5.011 m/s

Explanation:

Here we have the mass of the cannon and cart, m₁ = 4.65 kg

Velocity of cannon and cart, v₁ = 2.00 m/s

Mass of projectile, m₂ = 50.0 g = 0.05 kg

Velocity of projectile, v₂ = 647 m/s

Velocity of the launcher, v₃ = Required

Mass of cannon and cart, launcher after launching projectile m₃ = 4.65-0.05

= 4.6 kg

Therefore, from the principle of the conservation of linear momentum, we have

Total initial momentum = Total final momentum

m₁ × v₁ = m₂ × v₂ + m₃ × v₃

Substituting gives

4.65 kg × 2.00 m/s = 0.05 kg × 647 m/s + 4.6 kg × v₃

4.65 kg × 2.00 m/s - 0.05 kg × 647 m/s = 4.6 kg × v₃

-23.05 kg·m/s = 4.6 kg × v₃

$v_3 = \frac{-23.05 \, kg\cdot m/s}{4.6 \, kg} = \frac{-461}{92} m/s$

v₃ = -5.011 m/s.

3 0

3 years ago

Read 2 more answers

A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte

vaieri [72.5K]

Answer:

Part a)

$H = 44.1 m$

Part b)

$y = 13.48 m$

Part c)

$d = 8.86 m$

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

$t = \frac{v sin\theta}{g}$

now we have

$3 = \frac{vsin\theta}{9.8}$

$v sin\theta = 29.4 m/s$

Now maximum height above ground is given as

$H = \frac{v^2sin^2\theta}{2g}$

$H = \frac{29.4^2}{2(9.8)}$

$H = 44.1 m$

Part b)

Height of the fence is given as

$y = (vsin\theta) t - \frac{1}{2}gt^2$

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

$x = v_x t$

$97.5 = v_x (5.5)$

$v_x = 17.72 m/s$

now total time of flight is given as

$T = 3 + 3 = 6 s$

so range is given as

$R = v_x T$

$R = (17.72)(6)$

$R = 106.4 m$

so the distance from the fence is given as

$d = 106.4 - 97.5$

$d = 8.86 m$

7 0

4 years ago