What form of energy is a bonfire and a bunsen burner?

In chemistry, potential energy is usually considered to be the energy _____________.

Lunna [17]

Answer:

The answer is option D. Stored in the chemical bonds.

Hope it helps............

4 0

3 years ago

A scientist compares two samples of white powder. One powder was present at the beginning of an experiment. The other powder was

Jet001 [13]

Density of powder 1 = 0.5 g / 45 cm^3 = 1/90 g/cm^3
Density of powder 2 = 1.3 g / 65 cm^3 = 1/50 g/cm^3

Therefore the densities of the two powders are different, hence chemical reaction has occurred.

(note: none of the other choices make sense. In fact, a different density does not necessarily indicate a chemical change, see paragraph below).

Density of powders are not definitive unless they are each of the same size and texture. For example, granular sugar, rock sugar, and icing sugar all have different densities. I would conclude that this experiment does not lend to a reliable answer.

4 0

3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

Mariana [72]

a) $F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

$R-Fsin \theta - mg = 0$ (1)

where

R is the normal reaction of the floor (upward)

$F sin \theta$ is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

$F cos \theta - \mu_k R = 0$ (2)

where

$F cos \theta$ is the horizontal component of the force F (forward)

$\mu_k R$ is the force of friction (backward)

From (1) we get

$R=Fsin \theta +mg$

And substituting into (2)

$F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

b) $\mu_s=cot(\theta)$

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

$R-Fsin \theta - mg = 0$ (1)

$F cos \theta - \mu_s R = 0$ (2)

In this second case, we want to find the critical value of $\mu_s$ such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, $F cos \theta$ .

Therefore, using the same procedure as before,

$R=Fsin \theta +mg$

$F cos \theta - \mu_s (Fsin \theta +mg) = 0$

And solving for $\mu_s$ ,

$F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}$

Now we analyze the expression that we found. We notice that if the force applied F is very large, $F sin \theta >> mg$ , therefore we can rewrite the expression as

$\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)$

So, this is the critical value of the coefficient of static friction.

8 0

3 years ago

What is a moon node? have a great day

RideAnS [48]

Answer:

lunar node in other words is either of the two orbital nodes of the Moon, that is, the two points at which the orbit of the Moon intersects the ecliptic. The ascending node is where the Moon moves into the northern ecliptic hemisphere, while the descending node is where the Moon enters the southern ecliptic hemisphere

3 0

3 years ago

Read 2 more answers

Brainliest if correct Question 2 of 10

Troyanec [42]

Answer:

b and d

Explanation:

7 0

3 years ago