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3 years ago

Which of the follow represent a chemical changes

Physics

1 answer:

mixer [17]3 years ago

5 0

Answer:

C) steel turning to rust in salt air

Explanation:

The missing options are:

A) ice melting to form liquid water

B) water boiling to form steam

C) steel turning to rust in salt air

D) sugar dissolving into hot coffee

In a chemical change the atoms of the reacting compounds are reordered forming new compounds. In a chemical change, new compounds appear, but in a physical change not.

Then, change of states like ice melting and water boiling are not chemical changes.

During steel rust, components of steel, like iron, are oxidized, that is, reacts with oxygen forming oxides.

The dissolution of sugar into hot coffee is a physical change in which sugar molecules get further apart in the coffee, but they don't change.

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A box is at rest on a table. What can you say about the forces acting on the box?

Nikitich [7]

You can tell a lot about an object that's not moving,
and also a lot about the forces acting on it:

==> If the box is at rest on the table, then it is not accelerating.

==> Since it is not accelerating, I can say that the forces on it are balanced.

==> That means that the sum of all forces acting on the box is zero,
and the effect of all the forces acting on it is the same as if there were
no forces acting on it at all.

==> This in turn means that all of the horizontal forces are balanced,
AND all of the vertical forces are balanced.

Horizontal forces:
sliding friction, somebody pushing the box

All of the forces on this list must add up to zero. So ...

(sliding friction force) = (pushing force), in the opposite direction.

If nobody pushing the box, then sliding friction force = zero.

Vertical forces:
gravitational force (weight of the box, pulling it down)
normal force (table pushing the box up)

All of the forces on this list must add up to zero, so ...

(Gravitational force down) + (normal force up) = zero

(Gravitational force down) = -(normal force up) .

6 0

2 years ago

Read 2 more answers

Could the half of the moon that faces the earth ever be completely dark in any of these diagrams

docker41 [41]

Yes well maybe but I think yes

6 0

3 years ago

A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road

m is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

In this problem,

r = 750 m is the radius

$v=85 mph \cdot \frac{1609}{3600}=38.0 m/s$ is the speed

And solving for $\mu$ , we find the coefficient of friction required to keep the car in circular motion:

$\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

2 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v remain same at both side. so we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let final pressure is P and temp $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide 2 equation by 3rd equation

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, temperature on left side is 1.48 times the temperature on right

6 0

2 years ago