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3 years ago

In what ways is thermal energy transferred during physical and chemical changes

Physics

1 answer:

Sphinxa [80]3 years ago

3 0

Explanation:

chemical changes

>produce new substances w/ different chemical structure/ properties

> thermal energy can be used to BREAK bonds (endothermic)

>thermal energy can be transferred to a chem rxn to help form new products (endothermic)

> heat can be released by an exothermic chemical reaction (excess energy- the total energy used to break bonds is less than energy released in forming new bonds)

physical changes

>does not result in new substances (typically changes in state- solid, liquid, gas)

> thermal energy can be transferred to EVAPORATE water (add thermal E... water (liquid) to steam (gas)) or FREEZE water (remove thermal E... water (liquid) to ice (solid))

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During its lifespan, what characteristics of the sun will change

4vir4ik [10]

During its lifepsan, the sun's core would keep contracting and heating up.

The temperature will keep increasing to the point where the temperature outside the core will get to hydrogen fusion temperatures.
The sun will grow in surface and eventually became the Red Giant

6 0

3 years ago

Read 2 more answers

A typical mattress has a network of springs that provide support. If you sit on a mattress, the springs compress. A heavier pers

GenaCL600 [577]

Answer:

Explanation:

Spring has a tendency to store energy in them and deform its shape when force is applied on it. Once the applied force is removed it regains its original shape and size.

It is in helical shape and is used in mattress to give structure and support. Spring have elastic nature and follows spring forces, F = k * x

where is the applied force, k is the spring constant and x is the amount of extension.

When a heavier person sits on a mattress, more weight is applied on springs and they form coils, as weight is removed they regains its shape again.

4 0

3 years ago

A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material

My name is Ann [436]

Answer:

a) C = 4,012 10⁻¹⁴ F, b) Q = 1.6 10⁻¹¹ C , c) U = 3.21 10⁻¹¹ J

Explanation:

a) The capacitance of a capacitor is

C = k e₀ A / d

Let's calculate

C = 4 8.85 10⁻¹² 17 10⁻⁴ / 0.150 10⁻²

C = 4,012 10⁻¹⁴ F

b) let's look the charge

C = Q / ΔV

Q = C ΔV

Q = 4,012 10⁻¹⁴ 400

Q = 1.6 10⁻¹¹ C

c) The stored energy

U = ½ C ΔV²

U = ½ 4,012 10⁻¹⁴ 400²

U = 3.21 10⁻¹¹ J

4 0

3 years ago

14.A 90 kg quarterback gets tackled by being hit by a 120 kg lineman backwards

quester [9]

The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

F = Ma.

The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

So, the acceleration can be calculated using the formula,

500N = 90kg x a

a = 5.55 m/s².

Now, the weight if the lineman is 120kg, the force applied by him is 500N.

So, from the formula, his acceleration A will be,

500N = 120Kg x A

A = 4.16 m/s².

both of them will have acceleration in the same direction,

To know more about Force, visit,

brainly.com/question/25239010

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8 0

9 months ago

A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the

Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge $15\times10^-6$ C is placed which is the origin.

Let B be the point where the charge $9\times 10^-6$ C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, $\frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2}$ ,