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2 years ago

What is the difference between chemical change and a physical change?

Physics

1 answer:

NeX [460]2 years ago

3 0

Answer:

Explanation:

There are several differences between a physical and chemical change in matter or substances. A physical change in a substance doesn't change what the substance is. In a chemical change where there is a chemical reaction, a new substance is formed and energy is either given off or absorbed.

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A cardboard box sits on top of a concrete sidewalk where the coefficient of friction between the surfaces is 0.4. The mass of th

Sloan [31]

Answer:

Fg = 98.1 [N]; N = 98.1 [N]; Ff = 39.24 [N]; a = 2.076[m/^2]

Explanation:

To solve this problem, we must make a free body diagram and interpret each of the forces acting on the box. In the attached diagram we can find the free body diagram.

The gravitational force is equal to:

Fg = (10 * 9.81) = 98.1 [N]

Now by summing forces on the Y axis equal to zero, we can find the normal force exerted by the surface.

N - Fg = 0

N = Fg

N = 98.1 [N]

The friction force is defined as the product of normal force by the coefficient of friction.

Ff = N * μ

Ff = 98.1 * 0.4

Ff = 39.24 [N]

By the sum forces on the x-axis equal to the product of mass by acceleration (newton's second law), we can find the value of acceleration.

60 - Ff = m * a

60 - 39.24 = 10 * a

a = 2.076[m/^2]

6 0

2 years ago

If a ball is dropped from a height of 10 m, hits the ground and bounces back up to a maximum height of 7.5 m, describe how the E

Greeley [361]

Answer:hmmmm

Explanation:

8 0

2 years ago

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$