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3 years ago

Experiment: Limestone

1 answer:

7 0

When these bonds are destroyed, a reaction occurs. ... Vinegar reacting with limestone breaks the bonds of calcium carbonate and acetic acid.

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Complete each statement about the sign of the work done on a baseball. Carlton catches a baseball and his hand moves backward as

dezoksy [38]

Answer and Explanation:

The work done is negative as the force is applied in the forward direction and the displacement of the hand is in the backward direction.
The work done is positive as the work is done on the baseball by Carlton. Moreover, both the force applied and the displacement of Carlton are in a straight line in the forward direction.
The work done in this case is zero as Carlton rides with constant speed, the Kinetic energy too remains constant and hence net work done is zero.
The work done in this case is positive as the work is done on the baseball by Carlton and as the velocity changes its Kinetic energy also varies wit it.
Also the force applied and the displacement of Carlton in this case are in the directed to the same path.

8 0

3 years ago

A body of mass m, = 0.050 kg and initial velocity v₁ = 9 m / s turned to the right collides with a body of mass m₂ = 0.030 g wit

Ugo [173]

Answer:

Suppose two objects of different masses are moving with different velocities in the same direction on a straght-line before collision. After collision, they stick together and move with common (the same) velocity

6 0

2 years ago

A 75 N box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each

shusha [124]

Answer:

μk = 0.26885

Explanation:

Conceptual analysis

We apply Newton's second law:

∑Fx = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

a= -0.9 m/s²,

g = 9.81 m/s² : acceleration due to gravity

W= 75 N : Block weight

W= m*g

m = W/g = 75/9.8= 7.65 kg : Block mass

Friction force : Ff

Ff= μk*N

μk: coefficient of kinetic friction

N : Normal force (N)

Problem development

We apply the formula (1)

∑Fy = m*ay , ay=0

N-W-25 = 0

N = 75 +25

N= 100N

∑Fx = m*ax

20-Ff= m*ax

20-μk*100 = 7.65*(-0.90 )

20+7.65*(0.90) = μk*100

μk = ( 20+7.65*(0.90)) / (100)

μk = 0.26885

4 0

3 years ago

A rock is dropped from the top of a tower. When it is 40 meters above the ground velocity of 17 m/s. When its velocity is 24 m/s

eimsori [14]

Answer:

Option B is the correct answer.

Explanation:

Let us consider 40 meter above ground as origin.

Initial velocity = 17 m/s

Final velocity = 24 m/s

Acceleration = 9.81 m/s

We have equation of motion v² = u² + 2as

Substituting

24² = 17² + 2 x 9.81 x s

s = 14.63 m

Distance traveled by rock = 14.63 m down.

Height of rock from ground = 40 - 14.63 = 25.37 m = 25.4 m

Option B is the correct answer.

4 0

3 years ago

A 100-g toy car moves along a curved frictionless track. At first, the car runs along a flat horizontal segment with an initial

Sliva [168]

Answer:

The final velocity of the car is 1.85 m/s

Explanation:

Hi there!

The initial kinetic energy of the toy car can be calculated as follows:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass.

v = velocity.

KE = 1/2 · 0.100 kg · (2.66 m/s)² = 0.354 J

The gain in altitude produces a gain in potential energy. This gain in potential energy is equal to the loss in kinetic energy. So let´s calculate the potential energy of the toy car after gaining an altitude of 0.186 m.

PE = m · g · h

Where:

PE = potential energy.

m = mass.

g = acceleration due to gravity.

h = height.

PE = 0.100 kg · 9.8 m/s² · 0.186 m = 0.182 J

The final kinetic energy will be: 0.354 J - 0.182 J = 0.172.

Using the equation of kinetic energy, we can obtain the velocity of the toy car after running up the slope:

KE = 1/2 · m · v²

0.172 J = 1/2 · 0.100 kg · v²

2 · 0.172 J / 0.100 kg = v²

v = 1.85 m/s

The final velocity of the car is 1.85 m/s

3 0

3 years ago