A block with a mass of 55 kg is pulled with a force of 220 N to the right. Assuming a frictionless surface, what will be the acceleration of the block?

To calculate the acceleration of the block, we can use Newton's second law of motion, which states:

\[ F = m \cdot a \]

Where:

• \( F \) is the net force applied,
• \( m \) is the mass of the object,
• \( a \) is the acceleration.

Given:

• \( F = 220 , \text{N} \)
• \( m = 55 , \text{kg} \)

We can rearrange the equation to solve for acceleration \( a \):

\[ a = \frac{F}{m} \]

Now, substitute the values into the equation:

\[ a = \frac{220 , \text{N}}{55 , \text{kg}} \]

Calculating this:

\[ a = 4.0 , \text{m/s}^2 \]

Thus, the acceleration of the block is \( 4.0 , \text{m/s}^2 \).

The correct answer is:

a. 4.0 m/s²

During an experiment in which you are investigating the acceleration changes due to force changes, what value must stay constant during these trials?

a
Force
b
Velocity
c
Acceleration
d
Mass

In an experiment where you are investigating the changes in acceleration due to changes in force, the value that must stay constant during these trials is:

d. Mass

By keeping the mass constant, you can clearly observe how changes in force affect acceleration, according to Newton's second law of motion \( (F = m \cdot a) \). If the mass is not constant, it would introduce additional variables that could confound your results.

Which of the following would be an example of Newton's Third Law forces?

a
A horse pulling a horse cart
b
A hammer striking a nail
c
A soccer ball bouncing off a wall
d
All of the above

Newton's Third Law states that for every action, there is an equal and opposite reaction. This means that forces come in pairs; for one object exerting a force on another, the second object exerts an equal force in the opposite direction on the first object.

Let's analyze the options:

a. A horse pulling a horse cart - While the horse exerts a force on the cart, the cart also exerts an equal and opposite force on the horse.

b. A hammer striking a nail - When the hammer strikes the nail, it exerts a force on the nail, and the nail exerts an equal and opposite force back on the hammer.

c. A soccer ball bouncing off a wall - The ball exerts a force on the wall, and the wall exerts an equal and opposite force back on the ball.

Since all the provided options illustrate the principle of action and reaction described by Newton's Third Law, the correct answer is:

d. All of the above

Why is it generally easier to push a heavy object that is already moving instead of the same heavy object that is stationary

a
The force of gravity on the object is different if it is stationary or not.
b
The force of friction on the object is different if it is stationary or not
c
The force of the push on the object is different if it is stationary or not
d
The normal force on the object is different if it is stationary or not

The reason it is generally easier to push a heavy object that is already moving rather than one that is stationary is because:

b. The force of friction on the object is different if it is stationary or not.

When an object is stationary, the frictional force that needs to be overcome to start the motion is static friction. Static friction is usually greater than kinetic friction, which is the friction that acts when the object is already in motion. Thus, once the object is moving, you only need to overcome the lower kinetic friction, making it easier to push.

224/88 Ra arrow 4/2 He + A/Z X

What is the atomic mass of the new atom?

What is the atomic number of the new atom?

What is the symbol of the new atom?

What type of decay is demonstrated here?

To analyze the equation \( \frac{224}{88} \text{Ra} \rightarrow \frac{4}{2} \text{He} + \frac{A}{Z} \text{X} \), we will first interpret the components of the equation.

• Left Side:

  • Radium-224 (\( \frac{224}{88} \text{Ra} \)) has:
    • Atomic mass (A) = 224
    • Atomic number (Z) = 88

• Right Side:

  • Helium-4 (\( \frac{4}{2} \text{He} \)) has:
    • Atomic mass = 4
    • Atomic number = 2

The reaction can be interpreted as Radium-224 undergoing a decay process to produce a helium atom and a new atom (\( \frac{A}{Z} \text{X} \)).

Step 1: Conservation of Mass and Charge

In nuclear reactions, both atomic mass and atomic number must be conserved. Thus, we set up the equations:

1. For atomic mass (A): \[ 224 = 4 + A \Rightarrow A = 224 - 4 = 220 \]

2. For atomic number (Z): \[ 88 = 2 + Z \Rightarrow Z = 88 - 2 = 86 \]

Summary of the New Atom

Now we can summarize the new atom \( \frac{A}{Z} \text{X} \):

• A (Atomic mass) = 220
• Z (Atomic number) = 86
• The element with atomic number 86 is Radon (Rn).

Final Answers

1. Atomic mass of the new atom: 220
2. Atomic number of the new atom: 86
3. Symbol of the new atom: \( \frac{220}{86} \text{Rn} \)
4. Type of decay demonstrated here: This is an example of alpha decay, as a helium nucleus (\( \frac{4}{2} \text{He} \)) is emitted from the radium atom.

Balance the following the chemical equation:

_ Na2O + _ CaBr2 arrow _ NaBr + _ CaO

a
2:1:1:1
b
1:1:1:2
c
1:2:1:1
d
1:1:2:1

To balance the chemical equation \( \text{Na}_2\text{O} + \text{CaBr}_2 \rightarrow \text{NaBr} + \text{CaO} \), we will determine the number of each type of atom on both sides of the equation and adjust the coefficients accordingly.

Step 1: List the number of atoms for each element.

Reactants:

• Sodium (Na): 2 from \( \text{Na}_2\text{O} \)
• Oxygen (O): 1 from \( \text{Na}_2\text{O} \)
• Calcium (Ca): 1 from \( \text{CaBr}_2 \)
• Bromine (Br): 2 from \( \text{CaBr}_2 \)

Total for reactants:

• Na: 2
• O: 1
• Ca: 1
• Br: 2

Products:

• Sodium (Na): 1 from \( \text{NaBr} \)
• Bromine (Br): 1 from \( \text{NaBr} \)
• Calcium (Ca): 1 from \( \text{CaO} \)
• Oxygen (O): 1 from \( \text{CaO} \)

Total for products:

• Na: 1
• O: 1
• Ca: 1
• Br: 1

Step 2: Balance the equation.

• Balance Sodium (Na): Since we have 2 Na on the left, we need 2 Na on the right: \[ \text{Na}_2\text{O} + \text{CaBr}_2 \rightarrow 2\text{NaBr} + \text{CaO} \]

• Now let's recount:

  • Reactants:
    • Na: 2, O: 1, Ca: 1, Br: 2
  • Products:
    • Na: 2, O: 1, Ca: 1, Br: 2 (after adding the coefficient).

This shows that the equation is now balanced.

Final Coefficients

The coefficients for each compound are:

• \( 1 \) for \( \text{Na}_2\text{O} \)
• \( 1 \) for \( \text{CaBr}_2 \)
• \( 2 \) for \( \text{NaBr} \)
• \( 1 \) for \( \text{CaO} \)

Thus, the balanced equation is: \[ 1 \text{ Na}_2\text{O} + 1 \text{ CaBr}_2 \rightarrow 2 \text{ NaBr} + 1 \text{ CaO} \]

Therefore, the answer is:

d. 1:1:2:1