4. Which statement summarizes the excerpt? *

A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of

Sunny_sXe [5.5K]

Answer:

$5.8\; {\rm kg\cdot m \cdot s^{-1}}$ .

Explanation:

If the mass of an object is $m$ and the velocity of that object is $v$ , the linear momentum of that object would be $m\, v$ .

Assume that the initial velocity of the mass is positive ( $6.0\; {\rm m\cdot s^{-1}}$ .) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be $(-4.0\; {\rm m\cdot s^{-1}})$ .

Thus, the change in the velocity of the mass would be:

$\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}$ .

The change in the linear momentum of the mass would be:

$\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\ =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}$ .

Thus, the magnitude of the change of the linear momentum would be $5.8\; {\rm kg \cdot m \cdot s^{-1}}$ .

7 0

3 years ago

3. If I have a block of solid gold and I continually add energy to it, will it continually increase in

zloy xaker [14]

The gold was heated at rates too fast for the electrons absorbing the light energy to collide with surrounding atoms and lose energy.

7 0

3 years ago

Is a single property such as density sufficient to always correctly identify a substance?

bazaltina [42]

Not at all. Density on its own is not sufficient

7 0

3 years ago

Two students push on a 5-kg cart from opposite sides

sweet-ann [11.9K]

Answer:

nothing will happen the cart will be broken or as it is

4 0

3 years ago

Read 2 more answers

A certain heat engine takes in 300 J of energy from a hot source and then transfers 200 J of that energy to a colder object. Wha

Greeley [361]

Answer:

The efficiency is 0.33, or 33%.

Explanation:

From the thermodynamics equations, we know that the formula for the efficiency of a heat engine is:

$\eta=1-\frac{Q_2}{Q_1}$

Where η is the efficiency of the engine, Q_1 is the heat energy taken from the hot source and Q_2 is the heat energy given to the cold object. So, plugging the given values in the formula, we obtain:

$\eta=1-\frac{200J}{300J}=0.33$

This means that the efficiency of the heat engine is 0.33, or 33% (The efficiency of an engine is dimensionless).

5 0

3 years ago