An object moving in the xy-plane is acted on by a conservative force described by the potential-energy function

A 9.8-g bullet is fired into a stationary block of wood having mass m = 5.04 kg. The bullet imbeds into the block. The speed of

pogonyaev

Answer:

313.8 m/s

Explanation:

Momentum p must be conserved.

The momentum before the collision:

$p=m_{bullet}v_{bullet} + m_{block}v_{block}$

The momentum after the collision:

$p=(m_{bullet}+m_{block})v_{bullet+block}$

Solving both equations:

$v_{bullet}=\frac{(m_{bullet}+m_{block})v_{bullet+block}}{m_{bullet}}$

6 0

4 years ago

Which product forms from the synthesis reaction between calcium (Ca) and

cestrela7 [59]

Answer:

CaCl2

Explanation:

5 0

3 years ago

Read 2 more answers

Can you change your mass without changing your weight?

Ratling [72]

No, because mass is the amount of matter in something and weight is the pull of gravity on a object, for example you would weigh 65 pounds.. and have the mass of 1058..if you were in space your weight would change but your mass will always stay the same

6 0

4 years ago

Identify the type of chemical reaction that is described.

Anastasy [175]

Water breaks down into hydrogen and oxygen: It is a decomposition reaction as a single substance decomposes to give two products.

$H_2O \rightarrow H_2 +\frac {1}{2}O_2$

Leaves make starch using chlorophyll and carbon dioxide: Synthesis reaction: as the synthesis reaction involves two or more than two reactants which join together to result into a single main product along with the formation of simple by products.

$6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2$

Food burns in oxygen gas and releases a lot of energy: Combustion: Combustion process involves the use of oxygen to give products along with release of energy.

$C_6H_{12}O_6 + 6O_2\rightarrow 6CO_2 + 6H_2O$

Adding vinegar (acid) to baking soda (alkali) gives a product that is neither acidic nor alkaline: Neutralization: acetic acid in vinegar reacts with soda (base) to give salt (neutral) and .

$CH_3COOH + NaHCO_3\rightarrow CH_3COONa +H_2O + CO_2$

6 0

4 years ago

A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.

Alexxandr [17]

Answer:

Kinetic energy of the projectile at the vertex of the trajectory: $900\; {\rm J}$ .

Work done when firing this projectile: $2500\; {\rm J}$ .

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity $v_{x}$ of this projectile would stay the same (at $30\; {\rm m\cdot s^{-1}}$ ) throughout the flight.

The vertical velocity $v_{y}$ of this projectile would be $0\; {\rm m\cdot s^{-1}}$ at the vertex (highest point) of its trajectory. (Otherwise, if $v_{y} > 0$ , this projectile would continue moving up and reach an even higher point. If $v_{y} < 0$ , the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be $v = 30\; {\rm m\cdot s^{-1}}\!$ when it is at the top of the trajectory. The kinetic energy $\text{KE}$ of this projectile (mass $m = 2.0\; {\rm kg}$ ) at the vertex of its trajectory would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}$ .

Apply the Pythagorean Theorem to find the initial speed of this projectile:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Hence, the initial kinetic energy $\text{KE}$ of this projectile would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}$ .

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be $2500\; {\rm J}$ .

7 0

2 years ago