Which is an example of a chemical change?

A 20-kg child is coasting at 3.3 m/s over flat ground in a 4.0-kg wagon. The child drops a 1.0-kg ball out the back of the wagon

Shalnov [3]

Answer:

The final velocity is $u_f = 3.44 \ m/s$

Explanation:

From the question we are told that

The mass of the child is $m_1 = 20 \ kg$

The initial speed of the child is $u_1 = 3.3 \ m/s$

The mass of the wagon is $m_w = 4.0 \ kg$

The initial speed of the wagon is $u_w = 3.3 \ m/s$

The mass of the ball is $m_2 = 1.0 \ kg$

The initial speed off the ball is $u_2 = 3.3 \ m/s$

Generally the initial speed of the system (i.e the child , wagon , ball) is

$u_1 = u_w = u_2 = u =3.3 \ m/s$

Generally from the law of linear momentum conservation

$p_i = p_f$

Here $p_i$ is the momentum of the system before the ball is dropped which is mathematically represented as

$p_i = ( m_1 + m_2 + m_3 ) * u$

=> $p_i = ( 20 + 4 + 1 ) * 3.3$

=> $p_i = 82.5 \ kg \cdot m/s$

and

$p_f$ is the momentum of the system after the ball is dropped which is mathematically represented as

$p_f = ( m_1 + m_w ) * u_f$

=> $p_i = ( 20 + 4 ) * u_f$

So

$82.5 = 24 * u_f$

=> $u_f = 3.44 \ m/s$

4 0

3 years ago

A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block

kotegsom [21]

Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The velocity at the bottom is $v = 11.76 \ m/ s$

Explanation:

From the question we are told that

The total distance traveled is $d = 1.2 \ m$

The mass of the block is $m_b = 0.3 \ kg$

The height of the block from the ground is h = 0.60 m

According the law of energy

$PE = KE$

Where PE is the potential energy which is mathematically represented as

$PE = m * g * h$

substituting values

$PE = 3 * 9.8 * 0.60$

$PE = 17.64 \ J$

So

KE is the kinetic energy at the bottom which is mathematically represented as

$KE = \frac{1}{2} * m v^2$

So

$\frac{1}{2} * m* v ^2 = PE$

substituting values

=> $\frac{1}{2} * 3 * v ^2 = 17.64$

=> $v = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }$

=> $v = 11.76 \ m/ s$

4 0

4 years ago

The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a vel

Brilliant_brown [7]

Answer:

Explanation:

We shall write the velocities given in vector form to make the solution easy.

The velocity of water with respect to earth that is waV(e) makes 30 degree with north or 60 degree with east so in vector form

waV(e) = 2.2 cos 60 i + 2.2 sin 60 j

waV(e) = 1.1 i + 1.9 j

Similarly , velocity of wind with respect to earth that is wiV(e) , is making 50 degree with west or - ve of x axes so we cal write it in vector form as follows

wiV(e) = - 4.5 cos 50 i - 4.5 sin 50 j

wiV(e) = - 2.89 i - 3.45 j

Now we have to calculate velocity of wind with respect to water that is

wiVwa

wiV( wa) = wiV ( e)+ eV(wa)

= wiV( e)- waV(e)

- 2.89 i - 3.45 j - 1.1 i - 1.9 j

= - 3.99 i - 5.35 j

Magnitude of this relative velocity

D² = 3.99² + 5.35²

d = 6.67 m /s

6 0

4 years ago

An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.

SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be $v_{AT}=v_A-v_T$ (the train must see the car advancing at a lower speed), where $v_A$ is the speed of the automobile and $v_T$ the speed of the train.

So we have $v_{AT}=(95km/h)-(75Km/h)=20Km/h$ .

So the train (anyone in fact) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s$

And in that time the car would have traveled (relative to the ground):

$d=v_At=(95Km/h)(0.065h)=6.175Km$

If they are traveling in opposite directions, we have to do all the same but using $v_{AT}=v_A+v_T$ (the train must see the car advancing at a faster speed), so repeating the process:

$v_{AT}=(95km/h)+(75Km/h)=170Km/h$

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s$

$d=v_At=(95Km/h)(0.00765h)=0.73Km$

5 0

3 years ago

N iron nail is made up of particles. What is true about the particles?

kirill [66]

They stay in place and vibrate.

7 0

3 years ago