Plants need to....................synthesize

7nadin3 [17]

Answer:

<h2>C. 0.55 m/s towards the right</h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

x is the final velocity of the 0.25kg ball

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at 0.55 m/s towards the right

4 0

3 years ago

A 2.5-kg ball and a 5.0-kg ball have an elastic collision. Before the collision, the 2.5-kg ball was at rest and the other ball

lesantik [10]

The kinetic energy of 2.5 kg ball after collision is 27.09 J.

Answer:

Explanation:

In elastic collision, the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision.

We know that momentum is the product of mass and velocity acting on any object.

So, the conservation of energy in elastic collision leads to following equation:

$M_{1} u_{1} +M_{2} u_{2}=M_{1} v_{1}+M_{2} v_{2}$

Since, the momentum is conserved ,the kinetic energy will also be conserved in elastic collision. So

$M_{1} u_{1} ^{2}+M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+ M_{2}v_{2} ^{2}$

Since initial velocity for M1 ball is zero, then

$M_{2} u_{2}=M_{1} v_{1}+M_{2} v_{2}$

and

$M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+ M_{2}v_{2} ^{2}$

So, on solving all the above equation, we get an equation for velocity and that is

$\frac{2M_{2}u_{2} }{(M_{1}+M_{2} }$ =final velocity of ball with mass 2.5 kg

$v = \frac{2(5*3.5)}{2.5+5}=4.67 m/s$

So kinetic energy will be 1/2 mv2

Kinetic energy of 2.5 kg ball is $\frac{1}{2}*2.5*(4.67)^{2} =27.09 J$

So the kinetic energy of 2.5 kg ball after collision is 27.09 J.

6 0

3 years ago

Consider three axes of rotation for a pencil: along the lead, at right angles to the lead at the middle, and at right angles to

tensa zangetsu [6.8K]

Answer:

along the lead, at right angles to the lead at the middle, and at right angles to the lead at one end.

4 0

3 years ago

The first few steps in deriving the quadratic formula are shown. Which best explains why is not added to the left side of the eq

bija089 [108]

The term b^2 / 4a^2 is not added to the left side of the equation, because the term that was added to the right was not either b^2 / 4 a^2.

As you can see the ther b^2 / 4a^2 that appears in the last step of the table is inside a parenthesis, which is preceded by factor a.

Then, you need to apply the distributive property to know the term that you are really adding to the right side, i.e. you need to mulitply b^2 / 4a^2 * a which is b^2 / 4a.

That means that you are really adding b^2 / 4a to the right, so that is the same that you have to add to the left, which is what the last step of the table shows.

That situation is reflected by the statement "The distributive property needs to be applied to determine the value to add to the left side of the equation to balance the sides of the equation". That is the answer.

6 0

3 years ago

Read 2 more answers

A compound consisting of atoms of small atomic mass is more likely to require

MArishka [77]

A lower temp to liquefy

4 0

3 years ago