I need the answers to these questions

If a ping pong ball and a golf ball are both moving in the same direction with the same amount of kinetic energy, the speed of t

Liono4ka [1.6K]

If the kinetic energy of each ball is equal to that of the other,
then

(1/2) (mass of ppb) (speed of ppb)² = (1/2) (mass of gb) (speed of gb)²

Multiply each side by 2:

(mass of ppb) (speed of ppb)² = (mass of gb) (speed of gb)²

Divide each side by (mass of gb) and by (speed of ppb)² :

(mass of ppb)/(mass of gb) = (speed of gb)²/(speed of ppb)²

Take square root of each side:

√ (ratio of their masses) = ( 1 / ratio of their speeds)²

By trying to do this perfectly rigorously and elegantly, I'm also
using up a lot of space and guaranteeing that nobody will be
able to follow what I have written. Let's just come in from the
cold, and say it the clear, easy way:

If their kinetic energies are equal, then the product of each
mass and its speed² must be the same number.

If one ball has less mass than the other one, then the speed²
of the lighter one must be greater than the speed² of the heavier
one, in order to keep the products equal.

The pingpong ball is moving faster than the golf ball.

The directions of their motions are irrelevant.

5 0

3 years ago

Constructive interference leads to _________.

Minchanka [31]

D
the waves are in phase. crests come together to form bigger crests hence higher amplitude

6 0

3 years ago

20POINTS, this is not free, answer ASAP<br> Best answer gets BRAINLIEST

Nataly [62]

Answer: Well it's not sinking. It is also not flooring or floating. so my guess it must be rising up.

Explanation: When something is impacted, the pressure of the force of energy is released and lands directly on a other object. So what does Jell-O do when you shake it? It wobbles right? what happens when you put your hand in a bucket of water? It makes the water rise up. So That's why I think it is rising up.

8 0

3 years ago

Read 2 more answers

A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T

Serga [27]

Answer:

a) $k_{e}$ = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m, e) v = 15.23 m / s

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

$k_{e}$ = ½ k x²

$k_{e}$ = ½ 2900 0.80²

$k_{e}$ = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

U = m h and

U = 8 9.8 (-0.80)

U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

K = ½ m v²

K = 0

d) write the energy at two points, maximum compression and maximum height

Em₀ = ke = ½ m x²

$E_{mf}$ = mg y

Emo = $E_{mf}$

½ k x² = m g y

y = ½ k x² / m g

y = ½ 2900 0.8² / (8 9.8)

y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

Y = y- 0.80

Y = 11.8367 -0.80

Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

Emo = ½ k x² = 928 J

$E_{mf}$ = ½ m v²

Emo = $E_{mf}$

Emo = ½ m v²

v =√ 2Emo / m

v = √ (2 928/8)

v = 15.23 m / s

8 0

4 years ago

A crow is flying horizontally with a constant speed of 2.70m/s when it releases a claim from its beak. The clan lands on the roc

jenyasd209 [6]

Given:

Speed = 2.70 m

Time, t = 2.10 seconds

Let's solve for the following:

• (a) The horizontal component of the velocity.

To find the horizontal component, apply the formula:

$V_{ox}=V_o\cos \theta$

Where:

Vo is the initial speed = 2.70 m

θ = 0 degrees

Hence, we have:

$\begin{gathered} V_{ox}=2.70\cos 0 \\ \\ V_{ox}=2.7\text{ m/s} \end{gathered}$

The horizontal component of the velocity just before it lands is 2.70 m/s.

• (b) The vertical component of the velocity.

To find the vertical component, apply the formula:

$V_{oy}=V_{0y}-gt=\text{V}_{oy}\text{ sin}\Theta-gt$

Where:

g is the acceleration due to gravity = 9.8 m/s²

t is the time = 2.10 s

Hence, we have:

$\begin{gathered} V_{oy}=V_{oy}\sin \theta-gt \\ \\ V_{oy}=2.70\sin 0-9.8(2.10) \\ \\ V_{oy}=0-20.58 \\ \\ V_{oy}=-20.58\text{ m/s} \end{gathered}$

The vertical component of the velocity just before it lands is -20.58 m/s.

(c) Here, the initial speed is equal to the constant horizontal speed.

Therefore, in part (a) the horizontal component will increase in the x-direction if the speed of the crow is increased.

The initial vertical velocity is 0 m/s in both cases.

Therefore, in part (b) the vertical component will remain constant.

ANSWER:

(a) 2.70 m/s

(b) -20.58

(c) In part (a) the horizontal component will increase, while in part (b) the vertical component will remain constant.

6 0

1 year ago