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Gelneren [198K]

3 years ago

5

Use your answers from questions 1–3 as the basis for the first section of your lab report. This section provides your reader wit

h background information about why you conducted this experiment and how it was completed. Outline the steps of the procedure in full sentences. It also provides potential answers (your hypothesis/es) relative to what you expected the experiment to demonstrate. This section should be 1–3 paragraphs in length.

1 answer:

kodGreya [7K]3 years ago

4 0

Answer:

Do you have a picture for this one I need to see the question.

Explanation:

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Could someone explain to me how to got the answer B, thank you very much

Mnenie [13.5K]

Answer:

since -6 lasted for 5 seconds, multiplying both would result in -30

3 lasted for 10 seconds, so multiplying both would give +30

average = ( 30 + (-30) ) / 2

30 -30 is already equal to zero, so the answer should be 0

4 0

3 years ago

A roller coaster car rapidly picks up speed as it

TiliK225 [7]

Acceleration = (change of speed) / (time for the change)

Change in speed = (22 - 4) = 18 m/s.
Time for the change = 3 sec.

Acceleration = 18/3 = 6 m/s per second.

8 0

3 years ago

A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa

Nataly_w [17]

Answer:

$V_p = 267.258 m/s$

$V_n = 38.375 m/s$

Explanation:

using the law of the conservation of the linear momentum:

$P_i = P_f$

where $P_i$ is the inicial momemtum and $P_f$ is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

$P_i = m(270 m/s)$

$P_f = mV_P + M_nV_n$

where m is the mass of the proton and $V_p$ is the velocity of the proton after the collision, $M_n$ is the mass of the nucleus and $V_n$ is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m $V_p$ + 14m $V_n$

then, m is cancelated and we have:

270 = $V_p$ + $14V_n$

This is a elastic collision, so the kinetic energy K is conservated. Then:

$K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2$

and

Kf = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

then,

$\frac{1}{2}m(270)^2$ = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

here we can cancel the m and get:

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

now, we have two equations and two incognites:

270 = $V_p$ + $14V_n$ (eq. 1)

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

in the second equation, we have:

36450 = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$ (eq. 2)

from this last equation we solve for $V_n$ as:

$V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

and replace in the other equation as:

270 = $V_p +$ 14 $\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

so,

$V_p = -267.258 m/s$

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

$V_n = \frac{270+267.258}{14}$

$V_n = 38.375 m/s$

3 0

3 years ago

How do storms on Jupiter differ from storm systems on Earth?

Vladimir [108]

Answer:

As, Jupiter are one of the largest planet in the solar system and the large amount of the mass of the jupiter are consisted with the gases. The storm tracks in the symmetrical path at the proper latitude in the system. But the storm tracks on the earth in the system where planet are highly variable.

8 0

3 years ago

A 0.54 kg air hockey puck is initially at rest. What will it's kinetic energy energy be after a net force of 0.56 N acts on it f

Rufina [12.5K]

Answer:

Kf = 470 mJ

Explanation:

According the work-energy theorem, the change in the kinetic energy of one object, is equal to the net work done on it.
Since the puck is initially at rest, the change is kinetic energy is just the final kinetic energy of the puck.
Assuming that the net force is horizontal, and causes a horizontal displacement also, we can find the net work on the puck as follows:

$W_{net} = F_{net} * \Delta X = 0.56 N * 0.84 m = 0.47 J = 470 mJ (1)$

As we have already said, (1) is equal to the final kinetic energy of the puck:
⇒ Kf = 470 mJ (2)

8 0

3 years ago