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musickatia [10]

3 years ago

5

Mrs. Davis has asked Kyle to retrieve the compound mixture from the chemical cabinet. Which of the following should Kyle retriev

e?
A) Water

B) Oxygen

C) Chlorine

D) Aluminum

1 answer:

nignag [31]3 years ago

3 0

Answer:

water

Explanation:

coz out of the choices water is the only one that is not an element.

water formula is H2O

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In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vi

irinina [24]

To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where,

$m_{1,2}$ = Mass of each object

$v_{1,2}$ = Initial velocity of each object

$v_f$ = Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

$m_2v_2 = (m_1+m_2)v_f$

$(0.42)(21) = (90+0.42)v_f$

$v_f = 0.0975m/s$

Therefore the velocity right after catching the ball is 0.0975m/s

8 0

3 years ago

Light intensity is?

Fittoniya [83]

The correct answer is:
<span>The rate at which a waves energy flows through a given unit of area

In fact, light intensity is defined as the light power per unit of area:
</span> $I= \frac{P}{A}$
<span>but the power is the energy carried by the light per unit of time:
</span> $P= \frac{E}{t}$
<span>this means that the intensity can be rewritten as
</span> $I= \frac{E}{tA}$ <span>
So, it's basically the rate of energy (per unit of time) through a given surface.</span>

5 0

3 years ago

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2. A rock is shot straight up into the air with a slingshot that had been stretched 0.30 m. Assume

Luba_88 [7]

Answer:

e

Explanation:

4 0

3 years ago

Read 2 more answers

1. A diver dives off of a raft - what happens to the diver? The raft? How does this relate to Newton's Third Law? Action Force:

Natali [406]

The Action Force of this scenario is the pushing force of the Diver. The Reaction Force is the raft pushing back on the diver.

The Third Law of Motion states that "For every action, there is an equal and opposite reaction." Now when the diver dives off the raft, the raft is also pushing the same amount of force as the diver did as he dives off. The diver will then move forward and the raft on the other hand will move backwards.

The movement of the raft shows the opposite force. It will move backwards depending on how strong the diver will push off on the raft. And the amount of force he pushes on it, the raft will exert the same force so the stronger the force of the diver, the farther he will go because the raft will push him in that same direction as it goes backwards.

7 0

3 years ago

A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally

Ivenika [448]

(a) +2.60 m/s

The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, at constant speed

- a vertical motion, at constant acceleration (acceleration of gravity, $g=-9.8 m/s^2$ , downward)

In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):

$v_x = +2.60 m/s$

(b) -17.2 m/s

The vertical component of the fish's velocity instead follows the equation:

$v_y = u_y +gt$

where

$u_y = 0$ is the initial vertical velocity, which is zero

$g=-9.8 m/s^2$ is the acceleration of gravity

t is the time

Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:

$v_y = 0+(-9.8)(1.75)=-17.2 m/s$

where the negative sign indicates the direction (downward).

(c)

The horizontal component of the fish's velocity would increase

The vertical component of the fish's velocity would stay the same.

As we said from part (a) and (b):

- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too

- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.

4 0

3 years ago