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3 years ago

If the change in velocity increases, what happens to the acceleration during the same time period?

Physics

2 answers:

4vir4ik [10]3 years ago

4 0

When the velocity is increasing the acceleration increases too

maksim [4K]3 years ago

4 0

When the velocity is increasing the acceleration increases tooo

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A type of Lead (Pb) ion has a + 4 oxidation number. Sulfur (S) has a — 2 oxidation number. What would be the chemical formula fo

serious [3.7K]

Answer: The chemical formula for the compound of these two elements is $PbS_2$

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here metal lead is having an oxidation state of +4 called as $Pb^{4+}$ cation and sulphur non metal has oxidation state of -2 called as $S^{2-}$ . Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $PbS_2$

The chemical formula for the compound of these two elements is $PbS_2$

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3 years ago

In the doorknob shown above, when the handle is rotated a distance of 189 millimeters, the spindle is rotated a distance of 27 m

Alisiya [41]

If my math is right its A) 7

because 189 divided by 27 is 7

7 0

4 years ago

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What is the range of a ball thrown horizontally at 12 m/s if its time of flight is 3.0 s?

vladimir1956 [14]

Answer:

Explanation:

because I did this assignment, :) your welcome

Next time do it by yourself, but here's the answer kid

7 0

3 years ago

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A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to acc

Jlenok [28]

Answer:

a is the answer

Explanation:

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3 years ago

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Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction

Norma-Jean [14]

Answer:

The ball would hit the floor approximately $0.55\; \rm s$ after leaving the table.

The ball would travel approximately $5.5\; \rm m$ horizontally after leaving the table.

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Let $\Delta h$ denote the change to the height of the ball. Let $t$ denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let $v_0(\text{vertical})$ denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible: $\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t$ .

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was $v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}$ .

The height of the table was $1.5\; \rm m$ . Therefore, after hitting the floor, the ball would be $1.5\; \rm m \!$ below where it was before leaving the table. Hence, $\Delta h = -1.5\;\rm m$ .

The equation becomes:

$\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}$ .

Solve for $t$ :

$\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55$ .

In other words, it would take approximately $0.55\; \rm s$ for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at $v(\text{horizontal}) =10\; \rm m \cdot s^{-1}$ ) until the ball hits the floor.

The ball was in the air for approximately $t = 0.55\; \rm s$ and would have travelled approximately $v(\text{horizontal})\cdot t \approx 5.5\;\rm m$ horizontally during the flight.

4 0

3 years ago