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3 years ago

Explain how do batteries work?

1 answer:

4 0

This path is called a circuit. Batteries have three parts, an anode (-), a cathode (+), and the electrolyte. ... The chemical reactions in the battery causes a build up of electrons at the anode. This results in an electrical difference between the anode and the cathode.

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A 59kg child starting from rest slides down a water slide with a vertical height of 5.0m. what is the child's speed halfway down

KIM [24]

EP (potential energy) = mgy -> (59)(9.8)(-5) = -2,891
EP + EK (kinetic energy) = 0; but rearranging it for EK makes it EK = -EP, such that EK = 2891 when plugged in.
EK = 0.5mv^2, but can also be v = sqrt(2EK/m).
Plugging that in for sqrt((2 * 2891)/59), we get 9.9 m/s^2 with respect to significant figures.

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3 years ago

Can some one please help me with 7 and 8 I can’t fail this I’ll mark brainless .

Musya8 [376]

I think the first question is talking about the ionic compound Sodium sulfide and it’s formula is Na2S.
And for the second question, i’m pretty sure it’d be a positive charge

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3 years ago

Which of the following cannot be determined by analyzing the periodic table?

dybincka [34]

Your answer is C) The year the element was discovered :)

4 0

3 years ago

Read 2 more answers

HI PLEASE HELP, I ONLY HAVE 10 MINUTES

Whitepunk [10]

Answer:

um ok

?????

i dont like you get rejected

jk u gud

Explanation:

5 0

3 years ago

a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^

posledela

Answer:

Approximately $0.8\; \rm s$ (assuming that air resistance is negligible.)

Explanation:

Let $v_0$ denote the initial velocity of this ball. Let $\theta$ denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

Initial vertical velocity: $v_0(\text{vertical}) = v_0 \cdot \sin(\theta)$ .
Initial horizontal velocity: $v_0(\text{vertical}) = v_0 \cdot \cos(\theta)$ .

If air resistance on this ball is negligible, $v_0(\text{vertical})$ alone would be sufficient for finding the time of flight of this ball.

Calculate $v_0(\text{vertical})$ given that $v_0 = 8 \; \rm m \cdot s^{-1}$ and $\theta = 30^\circ$ :

$\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}$ .

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since $v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}$ , the vertical velocity of this ball right before landing would be $v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}$ .

Calculate the change to the vertical velocity of this ball:

$\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}$ .

In other words, the vertical velocity of this ball should have change by $8\; \rm m \cdot s^{-1}$ during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is $g = 10\; \rm m \cdot s^{-2}$ . In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by $10\; \rm m\cdot s^{-1}$ each second. What fraction of a second would it take to change the vertical velocity of this ball by $8\; \rm m \cdot s^{-1}$ ?

$\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}$ .

In other words, it would take $0.8\; \rm s$ to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be $0.8\; \rm s\!$ .

5 0

3 years ago