13. How many feet high do you think<br> you could jump on the moon?

You push on a 30 kg box with a force of 120 N. What is the acceleration of the box2

kumpel [21]

Answer:

6

Explanation:

120/30=6

8 0

3 years ago

In an electrolytic cell, to which electrode will a positive ion migrate and undergo reduction? the anode, which is negatively ch

ahrayia [7]

The reaction of reduction always undergoes with the cathode, the positive ion will migrate towards the cathode with the negative charge whilst the anode always has oxidation reaction. These two types of reaction does not change.

4 0

3 years ago

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When the speed of the bottle is 2 m/s, the KE is kg m2/s2. When the speed of the bottle is 3 m/s, the KE is kg m2/s2. When the s

d1i1m1o1n [39]

mass of the bottle in each case is M = 0.250 kg

now as per given speeds we can use the formula of kinetic energy to find it

1) when speed is 2 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(2)^2 = 0.5 J$

2) when speed is 3 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(3)^2 = 1.125 J$

3) when speed is 4 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(4)^2 = 2 J$

4) when speed is 5 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(5)^2 = 3.125 J$

5) when speed is 6 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(6)^2 = 4.5 J$

3 0

4 years ago

Read 2 more answers

In a clock, the average speed is maximum for the tip of<br />a) Seconds hand b) Hours hand c) Minutes hand

Simora [160]

A, hope this helped! I didn’t really get it but I think it’s correct?

3 0

4 years ago

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With two identical light bulbs and two identical batteries how would you arrange in order to get the maximum total power to the

Norma-Jean [14]

Answer:

The batteries would be connected in series while the bulbs would be connected in parallel

Explanation:

Power (W) = VI

where V = voltage, I = current and R = resistance

from V = IR , I = V/R

Power (W) now becomes = V (V/R) = $\frac{V^{2} }{R}$

Power (W) = $\frac{V^{2} }{R}$

from the above equation, power is directly dependent on voltage, hence the voltage has to be high for the power to be high and the power is also inversely dependent on the resistance (in this case the bulbs which act as the load)

We have to batteries, when batteries are connected in series the total voltage becomes the summation of the two voltages hence giving a higher voltage and when they are connected in parallel their voltage remains the same. Since we want to get higher voltage we will connect the two batteries in series.
we have two bulbs which are the resistance here, from the equation above the power is inversely dependent on the resistance so we would need its value to be minimal. When resistance is connected in series the resistance individual will be added to get the total resistance, hence the total resistance will be high but when the resistors are arranged in parallel you get the total resistance by applying the formula $\frac{R1R2}{R1+R2}$ which will give us a lower resistance. Hence we would connect the bulbs in parallel.

Take note that the power from this connection should not exceed the bulbs power rating so as to avoid damage of the bulbs.

4 0

4 years ago

13. How many feet high do you think you could jump on the moon?