Is it possible for a population with a high birth rate to decrease in size.

A student examines the effect of the number of D batteries in a closed circuit on the brightness of a light bulb. In the experim

xxTIMURxx [149]

Answer:

Placing cells in series increases the

voltage in the circuit by 1.5 V for each

cell. Increasing the voltage increases the

brightness of the bulb

Explanation:

I. MAKING THE CONNECTION

How many terminals are located on the battery? 2

How many terminals are located on the bulb? 2

II. PLACING CELLS IN SERIES

What is the effect on the brightness of the bulb by increasing the number of cells?

The bulbs become brighter when increasing the number of cells.

What changes occur in the current in the circuit when increasing the number of cells?

Increasing the number of cells increases the current in the circuit.

What changes occur in the voltage in the circuit when increasing the number of cells?

Increasing the number of cells increases the voltage (for cells in series the voltage is additive).

What changes occur in the resistance in the circuit when increasing the number of cells?

The resistance is determined by the number of bulbs. The resistance in the circuit remains unchanged.

III. PLACING BULBS IN SERIES

What is the effect on the brightness of the bulbs by increasing the number of bulbs?

Increasing the number of bulbs decreases the brightness of the bulbs.

What changes occur in the resistance in the circuit as more bulbs are added?

The resistance increases. In a series circuit, adding bulbs increases the resistance in the circuit.

What changes occur in the current in the circuit as more bulbs are added?

Increasing the resistance decreases the current.

Observations on unscrewing one bulb. Explain your observations.

A complete circuit requires the electrons to move from the negative terminal of the battery to the

positive terminal. When one bulb is unscrewed it opens the circuit preventing a complete circuit and

the electrons cannot return to the battery.

IV. PLACING BULBS IN PARALLEL

Compare the brightness with two bulbs in parallel with two bulbs in series.

Two bulbs in parallel are brighter than two bulbs in series.

How does increasing the number of circuits (bulbs) change the current and resistance?

In a parallel circuit each bulb is in its own circuit. As bulbs are added the resistance in the circuit

decreases since each circuit is another pathway for electrons to move from one end of the circuit

4 0

4 years ago

(Please Help, Will Give BRAINLIEST Answer)

Xelga [282]

Answer:

A

Explanation:

Initial gravitational energy = final kinetic energy + heat

mgh = KE + Q

(50 kg) (9.81 m/s²) h = 78400 J + 884000 J

h = 1960 meters

4 0

3 years ago

Mastering Problems

lesya692 [45]

Answer:

C , E , A , D , B

Explanation:

We evaluate the accelerations for each case, using the formula: a = (vf - vi) / t

A) a = (10.3 - 0.5 ) / 1 = 9.8 m/s^2 --> magnitude: 9.8 m/s^2

B) a = (0 - 20) / 1 = - 20 m/s^2 --> magnitude : 20 m/s^2

C) a = (0.02 - 0.004) / 1 = 0.016 m/s^2 --> magnitude : 0.016 m/s^2

D) a = (4.3 - 0) / 0.4 = 10.75 m/s^2 --> magnitude : 10.75 m/s^2

E) a = (1 - 2) / 8.3 = - 0.12 m/s^2 --> magnitude: 0.12 m/s^2

Then, comparing magnitudes from least to greatest:

C , E , A , D , B

7 0

3 years ago

A substance of unknown mass absorbs 138 kilojoules of energy, going from 298 to 303 kelvin. if the specific heat of the substanc

AveGali [126]

Heat absorbed to raise the temperature is given by

$Q = mc\Delta T$

here

m = mass

c = 7.11 J/g C

$\Delta T = 303 - 298 = 5 kelvin$

Q = 138 KJ

Now by using the above formula we will have

$138 * 10^3 = m* 7.11 * 5$

$m = 3881 gram$

$m = 3.88 kg$

So the amount of the substance will be 3.88 kg

4 0

4 years ago

The Earth is 81.25 times as massive as the Moon and the radius of the Earth is 3.668 times the radius of the Moon. If a simple p

Alexeev081 [22]

Answer:

option (b)

Explanation:

Mass of moon = m

Mass of earth = 81.25 x mass of moon = 81.25 m

Radius of moon = r

radius of earth = 3.668 x radius of moon = 3.668 r

Frequency on earth = 2 Hz

Let the frequency on moon is f.

The formula for the frequency is given by

$f = \frac{1}{2\pi }\times \sqrt{\frac{g}{l}}$

The value of acceleration due to gravity on earth is g.

ge = G Me / Re^2

ge = G x 81.25 m / (3.668 r)^2

ge = 6.039 x G m / r^2 = 6.039 x gm

ge / gm = 6.039

Now use the formula for frequency

$\frac{fe}{fm} = \sqrt{\frac{ge}{gm}}$

$\frac{2}{fm} = \sqrt{\frac{6.039 gm}{gm}}$

$\frac{2}{fm} = 2.46$

fm = 0.814 Hz

4 0

3 years ago