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3 years ago

What do electrons carry and then deposit when it goes from a battery to a light bulb?

Physics

1 answer:

Nataliya [291]3 years ago

7 0

Answer:

I think the answer is a negative charge because electrons move through the filament which means they lose a lot of energy. I hope this helps

Explanation:

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Can I get answers 14 through 18 please? Plz show work number 18!!!

kicyunya [14]

14). c
15). c
16). a little less than d
17). a

18).
-- The parallel combination of 4Ω, 6Ω, and 10Ω is 1.935Ω .

-- The battery sees 1.935Ω in series with 2Ω. That's 3.935Ω .

-- The voltage across the 3 parallel resistors is (1.935/3.935)
of the total battery voltage. That's about 5.9 volts.

-- I = V/R .
The current in the 10Ω resistor is (5.9v/10Ω) = 0.59 Amp. (choice-a)

8 0

3 years ago

The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee

kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2> $v_f = 35.3 m/s$ </h2>

Explanation:

Part a)

As we know that drag force is given as

$F = \frac{C_d \rho A v^2}{2}$

$C_d = 0.35$

$A = \frac{\pi d^2}{4}$

$A = \frac{\pi(0.074)^2}{4}$

$A = 4.3 \times 10^{-3} m^2$

$v = 40.2 m/s$

so we have

$F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}$

$F = 1.46 N$

So acceleration of the ball is

$a = \frac{F}{m}$

$a = \frac{1.46}{0.145}$

$a = 10.1 m/s^2$

Part B)

As per kinematics we know that

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 40.2^2 = 2(-10.1)(18.4)$

$v_f = 35.3 m/s$

4 0

4 years ago

A student builds an electromagnet using a battery, an iron nail, and some insulated wire. The wire is wrapped around the nail 50

dedylja [7]

Answer:

Explanation:

its right

5 0

3 years ago

Read 2 more answers

Imagine a car tire that contains 5.1 moles of air when at a gauge pressure of 2.1×10^5N/m2 (the pressure above atmospheric press

True [87]

To solve this problem it is necessary to use the concepts given through the ideal gas equation.

For this it is defined that

$PV = nRT$

Where,

P = Pressure

V = Volume

R = Gas ideal constant

T = Temperature

n = number of moles.

In this problem we have two states in which the previous equation can be applied, so

$1) P_1V_1 = n_1RT_1$

$2) P_2V_2 = n_2RT_2$

From the first state we can calculate the Volume

$V_1 = \frac{n_1RT_1}{P_1}$

Replacing

$V_1 = \frac{5.1*8.314*300.15}{3.1*10^5}$

$V_1 = 0.041m^3$

From the state two we can calculate now the number of the moles, considering that there is a change of 0.8 from Volume 1, then

$n_2 = \frac{P_2(0.8*V_2)}{RT_2}$

$n_2 = \frac{2.6*10^5(0.8*0.041)}{8.314*310.15}$

$n_2 = 3.3moles$

Therefore there are 3.3moles of air left in the tire.

8 0

4 years ago

Smoke detectors fall into two major classes. Ionization detectors, the most common units, contain two parallel electrodes that a

bonufazy [111]

Answer:

1720 years is the amount of time that will pass.

Option a) 1720 years is the correct answer

Explanation:

Given the data in the question;

Number of nuclei initially N₀ = 60 million = 60,000,000

After time t, Number of nuclei remaining N $_{rem$ = 3.75 × 10⁶

Also given that; half-life of radioactive americium $t_{1/2$ = 430 years.

so;

λ = ln2 / $t_{1/2$

we substitute

λ = ln2 / 430 years

N $_{rem$ = N₀e^(-λt)

solve for t

t = 1/λ × ln( N₀/N $_{rem$ )

so we substitute

t = 1 / (ln2 / 430 years) × ln( 60,000,000 / ( 3.75 × 10⁶ ) )

t = ( 430 years / ln2 ) × ln( 60,000,000 / ( 3.75 × 10⁶ ) )

t = ( 430 years / 0.693147 ) × ln( 16 )

t = 620.359 years × 2.7725887

t = 1720.0003 ≈ 1720 years

Therefore, 1720 years is the amount of time that will pass.

Option a) 1720 years is the correct answer

7 0

3 years ago