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snow_lady [41]
3 years ago
9

The total charge a battery can supply is rated in mA⋅h , the product of the current (in mA ) and the time (in h ) that the batte

ry can provide this current. A battery rated at 1000 mA⋅hr can supply a current of 1000 mA for 1.0 h , 500 mA current for 2.0 h , and so on. A typical AA rechargeable battery has a voltage of 1.2 V and a rating of 1800 mA⋅h . Part A For how long could this battery drive current through a long, thin wire of resistance 34 Ω ? Express your answer in hours. Δt Δ t = nothing h SubmitRequest Answer Provide Feedback Next
Physics
1 answer:
natita [175]3 years ago
4 0

Answer: 0.2  hours

Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA  in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .

Besides, this battery has a voltage of 12 V

so by using the Ohm law we also know that V=R*I,

Fron this we can obtain:

I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA

then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA  in x time in the form:

1hour------- 1800 mA

x hour--------350 mA

time= 350/1800= 0.2 hour

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Does someone know how to do math with that equation
mestny [16]

Answer:

f = 5 cm

Explanation:

using the thin lens equation, given as follows:

\frac{1}{f} = \frac{1}{d_{o}}+\frac{1}{d_{i}}\\

where,

f = focal length = ?

do = the distance of object from lens = 20 cm

di = the distance of image from lens = 6.6667 cm

Therefore,

\frac{1}{f} = \frac{1}{20\ cm}+\frac{1}{6.6667\ cm}\\\\\frac{1}{f} =  0.199999\ cm^{-1}\\\\f = \frac{1}{0.199999\ cm^{-1}}\\\\

<u>f = 5 cm</u>

8 0
3 years ago
An iron block of mass 10 kg rests on a wooden plane inclined at 30° to the horizontal. It is found
Kaylis [27]

I assume the 100 N force is a pulling force directed up the incline.

The net forces on the block acting parallel and perpendicular to the incline are

∑ F[para] = 100 N - F[friction] = 0

∑ F[perp] = F[normal] - mg cos(30°) = 0

The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.

Then

F[friction] = 100 N

F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N

If µ is the coefficient of static friction, then

F[friction] = µ F[normal]

⇒   µ = (100 N) / (84.9 N) ≈ 1.2

5 0
2 years ago
You shake a bottle of soda and take off the cap. If the soda shoots out of the
kicyunya [14]

Answer:

A. 11.5 m

Explanation:

Given,

The initial velocity of the soda cap, u = 15 m/s

The soda bottle cap is projected vertically upwards,

Hence, the angle formed with the ground, Ф = 90°

The maximum height of the projectile is given by the formula,

                                    h_{max} =\frac{u^{2}sin^{2}\phi}{2g}

Substituting the given values in the above equation

                                   h_{max} =\frac{15^{2}sin^{2}90}{2X9.8}

                                                        = 11.5 m

Hence, the maximum height of the cap is h = 11.5 m

8 0
3 years ago
18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de
Alona [7]

Answer:

Definimos momento como el producto entre la masa y la velocidad

P = m*v

(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)

Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.

Peso = m*9.8m/s^2 = 175N

m = (175N)/(9.8m/s^2) = 17.9 kg

Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.

Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:

Vx = 36m/s

Mientras que para la velocidad vertical, usamos la conservación de la energía:

E = U + K

Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)

Entonces al principio solo hay energía potencial:

U = m*g*h

donde:

m = masa

g = aceleración gravitatoria

h = altura  

Sabemos que la altura inicial es 60m, entonces la energía potencial es:

U = 175N*60m = 10,500 N

Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:

10,500N = (m/2)*v^2

De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.

√(10,500N*(2/ 17.9 kg)) = 34.25 m/s

La velocidad vertical es 34.25 m/s

Entonces el vector velocidad se podrá escribir como:

V = (36 m/s, -34.25 m/s)

Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.

Reemplazando esto en la ecuación del momento obtenemos:

P = 17.9kg*(36 m/s, -34.25 m/s)  

P = (644.4 N, -613.075 N)

6 0
3 years ago
12. The bond formed when two atoms share a pair of electrons is called a
givi [52]

Answer:A covalent bond, also called a molecular bond, is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

Explanation:

4 0
3 years ago
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