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IRISSAK [1]
3 years ago
13

A typical AA size rechargeable NiMH battery can store 1100-2100 mAh of electric charge. The small print on the battery in your h

and says 1390 mAh. Can you express this amount of electric charge using the usual coulomb units?
Physics
1 answer:
Anni [7]3 years ago
8 0

Answer:

The electric charge, q (in coulomb units) = 5004 C  

Given:

The charge stored as printed on NiMH battery, q = 1390 mAh  

Solution:

To express the amount of electric charge printed on the battery in milli-ampere-hour (mAh) in coulomb, we will do simple conversion of milli amperes in ampere and hours in seconds:

1 mA = 1\times 10^{-3}

1 hour = 60\times 60 = 3600 s

Also, we know that the rate of flow of charge is electric current, I:

I = \frac{q}{t}

⇒ q = [tex]I\times t[tex]                                   (1)

where

q = electric charge

I = current

t = time taken for flow of current

Using eqn(1), we get:

q = [tex]1390\times 10^{-3}\times 60\times 60[tex]

q = 5004 A-s = 5004 C

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P6: An object of mass m sits on a spring of constant k in an elevator that is accelerating upwards with acceleration a. a) In te
tankabanditka [31]

Answer:

(a). The spring compressed is \dfrac{ma+mg}{k}.

(b). The acceleration is 1.5 g.

Explanation:

Given that,

Acceleration = a

mass = m

spring constant = k

(a). We need to calculate the spring compressed

Using balance equation

kx-mg=ma

x=\dfrac{ma+mg}{k}....(I)

The spring compressed is \dfrac{ma+mg}{k}.

(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,

The compression is given by

x=2.5\times x_{0}

Here, acceleration is zero

So, x=2.5\times\dfrac{mg}{k}

We need to calculate the acceleration

Put the value of x in equation (I)

2.5\times \dfrac{mg}{k}=\dfrac{ma+mg}{k}

2.5\times\dfrac{mg}{k}=\dfrac{m}{k}(a+g)

a=2.5g-g

a=1.5g

Hence, (a). The spring compressed is \dfrac{ma+mg}{k}.

(b). The acceleration is 1.5 g.

8 0
2 years ago
Finderkeepers,ismadeormodel
ankoles [38]

Answer:

model lol

Explanation:

7 0
2 years ago
The angular velocity of a flywheel obeys the equa tion w(1) A Br2, where t is in seconds and A and B are con stants having numer
makkiz [27]

Answer:

A \to rad/s

B  \to rad/s^3

Explanation:

\omega_z(t)=A + Bt^2

Required

The units of A and B

From the question, we understand that:

\omega_z(t) \to rad/s

This implies that each of A and Bt^2 will have the same unit as \omega_z(t)

So, we have:

A \to rad/s

Bt^2 \to rad/s

The unit of t is (s); So, the expression becomes

B * s^2 \to rad/s

Divide both sides by s^2

B  \to \frac{rad/s}{s^2}

B  \to rad/s^3

5 0
2 years ago
An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact wi
hoa [83]

Answer:

a)       a = 3.72 m / s², b)    a = -18.75 m / s²

Explanation:

a) Let's use kinematics to find the acceleration before the collision

             v = v₀ + at

as part of rest the v₀ = 0

             a = v / t

Let's reduce the magnitudes to the SI system

              v = 115 km / h (1000 m / 1km) (1h / 3600s)

              v = 31.94 m / s

              v₂ = 60 km / h = 16.66 m / s

l

et's calculate

             a = 31.94 / 8.58

             a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

            I = Δp

            F Δt = m v_f - m v₀

            F = \frac{m ( v_f - v_o)}{t}

            F = m [16.66 - 31.94] / 0.815

            F = m (-18.75)

Having the force let's use Newton's second law

            F = m a

            -18.75 m = m a

             a = -18.75 m / s²

4 0
2 years ago
Un comerciante compra 300 litros de aceite a 3 L y los vendió 5 L qué beneficio obtuvo​
andre [41]
El beneficio que obtuvo es de 2 L
5 0
3 years ago
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