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Vanyuwa [196]
3 years ago
10

A 5.0 V battery storing 43.0 kJ of energy supplies 1.5 A of current to a circuit. How much energy does the battery have left aft

er powering the circuit for 1.0 h? A. 43 kJ B. 16 kJ C. 41 kJ D. 27 kJ
Physics
1 answer:
Free_Kalibri [48]3 years ago
3 0

Answer:

B. 16 kJ

Explanation:

Energy = VIt.............. Equation 1

Where V = Voltage, I = Current, t = time

Given: V = 5.0 V, I = 1.5 A, t = 1 h = 3600 s.

Substitute these values into equation 2

E = 5.0(1.5)(3600)

E = 27000 J

E = 27 kJ.

Amount of energy left = 43 kJ - 27 kJ

Amount of energy left = 16 kJ.

Hence the right option is B. 16 kJ

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What is the time constant of a 9.0-nm-thick membrane surrounding a 0.040-mm-diameter spherical cell? Assume the resistivity of t
yawa3891 [41]

Given Information:

Diameter of spherical cell = 0.040 mm

thickness = L = 9 nm

Resistivity =  ρ = 3.6×10⁷ Ω⋅m

Dielectric constant = k = 9.0

Required Information:

time constant = τ = ?

Answer:

time constant = 2.87×10⁻³ seconds

Explanation:

The time constant is given by

τ = RC

Where R is the resistance and C is the capacitance.

We know that resistivity of of any material is given by

ρ = RA/L

R =  ρL/A

Where area of spherical cell is given by

A = 4πr²

A = 4π(d/2)²

A = 4π(0.040×10⁻³/2)²

A = 5.026×10⁻⁹ m²

The resistance becomes

R =  (3.6×10⁷*9×10⁻⁹)/5.026×10⁻⁹

R = 6.45×10⁷ Ω

The capacitance of the cell membrane is given by

C = kεoA/L

Where k = 9 is the dielectric constant and εo = 8.854×10⁻¹² F/m

C = (9*8.854×10⁻¹²*5.026×10⁻⁹)/9×10⁻⁹

C = 44.5 pF

C = 44.5×10⁻¹² F

Therefore, the time constant is

τ = RC

τ = 6.45×10⁷*44.5×10⁻¹²

τ = 2.87×10⁻³ seconds

6 0
3 years ago
A 0.8 g object is placed in a 534 N/C uniform electric field. Upon being released from rest, it moves 12 m in 1.2 s. Determine t
Mice21 [21]

Explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

Electric field, E = 534 N/C

Distance, s = 12 m

Time, t = 1.2 s

We need to find the acceleration of the object. It can be solved as :

m a = q E.......(1)

m = mass of electron

a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

s=ut+\dfrac{1}{2}at^2

s=0+\dfrac{1}{2}at^2

a=\dfrac{2s}{t^2}

a=\dfrac{2\times 12\ m}{(1.2\ s)^2}

a = 16.67 m/s²

Now put the value of a in equation (1) as :

q=\dfrac{ma}{E}

q=\dfrac{0.0008\ kg\times 16.67\ m/s^2}{534\ N/C}

q = 0.0000249 C

or

q=2.49\times 10^{-5}\ C

Hence, this is the required solution.

5 0
3 years ago
Does gravity exist between a pencil on a coffee mug
DIA [1.3K]

Answer:

Yes it does

Explanation:

Gravity is pushing down on the pencil but the coffee mug is also pushing the pencil up with the same amount of force so they both don't move

5 0
3 years ago
3. What is the gravitational force between a 70 kg physics student and her 1 kg textbook, at a distance of 1 meter? (This number
Rina8888 [55]

ANSWER

\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}

EXPLANATION

Parameters given:

Mass of the student, M = 70 kg

Mass of the textbook, m = 1 kg

Distance, r = 1 m

To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

F=\frac{GMm}{r^2}

where G = gravitational constant

Therefore, the gravitational force acting between the student and the textbook is:

\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\  \\ F=4.67*10^{-9}\text{ }N \end{gathered}

That is the answer.

6 0
1 year ago
(i) 10 m (ii) 20 m (iii) 40 m (iv) 80 m
IRINA_888 [86]

Answer:

20m

420=80m

100

increases

increases then decreases

6 0
3 years ago
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