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WITCHER [35]
3 years ago
13

The manufacturer of a 9V dry-cell flashlight battery says that the battery will deliver 20 mA for 80 continuous hours. During th

at time the voltage will drop from 9 V to 6 V. Assume the drop in voltage is linear with time. How much energy does the battery deliver in this 80 h interval
Physics
1 answer:
Maksim231197 [3]3 years ago
5 0

Answer:

17280 J or 17.28 kJ

Explanation:

Given that the voltage drop,

U = U2 - U1

U = 9 - 6

U = 3V

Also, we're told that the current, I is equal to 20 mA with the discharge time, t being 80 hrs.

Converting the time from h oi urs to seconds, we have

t = 80 * 3600

t = 288000

Now, to find the energy needed, we're going to use the formula

w = pt, where p = U * I

p = 3 * 20*10^-3

p = 60*10^-3

w = 60*10^-3 * 288000

w = 17280 J or 17.28 kJ

Therefore, the total energy the battery delivers in the 80 hrs is 17.28 kJ

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Write down the addres of earth in as much detail as possible.
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8 0
3 years ago
A man pushes a 34 kg lawn mower using a handle inclined 45.0 ∘ from the horizontal. He pushes with 75 N directed along the handl
tresset_1 [31]

Answer:

333 N

Explanation:

The weight of an object is given by:

W=mg

where

m is the mass of the object

g is the acceleration of gravity

For the mower in the problem,

m = 34 kg

Also

g=9.8 m/s^2

So, the weight of the mower is

W=(34)(9.8)=333 N

6 0
3 years ago
In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of 48 N to the input pi
erik [133]

Answer:

Force that the output plunger applies to the car; F2 = 3888N

Explanation:

For a hydraulic device, the relationship between the force and the area using Pascal's principle is;

F1/A1 = F2/A2

Where;

F1 is force applied to the input piston

F2 is force that the output plunger applies to the car

A1 is Area of input piston

A2 is area of larger piston

We are given;

R2/R1 = 9

So,R2 = 9R1

F1 = 48N

Area of input piston;

A1 = π(R1)²

Area of output piston;

A2 = π(9R1)²

Since, (F1/A1) = (F2/A2)

Thus;

F1/(π(R1)²) = F2/(π(9R1)²)

If we simplify, π(R1)² will cancel out to give;

F1 = F2/9²

Thus;

F2 = 9² x F1

Plugging in 48N for F1, we have;

F2 = 9² x 48

F2 = 81 x 48

F2 = 3888N

8 0
3 years ago
A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
Kipish [7]

Answer:

C. 0.25J

Explanation:

Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;

L is the inductance

I is the current flowing in the inductor

Given parameters

L = 20mH = 20×10^-3H

I = 5A

Required

Energy stored in the magnetic field.

E = 1/2 × 20×10^-3 × 5²

E = 1/2 × 20×10^-3 × 25

E = 10×10^-3 × 25

E = 0.01 × 25

E = 0.25Joules.

Hence the energy stored in the magnetic field of this inductor is 0.25Joules

7 0
4 years ago
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