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alina1380 [7]
3 years ago
13

How are chargeable cells different from ordinary dry cells​

Physics
1 answer:
topjm [15]3 years ago
6 0
Ordinary cells can convert chemical energy to electrical energy only, but rechargeable cells can also store electrical energy into chemical energy and vice versa. You will study more about it in your higher classes. secondary cells can be recharged and used again but dry cells cannot be recharged.
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If a person is throwing darts at a target, and shoots three in a row in the same spot on the 20 section outer ring, but none in
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a) precise but not accurate. 

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A 4500 kg car accelerates from rest to 45.0
Llana [10]

The car undergoes an acceleration <em>a</em> such that

(45.0 km/h)² - 0² = 2 <em>a</em> (90 m)

90 m = 0.09 km, so

(45.0 km/h)² - 0² = 2 <em>a</em> (0.09 km)

Solve for <em>a</em> :

<em>a</em> = (45.0 km/h)² / (2 (0.09 km)) = 11,250 km/h²

Ignoring friction, the net force acting on the car points in the direction of its movement (it's also pulled down by gravity, but the ground pushes back up). Newton's second law then says that the net force <em>F</em> is equal to the mass <em>m</em> times the acceleration <em>a</em>, so that

<em>F</em> = (4500 kg) (11,250 km/h²)

Recall that Newtons (N) are measured as

1 N = 1 kg • m/s²

so we should convert everything accordingly:

11,250 km/h² = (11,250 km/h²) (1000 m/km) (1/3600 h/s)² ≈ 0.868 m/s²

Then the force is

<em>F</em> = (4500 kg) (0.868 m/s²) = 3906.25 N ≈ 3900 N

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4 years ago
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Frequency is deoted as hertz; hertz is a measurement of the _________ _____ __________ that a wave is occurring.
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4 years ago
Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0
3 years ago
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