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velikii [3]
2 years ago
9

Two identical cells are of negligible internal resistance are connected in series and in parallel. Find the ratio of current thr

ough a load resistance R in two cases
Physics
1 answer:
Vilka [71]2 years ago
3 0

The ratio of the current passing through the load resistance in the two cases is 2.

Two identical cells of negligible internal resistance, are connected in parallel and in series with each other.

Let the voltage of each cell be V.

In series, the voltage of the cells will be doubled i.e. 2V.

The current in the circuit is:

I = 2V/R

In Parallel connection, the voltage of cells will be the same i.e. V.

The current in this circuit

I' = V/R

Now,

Current in Series / Current in Parallel = I/I' = (2V/R)/(V/R) = 2

Hence, the ratio of currents through a load resistance R in the two cases is 2.

Learn more about current here:

brainly.com/question/1100341

#SPJ9

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3 years ago
What is the acceleration of the object?<br> m/s²
Orlov [11]

Answer:

-2.5m/s^2

Explanation:

10-40/12-0=-2.5

7 0
2 years ago
A force acts on a body of mass 13 kg initially at rest. The force
PtichkaEL [24]

Answer:

Force that acted on the body was F = 13 N

Explanation:

If once accelerated, the body covers 60 meters in 6 seconds, then its velocity is 60/6 m/s = 10 m/s

When the force was acting (for 10 seconds) the object accelerated from rest (initial velocity vi = 0) to 10 m/s (its final velocity). therefore we can use the kinematic equation for the velocity in an accelerated motion given by:

v_f=v_i+a*t

which in our case becomes;

10\,m/s=0+a*(10\,s)

and we can solve for the acceleration as:

a = 10/10  m/s^2 = 1 m/s^2

Therefore the force acting on the body, based on Newton's 2nd Law expression: F = m * a is:

F = 13 kg * 1 m/s^2 = 13 N

4 0
3 years ago
A force of 55N accelerates a 7.5kg wagon at 5.3 m/s^2 along a road. How large is the frictional force?
Varvara68 [4.7K]

Answer:

<h2>15.25 N</h2>

Explanation:

       A force of 55\text{ }N is acting on a wagon along the road. The wagon weights 7.5\text{ }kg. Acceleration of the wagon is given as 5.3\text{ }\frac{m}{s^{2}}.

       Consider the block as the system, the forces acting are Frictional force, Gravitational force, Normal reaction and External force applied by us.

       Gravitational Force and Normal Reaction cancel out each other.

       Net External Force = Mass of system/wagon \times Acceleration of wagon

       F_{ext}-F_{friction}=(7.5\text{ }kg)\times(5.3\text{ }\frac{m}{s^{2}})=39.75\text{ }N\\55\text{ }N-F_{friction}=39.75\text{ }N\\F_{friction}=15.25\text{ }N

F_{friction} has a negative sign because it opposes the motion of the wagon.

∴ Frictional Force = 15.25 N

4 0
3 years ago
Your friend says that if Newton’s third law is correct, no object would ever start moving. Here is his argument: You pull a sled
m_a_m_a [10]

Answer:

Explanation:

You pull a sled exerting a 50 N force on it , sled also exerts a force on you . These forces are action and reaction force , as per third law of Newton . These two forces are equal and  opposite . But they do not act on the same object so they do not cancel each other . They act on different objects , one on the sledge and the other on you . Due to force on sledge , sledge moves in the direction of force or towards you . You will start moving in opposite direction if frictional force of ground is nil or less .

6 0
3 years ago
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