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Paladinen [302]

2 years ago

13

When two forces are acting a body in opposite direction their resultant become 6N and when they are acting in the same direction

their resultant becomes 12N. What is the value of two forces?

1 answer:

Norma-Jean [14]2 years ago

5 0

Answer:

9 and 3 N

Explanation:

Forces in the same direction sum up to produce the resultant force;

One force subtract the other will give the resultant force when they are in opposite directions;

Lets say one direction is forwards and the opposite backwards;

We have one force, let's say force A, in the forwards direction and another force, force B, acting in the same (forwards) or opposite (backwards) direction;

If B is acting in the same direction, then the resultant force (in this case) will be as follows:

A + B = 12

If B is acting in the opposite direction, then the resultant force will be as follows:
A - B = 6

Summing the two equations will allow us to solve for A:

A + B + (A - B) = 12 + 6

2A = 18

A = 9

Substitute this into either of the above equations and we can solve for B:
(9) - B = 6

B = 9 - 6

B = 3

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An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec

jolli1 [7]

Answer:

$v_o=39\ m/s\\t_m=4\ s$

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

$\displaystyle h_m=H+\frac{v_o^2}{2g}$

The object referred to in the question is thrown from a height H=0 and the maximum height is hm=77.5 m.

(a)

To find the initial speed we solve for vo:

$\displaystyle v_o=\sqrt{2gh_m}$

$v_o=\sqrt{2\cdot 9.8\cdot 77.5}$

$v_o=39\ m/s$

(b)

The maximum time or the time taken by the object to reach its highest point is calculated as follows:

$\displaystyle t_m=\frac{v_o}{g}$

$\displaystyle t_m=\frac{39}{9.8}$

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7 0

2 years ago

A parallel-plate capacitor is connected to a battery. After it becomes charged, the capacitor is disconnected from the battery a

Inessa [10]

Answer:

The potential difference between the plates increases

Explanation:

As we know that the capacitance of the capacitor is given by:

$q = CV$ (1)

where

q = charge

C = capacitance

V = Voltage or Potential Difference

Also, the capacitance of a parallel plate capacitor is given as:

$C = \frac{\epsilon_{o}A}{D}$ (2)

where

$\epsilon_{o} = permittivity of free space or vacuum$

A = Area of the plates

D = Separation distance between the plates

Now, from eqn (1) and (2):

$V = \frac{qD}{A\epsilon_{o}}$

Now, from the above eqn we can say that:

Potential difference depends directly on the separation distance between the plates of the capacitor and is inversely dependent on the area of the plates of the capacitor.

Therefore, after disconnecting, if the separation between the plates is increased the potential difference across it also increases.

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3 years ago

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3 years ago

Two trains A and B of length 400 m each are moving on two parallel tracks with

Vinvika [58]

<u>Answer:</u>

<em>The initial distance between the trains is 1450 m. </em>

<u>Explanation:</u>

In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.

Train B’s acceleration is $1m/s^2$ and it accelerated for 50 seconds.

<em> $a=1 m/s^2$ </em>

<em>t=50 s </em>

<em>initial speed u=72km/h </em>

<em>we have to convert this speed into m/s </em>

<em> $u=72 \times 5/18=20 m/s$ </em>

<em>Distance covered in accelerating phase $S=ut+1/2 at^2$ </em>

<em> $=20 \times 50+1/2 \times 1 \times 50^2$ </em>

<em> $=1000+1250=2250 m$ </em>

If a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.

<em>Here length of train A+length of train $B=400+400=800 m$ </em>

<em>Hence the initial distance between the trains = $2250-800=1450 m$ </em>

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