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MaRussiya [10]
2 years ago
14

A tight knot can be easily opens by using a longer spanner. Give reasons

Physics
1 answer:
kobusy [5.1K]2 years ago
6 0

It is because the effort distance is greater than the load distance

Explanation:

As we know, Effort×effort distance = load × load distance

So when effort distance is increases,

The effort decreases

So when the spanner’s handle is long

A tight knot can easily be opened by less effrot

I hope it helped

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All waves carry a) energy B) light C) matter D) particles
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All waves carry energy.
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It is important to wear protective equipment in both practice and games.
Ede4ka [16]

Answer:

OT

It is important to wear protective equipments during games and practices.

Explanation:

Practice is a review of what is about to happen in some time to come and with practice ,the future can be trusted.

During practice it is very important to use protective gears or equipments because without them, injuries and accidents are liable to occur.

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4 0
3 years ago
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
natima [27]

Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

r=\sqrt{x_{2}^2+y_{2}^2}

Put the value into the formula

r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

r=0.0876\ m

We need to calculate the magnitude of the charge q₃

Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

Put the value into the formula

14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

q_{3}=40.46\ \mu C

Hence, The value of  charge q₃ is 40.46 μC.

5 0
3 years ago
The components of vector A are Ax = +2.2 and Ay = -6.9 , and the components of vector B are given are Bx = -6.1 and By = -2.2. W
Zina [86]
For simplicity, let's call vector B-A  vector C  Then C is
Cx = (-6.1 - 2.2)  
Cy = (-2.2 - (-6.9))  Or,
Cx = -8.3  Cy = 4.7
The magnitude is found with the Pythagorean theorem
||C|| = √(-8.3² + 4.7²) = 9.538
3 0
3 years ago
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