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2 years ago

A tight knot can be easily opened by using a long spanner. Give reason.

Physics

2 answers:

olga nikolaevna [1]2 years ago

5 0

Answer:

Torque

Explanation:

loosening loosening a boat requires a certain amount of torque which is the product of the force acting on the wrench and the moment arm, or distance from the center of rotation at which the force is applied. the longer the handle, the last force it takes to generate the same amount of torque

erma4kov [3.2K]2 years ago

4 0

Answer:

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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Answer:

rate of fission =5.89*10^3 1\Year

Explanation:

we know that

rate of fission is given as $rate of fission = \frac{0.69}{(T_{1/2})_{fission}} *\frac{ Mass* Avogardo\ number}{Molar\ mass}$

$(T_{1/2})_{fission} = 3*10^{17} y$

mass = 1.0 g

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molar mass of isotopes 235U =235

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2 years ago

For an object starting from rest and accelerating with constant acceleration, distance traveled is proportional to the square of

natali 33 [55]

The problem states that the distance travelled (d) is directly proportional to the square of time (t^2), therefore we can write this in the form of:

d = k t^2

where k is the constant of proportionality in furlongs / s^2

Using the 1st condition where d = 2 furlongs, t = 2 s, we calculate for the value of k:

2 = k (2)^2

k = 2 / 4

k = 0.5 furlongs / s^2

The equation becomes:

d = 0.5 t^2

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8 0

2 years ago

A mercury thermometer has a glass bulb of interior volume 0.100 cm3 at 10°c. the glass capillary 10) tube above the bulb has an

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Initial volume of mercury is
V = 0.1 cm³

The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.

Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³

The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
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h = ΔV/A
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Answer: 4.5 cm

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2 years ago

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