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3 years ago

At the particle level, the weakest of the fundamental forces is:

Physics

2 answers:

Anna [14]3 years ago

7 0

Answer:

GRAVITY

Explanation:

There are four fundamental forces in nature

1). Gravitational force

2). Electromagnetic force

3). Strong Nuclear Force

4). Weak Nuclear Force

So here out of all above forces the strongest force is nuclear force which only depends on the distance between the nucleons while the weakest force among all is gravitational force which depends on the mass of two bodies and distance between them.

Gravitational force is a long range force so it exist upto the large distance while nuclear force is a short range force and could not exist for large distance which is more than 1 fermi.

So here correct answer will be

Gravity

marshall27 [118]3 years ago

3 0

It should be gravity.

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An amusement park ride consists of airplane-shaped cars attached to steel rods. Each rod has a length of 15m and a cross-section

Oliga [24]

Explanation:

Given that,

Length of each rod, L = 15 m

Area of cross section of the rod, $A=8\ cm^2=0.0008\ m^2$

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Total weight of two people, F = 1900 N

We need to find the stretching in the rod when the ride is at rest. The formula of Young's modulus of a material is given by :

$Y=\dfrac{FL}{A\Delta L}$

Here, $\Delta L$ is the stretching in the rod

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3 years ago

slader A uniform cylindrical grinding wheel of mass 50.0 kg and diameter 1.0 m is turned on by an electric motor. The friction i

DiKsa [7]

Answer:

The value of torque on the rod is = 3.94 N - m

Explanation:

mass = 50 kg

Radius = 0.5 m

Moment of inertia ( I ) = $\frac{MR^{2} }{2}$

$I = 50 \frac{0.5^{2} }{2}$

I = 6.25 kg- $m^{2}$

Angular velocity $\omega$ = 120 $\frac{Rev}{min}$ = 12.6 $\frac{rad}{sec}$

Δ $\theta$ = No. of rev × 2 $\pi$

Δ $\theta$ = 20 × 2 $\pi$

Δ $\theta$ = 126 rad

From work energy theorem

Work done is equal to change in kinetic energy.

$W_{AB} = K_{B} - K_{A}$ -------- (1)

Where A & B represents the initial & final states.

Since $K_{A}$ = 0

$W_{AB} = K_{B}$

$W_{AB} = \frac{1}{2} I \omega^{2}$

$W_{AB} = \frac{1}{2} 6.25 (12.6)^{2}$

$W_{AB} =$ 496 J

We know that

$W_{AB} =$ T Δ $\theta$

Where $W_{AB} =$ work done

T = torque applied on the rod

Δ $\theta$ = Angular displacement of the rod.

496 = T × 126

T = 3.94 N - m

Therefore the value of torque on the rod is = 3.94 N - m

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