3 years ago

Newton's laws of motion

Physics

2 answers:

Kipish [7]3 years ago

8 0

Answer:second law

Acceleration of an object depends on the mass of the object and the amount of force applied

Strike441 [17]3 years ago

4 0

It’s newton’s 2nd law of motion

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Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

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3 years ago

What is the magnitude of the electric field strength between them, if the potential 7.95 cm from the zero volt plate (and 2.05 c

Dmitriy789 [7]

Answer:

ΔVab = Ed

ΔVab = Va-Vb = Va-V0 = Va

E = Va/ d

= 413V / 0.0795 m

= 5194.97 V/M

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A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to osc

ludmilkaskok [199]

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

$T=2\pi \sqrt{\frac{L}{g} }$

Where T is period

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g is acceleration due to gravity = $9.8m/s^{2}$

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this $L_{O}$

$L_{O} = 3/4 * 5.95m$

= 4.4625m

thus $T=2\pi \sqrt{\frac{L_{O} }{g} }$

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3 years ago

A 2.2 kgkg block slides along a frictionless surface at 1.2 m/sm/s . A second block, sliding at a faster 4.0 m/sm/s , collides w

aleksklad [387]

Answer:

0.6kg

Explanation:

the unknown here is the mass of the second block

applying the law of the conservation of momentum

m₁v₁ + m₂v₂ = (m₁ + m₂) v₃

where m₁=mass of first block=2.2kg

m₂=mass of colliding block= ?

v₁= velocity of first block=1.2m/s

v₂=velocity of colliding block=4.0m/s

v₃= final velocity of combined block=1.8m/s

applying the formula above

(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8

2.64 + 4m₂ = 3.96 + 1.8m₂

collecting like terms

4m₂ - 1.8m₂ = 3.96 - 2.64

2.2m₂=1.32

divide both sides by 2.2

m₂= 0.6kg

4 0

3 years ago