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3 years ago

What is the force on a 1000kg elevator that is falling freely at 9.8 m/sec^2

1 answer:

3 0

From Newton's second law, we know F = ma, where a is the acceleration and m is the mass in kg.

F = 1000kg * 9.8m/s = 9800N

F = 9800 N

Hope this helps!

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sergejj [24]

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Using newtons second law

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$\\ \sf\longmapsto Force=1(9.8)$

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4 years ago

An airplane accelerates down a run-way at 3.20m/s2 for 32.8s until is finally lifts off the ground. Determine the distance trave

natta225 [31]

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3 years ago

A particle with negative charge q and mass m = 2.65×10−15 kg is traveling through a region containing a uniform magnetic field B

Norma-Jean [14]

Answer:

q = 2,95 10-6 C

Explanation:

The magnetic force on a particle is described by the equation

F = q v x B

Where bold indicate vectors

Let's make the vector product

vxB = $\left[\begin{array}{ccc}i&j&k\\-3&4&12\\0&0&-0.13\end{array}\right]$

v x B = 1.20 106 [i ^ (4 0.130) - j ^ (3 0.130)]

vx B = 1.20 106 [0.52 i ^ - 0.39j ^]

As they give us the force module, let's use Pythagoras' theorem,

|v xB | =1.20 10⁶ √( 0.52² + 0.39²)

|v x B| = 1.20 10⁶ 0.65

v xB = 0.78 10⁶

Let's replace and calculate

2.30 = q 0.78 10⁶

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3 years ago

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