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3 years ago

How much work is done if a force of 20N is used to move an object 6 metres? pls help

2 answers:

8 0

I assume that the force of 20 N is applied along the direction of motion and was applied for the whole 6 meters, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula; Work = 20 N * 6 m * cos 0º.

Work = 20 N * 6 m * 1

Work = 120 Nm

Work = 120 joules

Hope this helps!

Ipatiy [6.2K]3 years ago

8 0

the answer is 120 as you just 20 x 6

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A 5 kg mass is oscillating on a spring with a time period of 2.8 seconds. What is the spring constant k of the spring?

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Answer:

Explanation:

<u>Simple Harmonic Oscillator</u>

It consists of a weight attached to one end of a spring being allowed to move forth and back.

If m is the mass of the weight and k is the constant of the spring, the period of the oscillation is given by:

$\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}$

If the period is known, we can find the value of the constant by solving for k:

$\displaystyle k=m\left(\frac{2\pi}{T}\right)^2$

Substituting the given values m=5 Kg and T=2.8 seconds:

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3 years ago

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NemiM [27]

Answer:

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The substances that are higher up in the series are more reactive in electrochemical cells.

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3 years ago

A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,

Romashka [77]

Answer:

$v_{y} = -104 m/s$

Explanation:

Using:

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$F=e*q$

Force = magnitude of charge * velocity * magnetic field * sin tither

$F_{x2}= |q|*v*B*sin \alpha$

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Force on particle due to magnetic field:

$F_{x2}= |q|*v*B*sin \alpha = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)$

$F_{x2}$ is in the positive x direction as $F_{x1}$ is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

$F_{xnet}- F_{x1 } = F_{x2 }$

$(6.13*10^{-3}) - (8.53*10^{-3} ) = (7.728*10^{-6})*(v)$

$v = (F_{xnet} - F_{x1}) / (F_{x2} )$

$=((6.13*10^{-3} ) - (8.53*10^{-3})) / (7.728*10^{-6})$

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3 years ago

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