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bixtya [17]
4 years ago
14

How much work is needed to lift a barbell that has a weight of 445 N a total of 2 meters off ground

Physics
1 answer:
AleksAgata [21]4 years ago
5 0

Answer:

Total work done to lift barbell = 890 J

Explanation:

Given:

Weight of barbell (F) = 445 N

Height (Distance) = 2 meter

Find:

Total work done to lift barbell = ?

Computation:

⇒ Work = Force(F) × Distance

⇒ Total work done to lift barbell = Weight of barbell × Distance

⇒ Total work done to lift barbell = 445 N × 2 meter

⇒ Total work done to lift barbell = 890 J

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Explanation:

As per Rayleigh criterion, the angular resolution is given as follows:

\theta=\frac{1.22 \lambda}{D}

From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.

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4 years ago
What do electric forces between charges depend on
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Answer:

On the magnitude of the charges, on their separation and on the sign of the charges

Explanation:

The magnitude of the electric force between two charges is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

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From the formula, we see that the magnitude of the force depends on the following factors:

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Moreover, the direction of the force depends on the sign of the two charges. In fact:

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3 years ago
We cannot consider something to be work unless
iren [92.7K]
<span>force applied causes movement of an object in the same direction as the applied force.</span>
5 0
3 years ago
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Is a perch a consumer or...
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8 0
4 years ago
A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns throu
anastassius [24]

Answer:

a. 3 s.

Explanation:

Given;

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time of wheel rotation, t = 4 s

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Apply the kinematic equation below,

\theta = \omega_1 t \ + \ \frac{1}{2} \alpha t^2\\\\80 = 4\omega_1 + \frac{1}{2}*4*4^2\\\\80 = 4\omega_1 + 32\\\\ 4\omega_1 = 48\\\\ \omega_1 = \frac{48}{4}\\\\ \omega_1 = 12 \ rad/s

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Apply the kinematic equation below;

\omega_1 = \omega_o + \alpha t_1\\\\12 = 0 + 4t\\\\4t = 12\\\\t = \frac{12}{4}\\\\t = 3 \ s

Therefore, the wheel had been in motion for 3 seconds.

a. 3 s.

8 0
3 years ago
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