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3 years ago

A man pushes a lawn mower on a level lawn with a force of 186.5 N. If 3.8% of this

Physics

1 answer:

Paladinen [302]3 years ago

4 0

Explanation:

The force is 186.5 N. 3.8% of the force is directed downward. So the vertical component of the force is:

Fᵧ = 0.038 (186.5 N)

Fᵧ = 7.087 N

The horizontal component is found with Pythagorean theorem:

F² = Fₓ² + Fᵧ²

(186.5 N)² = Fₓ² + (7.087 N)²

Fₓ = 186.4 N

So the work done is:

W = Fd

W = (186.4 N) (7.4 m)

W = 1379 J

Round as needed.

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3 years ago

Technician A says when diagnosing an overheating hydraulic system, be sure that the oil cooler is not plugged and the cooler’s f

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Answer:

Tech B is correct and Tech A is incorrect.

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Here Tech A is wrong because when diagnosing an overheating hydraulic system, it is not necessary to un plugg the oil cooler rather it should be plugged to for proper diagnosis.

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3 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

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2 years ago

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3 years ago

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