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3 years ago

How is work calculated when the force applied is not parallel to the displacement

Physics

2 answers:

Leno4ka [110]3 years ago

7 0

Pretty sure it’s Force*Distance*Cos(theta)

harkovskaia [24]3 years ago

6 0

Answer:

Workdone=horizontal component Of The force x displacement

Explanation:

If the force is not parallel to the displacement,you have to find the horizontal component Of The force using the given angle, after that you multiply the horizontal component Of The force by the displacement then you get work done

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A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark

Natali5045456 [20]

The total displacement of the toy car at the given positions is 0.

The given parameters;

First displacement of the car, = 5 cm left
Second displacement of the car, = 8 cm right
Third displacement of the car, = 3 cm to the left

The total displacement of the car is calculated as follows;

Let the left direction be "negative direction"
Let the right direction be "positive direction"

$\Delta x = - \ 5\ cm \ + \ 8 \ cm \ - \ 3 \ cm \\\\\Delta x = 0$

Thus, the total displacement of the toy car at the given positions is 0.

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2 years ago

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The water creates less friction between your foot and the ground

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3 years ago

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(15 Points)

oksian1 [2.3K]

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

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5 0

2 years ago

Can someone solve this problem and explain to me how you got it

evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

$F=9*10^{9}*\frac{0.041*0.029}{12^{2}}$

$F=74312.5N$

Question6:

Given: q1= $3*10^{-18}C$ , r=5 m, F= $4*10^{-11}N$

therefore by Coulumb's law,

$4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}$

⇒ $q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C$

4 0

3 years ago

The near point of a farsighted person's uncorrected eyes is 80 cm. what power contact lens should be used to move the near point

makvit [3.9K]

To solve this problem, we will get f and then we will use it to calculate the power.

So, for this farsighted person,
do = 25 cm and di = -80
Therefore:
1/f = (1/25) + (1/-80) = 0.00275 = 0.275 m

Power = 1/f = 1/0.275 = +3.6363 Diopeters.
This means that the lens is converging.

5 0

3 years ago