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3 years ago

¿Qué trabajo hace una fuerza de 110 N cuando mueve su punto de aplicación 20 mt en su misma dirección? *

Physics

1 answer:

lys-0071 [83]3 years ago

7 0

Answer:

W = 2.200 Joules

Explanation:

Datos (data):

Fuerza [force] (F) = 110 N
Metros [meters] (m) = 20 m
Trabajo [work] (W) = ?

Usar la fórmula (use formula):

$\boxed{\bold{W = F*d}}$

Reemplazar (replace):

$\boxed{\bold{W = 110\ N*20\ m}}$

Resolver la multiplicación, recuerda que 1 N * 1 m = 1 J (resolve the multiplication, remember that 1 N * 1 m = 1 J:

$\boxed{\boxed{\bold{W =2.200\ J}}}$

Greetings.

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Mass is the amount of matter in an object whereas weight is the force of gravity acting on the mass of an object. Different planets exert a different force of gravity on an object-meaning that an object's weight will change depending on the force of gravity acting on it, but it's mad will remain unchanged.

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3 years ago

Question 10 of 34

labwork [276]

Julia walks from the park, which is six blocks east of her house, to the store, which is three blocks east of her house. Julia walks for 5 minutes. This walk's average speed will be 1.2 blocks per minute. Option B is correct.

<h3>What is displacement?</h3>

Displacement is defined as the shortest distance between the two points. Distance is the horizontal length covered by the body. While displacement is the shortest distance between the two points.

Displacement is a vector quantity .its unit is m.

The average velocity on this walk will be;

$\rm v_{avg}= \frac{d}{t} \\\\ \rm v_{avg}= \frac{6 \ block+ 3 \ block }{5 \ minute } \\\\ v_{avg}=1.4 \ block /min$

Hence option B is correct.

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2 years ago

During which month is the bacteria population just under 1 million?

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3 years ago

Itzel travels south 4 miles and then goes east 2 miles and then north 10 miles. What is her displacement? Round your answer to t

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The displacement of Itzel according to the question is 6.3 miles SW

Displacement is defined as the distance moved by a body in a specified direction

Find the diagram attached

From the diagram given, we can see that AB is the displacement

To get the length AB, we will have to use the Pythagoras theorem:

$AB^2=2^2+ 6^2\\AB^2 ^2=4+36\\AB^2=40\\AB=\sqrt{40}\\AB= 6.3 miles\\$

From the diagram, we can also se that the direction of the displacement in the South West direction.

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3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

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3 years ago