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sveta [45]
3 years ago
11

A force Ě = F, î + Fy h acts on a particle

Physics
1 answer:
AVprozaik [17]3 years ago
8 0

Answer:

Pt 1: W=39J

Pt 2: θ = 19.7

Explanation:

Part 1:

W=FS

W=F_{_xs_x}+F{_ys_y}

imagine the "i"and the "j" with the hat

W=(10i-1j)(4i+3j)

W=(10*4)-(1*1)

W=40-1

W=39J

Part 2:

|F|=\sqrt{10^2+(-1)^2}

|F|=\sqrt{101}

|S|=\sqrt{4^2+1^2}

|S|=\sqrt{17}

W=|F||S|cosθ

39=(\sqrt{101})(\sqrt{17} )cosθ

θ = 19.7

Hope it helps

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1. 1500j of work was done to move a box 20m. What force was applied to the box ?
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Answer:

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Explanation:

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we are given that Work = 1,500J and Distance = 20m

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1,500 = Force x 20

Force = 1,500 ÷ 20 = 75N

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we are given that mass = 165 kg and change in height = 42m

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4 years ago
A boy kicks a soccer ball, giving it an initial speed of 7.5m/s at an angle of 27° above the horizontal(=ground). How high will
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When the initial speed given is 7.5m/s at an angle of 27° , ball will go

4.637 meters.

Assume no air opposition to the ball ;

Vertical component of ball is sin 27° =  0.453

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Form of motion experienced by an object or particle that is projected near surface of the earth and moves along a curve is called Projectile motion. Three types of projectile motion are Horizontal projectile motion. Oblique projectile motion and Projectile motion on an inclined plane.

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3 years ago
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Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of y_0=1500m, and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

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With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

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Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

t=\sqrt{\frac{2y_0}{g}}

Since we have y_0=1500m

t=\sqrt{\frac{2(1500)}{9.8}}

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Replacing in [2]

x = 688\ m/sec \ (17.5sec)

x = 12040\ m

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

6 0
3 years ago
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