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3 years ago

Calculate the final displacement from 0 if an object first moves 3 cm to the left and then 5 cm to the right.

2 answers:

5 0

So.. if it moves 3cm to the left and 5 cm to the right, you know that it takes 3cm to get it back to it's initial starting position, but it's been displaced.
So the final displacement is 2cm to the right.

antiseptic1488 [7]3 years ago

3 0

Answer: 2 cm to the right

Explanation:

The displacement is the shortest distance between initial and final position.

It can be calculated by subtracting the initial position from final position.

Let the sign be positive for direction towards right and negative for direction towards left.

The initial position is 0.

Firstly, object moves towards left = -3 cm

Then, the object moves towards right = 5 cm

Thus, the final position of the object is = -3 cm + 5 cm = +2 cm

Displacement = final position - initial position = + 2 cm - 0 = + 2 cm

Thus, the final displacement is 2 cm to right.

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A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 1

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Answer:

202.8m

Explanation:

Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.

First calculate the total time travelled by using the second equation of motion

h = Ut + 1/2gt^2

Let assume that u = 0

And h = 3.5

Substitute all the parameters into the formula

3.5 = 1/2 × 9.8 × t^2

3.5 = 4.9t^2

t^2 = 3.5/4.9

t^2 = 0.7

t = 0.845s

To know how far the cannonball travel, let's use the equation

S = UT + 1/2at^2

But acceleration a = 0

T = 2t

T = 1.69s

S = 120 × 1.69

S = 202.834 m

Therefore, the distance travelled by the cannon ball is approximately 202.8m.

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2 years ago

A 25 kg mass is accelerated by a force at a rate of 5 m/s2. What is the magnitude of the force that accelerates the man?

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3 years ago

The magnitude of a force vector is 89.6 newtons (N). The x component of this vector is directed along the +x axis and has a magn

insens350 [35]

Answer:

(a) θ = 33.86°

(b) Ay = 49.92 N

Explanation:

You have that the magnitude of a vector is A = 89.6 N

The x component of such a vector is Ax = 74.4 N

(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:

$A_x=Acos\theta$ (1)

Ax: x component of vector A

A: magnitude of vector A

θ: angle between vector A and the x axis

You solve the equation (1) for θ, by using the inverse of cosine function:

$\theta=cos^{-1}(\frac{A_x}{A})=cos{-1}(\frac{74.4N}{89.6N})\\\\\theta=33.86\°$

the angle between the A vector and the x axis is 33.86°

(b) The y component of the vector is given by:

$A_y=Asin\theta\\\\A_y=(89.6N)sin(33.86\°)=49.92N$

the y comonent of the vecor is Ay = 49.92 N

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3 years ago