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3 years ago

How might you describe the mathematical procedure of finding the displacement when an object travels in two opposite directions?

1 answer:

4 0

Displacement is a vector quantity. So, you incorporate the vector calculations when you try to determine the resultant vector. This is the shortest path from the starting point to the endpoint. If they are moving on one axis only, you use sign conventions. For motions moving to the left, use the negative sign. If it's moving to the right, then use the positive sign. Now, it the object moves 2 km to the left, and 2 km also to the right, the displacement is zero.

Displacement = 2 km - 2km = 0

Generally, the equation is:
<span>Displacement = Distance of motion to the right - Distance of motion to the left</span>

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Two pulses move in opposite directions on a string and are identical in shape except that one has positive displacements of the

Leni [432]

A single disturbance that travels via a transmission medium is referred to as a pulse. This medium might be formed of stuff or a vacuum, and it might be endlessly large or finite in size.

Consider two pulses that are identical in shape and proceed in opposite directions along a string, with the exception that one has positive displacements of the string's elements while the other has negative displacements.

On the string, the two pulses blend together completely.

The pulses completely balance one another out in terms of removing string elements from equilibrium, yet the string still moves. Shortly after the string is once again shifted, the pulses will have passed each other.

The correct option is (c)

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1 year ago

A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit

Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

$y = \frac{1}{2} at^2+v_0t+y_0$

Where,

a = acceleration

t = time

$v_0$ = Initial velocity

$y_0$ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

$v = \frac{x}{t}$

x = Displacement

t = time

With the data we have we can find the time as well

$v = \frac{x}{t}$

$t = \frac{x}{v}$

$t = \frac{18.6}{34}$

$t = 0.547s$

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

$y = \frac{1}{2} at^2+v_0t+y_0$

$y = \frac{1}{2} gt^2+0+0$

$y = \frac{1}{2} gt^2$

$y = \frac{1}{2} 9.8*0.547^2$

$y = 1.46m$

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

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3 years ago

in a one dimensional collision, a 4kg object and 6 kg onject have initial velocity. calculate the magnitude of impulse

Greeley [361]

in a one dimensional collision, a 4kg object with 5ms^1 and 6 kg object with 2ms^1 have initial velocity, the magnitude of impulse is 12 , 18

given,

mass 1 = 4kg

mass 2 = 6kg

velocity 1 = 5ms^1

velocity 2 = 2ms^1

impulse 1 = 4*(5-2)

= 12

Impulse 2 = 6*(5-2)

= 18

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8 0

1 year ago

A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia

Musya8 [376]

Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

When the tiger reaches the ground, y = 0:

0 = 6.00 m - 1/2 · 9.81 m/s² · t²

2 · (-6.00 m) / -9.81 m/s² = t²

t = 1.11 s

We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:

x = x0 + v0 · t

At t = 1.11 s, x = 2.00 m

x0 = 0

2.00 m = v0 · 1.11 s

2.00 m / 1.11 s = v0

v0 = 1.80 m/s

The initial speed of the tiger is 1.80 m/s

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3 years ago

Jane walked 5.00 meters on a road that inclines 13.0 degrees. How much distance did she cover horizontally?

Paladinen [302]

Any options available?

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3 years ago