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Hoochie [10]
3 years ago
6

Just wondering, what do you guys think the 5th dimension is? I've always believed it to be light.

Physics
1 answer:
zepelin [54]3 years ago
8 0

Answer: In the 5th dimension, they who claim to know, say that there is only one time, including the past and the future.

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Interactive Solution 6.39 presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a
mariarad [96]

(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

v_y^2 - u_y^2 = 2as

where, taking upward as positive direction:

v_y is the final vertical velocity

u_y = 0 is the initial vertical velocity (zero because the pebble is launched horizontally)

a=g=-9.8 m/s^2 is the acceleration of gravity

s = -25.0 m is the displacement

Solving for vy,

v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

v_x = 20.0 m/s

So, the final speed of the pebble as it strikes the ground is

v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

u_y = 20.0 m/s

So we can find the final vertical velocity using the same suvat equation as before:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

u_y = -20.0 m/s

Using again the same suvat equation:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

We notice that the final value of the speed is always the same in all the three parts, so it does not depend on the direction of launching. This is due to the law of conservation of energy: in fact, the initial mechanical energy of the pebble (kinetic+potential) is the same in all three cases (because the height h does not change, and the speed v does not change either), and the kinetic energy gained during the fall is also the same (since the pebble falls the same distance in all 3 cases), therefore the final speed must also be the same.

7 0
3 years ago
A little help please?
Tpy6a [65]
A. Made of the marble.

the mass remains constant when you drop the marble but the rest of the variables change as the marble is dropped, therefore, the only constant variable is its mass.
7 0
3 years ago
Can someone explain how the weight of the block is 10.26N, with reference to an appropriate law of motion?
spayn [35]

this process is called parellelogram method of resolving vectors.

5 0
3 years ago
6
hodyreva [135]

Answer:....

Explanation:.

6 0
3 years ago
A 7 kg object attached to a horizontal string moves with constant speed in a circle on a frictionless horizontal surface. The ki
Marizza181 [45]

Answer:

Explanation:

Given

mass of object m=7 kg

kinetic Energy k=36\ J

Tension in string T=326\ N

mass is moving in a horizontal circle so tension is providing the centripetal acceleration

therefore T=\frac{mv^2}{r}----1

where r=radius of circle

kinetic energy of particle k=\frac{1}{2}mv^2----2

divide 1 and 2 we get

\frac{T}{k}=\frac{\frac{mv^2}{r}}{\frac{1}{2}mv^2}

\frac{T}{k}=\frac{2}{r}

r=\frac{2k}{T}=\frac{2\times 36}{326}

r=0.2208\ m

   

8 0
3 years ago
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