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nata0808 [166]
3 years ago
10

A 8.1 kg object initially at rest is pushed down a 15.0 m tall hill. What is the speed of the object at the bottom of the hill?

Physics
1 answer:
topjm [15]3 years ago
8 0

Answer:

The velocity of the object at the bottom is, v = 17.15 m/s

Explanation:

Given data,

The initial velocity of the object, u = 0

The height of the hill, h = 15 m

Let 'S' be the distance of the slope of the hill and 'Ф' be the slope of the hill formed with the ground.

The acceleration due to gravity component along the slope is given by,

                                      a = g Sin Ф

The distance of the slope since height 'h' of the hill is given,

                                       s = h / Sin Ф

Using the III equation of motion,

                                      v² = 2 as                    (∵ u = 0)

                                      v² = 2 x g Sin Ф x h / Sin Ф

                                           = 2 gh

Therefore,

                                      <em> v = √(2gh)</em>

Substituting the given values,

                                       v = √(2x9.8x15)

                                          = 17.15 m/s

Hence, the velocity of the object at the bottom is, v = 17.15 m/s

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mylen [45]

Answer: The potential difference between the plates = 0.4061V

Explanation:

Given that the

Electric field strength E = 155 N/C

Distance d = 0.00262 m

From the definition of electric field strength, is the ratio of potential difference V to the distance between the plates. That is

E = V/d

Substitute E and d into the above formula

155 = V/0.00262

Cross multiply

V = 155 × 0.00262

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The potential difference between the plates is 0.4061 V

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A go-cart is traveling at a rate of 25 m/sec for 20 seconds. How far will the go cart travel?
notsponge [240]

Answer:

Distance travel by go-cart = 500 meter

Explanation:

Given:

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Time travel = 20 seconds

Find:

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Computation:

Distance = Speed x time

Distance travel by go-cart = Speed of go cart x Time travel

Distance travel by go-cart = 25 x 20

Distance travel by go-cart = 500 meter

4 0
3 years ago
From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the
Ivahew [28]

Answer:

The launching point is at a distance D = 962.2m and H = 39.2m

Explanation:

It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

           x = Vox t

           t = x / vox

           t = 7.1 / 340

           t = 2.09 10-2 s

In this same time the height of the window fell

           Y = Voy t - ½ g t²

Let's calculate the initial vertical speed, this speed is in the window

           Voy = (Y + ½ g t²) / t

           Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

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We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

             Vy = Voy - gt₂

             Vy = 0 -g t₂

             t₂ = Vy / g

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This is the time it also takes to travel the horizontal and vertical distance

            X = Vox t₂

            D = 340 2.83

            D = 962.2 m

           

            Y = Voy₂– ½ g t₂²

            Y = 0 - ½ g t2

            H = Y = - ½ 9.8 2.83 2

            H = 39.2 m

The launching point is at a distance D = 962.2m and H = 39.2m

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bixtya [17]

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