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dedylja [7]
3 years ago
15

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to c

ome near, the clown turns due east and runs 19.8 m to exit the arena. What is the magnitude of the clown's displacement?

Physics
1 answer:
IRINA_888 [86]3 years ago
5 0

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

<u>Explanation: </u>

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

In first step, the clown runs 7.7 m in north direction, so the image will be  as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}

\text { Distance covered by the clown in east direction }=(6.4 \times \cos 49.9)+19.8=23.9 \mathrm{m}

Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.

\text { Distance covered by the clown in north direction }=6.4 \sin 49.9+7.7=12.6 \mathrm{m}

Thus, the total displacement covered by the clown is

\text { Total displacement }=\sqrt{(23.9)^{2}+(12.6)^{2}}=\sqrt{571.21+158.76}=\sqrt{729.97}=27 \mathrm{m}

Thus, the total displacement by the clown is 27 m.

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Becomes older

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8 0
3 years ago
hydraulic lift is to be used to lift a 2100-kg weight by putting a weight of 25 kg on a piston with a diameter of 10 cm. Determi
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840 cm

Explanation:

Note: A hydraulic press operate based on pascal's principle.

From pascal's principle

W₁/d₁ = W₂/d₂...................... Equation 1

Where W₁ and W₂ are the first and second weight, and d₁ and d₂ are the  first and second diameter of the piston.

make d₁ the subject of the equation

d₁ = W₁×d₂/W₂................ Equation 2

Given: W₁ = 2100 kg, W₂ = 25 kg, d₂ = 10 cm = 0.1 m.

Substitute these values into equation 2

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The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.4cm apart with a 28kV potential d
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Explanation:

a) The electric force is

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     V = E d

     E = V / d

Let's replace

    F = e V / d

Let's calculate

    F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²

    F = 3.2 10⁻¹⁰ N

b) For this part we can use kinematics

          v² = v₀ + 2 a d

          v = √ 2 ad

Acceleration can be found with Newton's second law

        e V / d = m a

        a = e / m V / d

        a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²

        a = 3,516 10⁻¹⁷ m / s²

Let's calculate the speed

       v = √ (2 3,516 10¹⁷ 1.4 10⁻²)

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Let's replace

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Let's reduce the density to SI units

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