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nignag [31]
1 year ago
10

A ball is projected at an angle of 30° above the horizontal with a speed of 35 m/s. What will be its approximate horizontal rang

e across a level surface?
Physics
1 answer:
BaLLatris [955]1 year ago
3 0

37.5 m  will be its approximate horizontal range across a level surface.

we have given in the data

u (initial velocity) = 35 m/s

projected  angle = 30°

Horizontal range =?

range of horizontal distance=\frac{u^2 Sin 2 \theta}g

PUTTING all the values in the equation

horizontal range= \frac{35\cdot35\cdot sin60}{9.8}

by solving, we got

horizontal range=37.5 m

  • The horizontal distance traversed is the projectile's range,The thing moves in one direction until it hits the ground.
  • When an item is hurled at an angle of 90°, only vertical motion occurs. As a result, the projectile's range will be zero.
  • horizontal range variates with angle, initial velocity, and by the value of g.

To know more about  horizontal range visit : brainly.com/question/23827445

#SPJ1

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Explanation:

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We can calculate the potential difference between the two points by using the law of conservation of energy:

\Delta K+ \Delta U=0\\\Delta K + q\Delta V=0

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\Delta K=+1.6\cdot 10^{-18} J is the change in kinetic energy of the particle

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\Delta V =V_b-V_a is the potential difference

Re-arranging the equation, we can find the value of the potential difference:

\Delta V=V_b-V_a = -\frac{\Delta K}{q}=-\frac{1.6\cdot 10^{-18} J}{3.2\cdot 10^{-19} C}=-5 V

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Which description best defines blastocyst?
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A pillow with mass of 0.3 kg sits on a bed with a coefficient of static friction of 0.6. What is the maximum force of static fri
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The maximum force of static friction is the product of normal force (P) and the coefficient of static friction (c). In a flat surface, normal force is equal to the weight (W) of the body. 
 
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Solving for the static friction force (F), 
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2 years ago
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The peak emf generated by the coil is 2.67 V

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V₀ = 940 x (π x 0.24²/4) x 5 x 10⁻⁵ x (2π/0.005)

V₀ = 940 x 0.04524 x  5 x 10⁻⁵ x 1256.8

V₀ = 2.6723 V = 2.67 V

The peak emf generated by the coil is 2.67 V

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