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guajiro [1.7K]
3 years ago
15

A hockey puck slides off the edge of a horizontal platform with an initial velocity of 28.0 m/shorizontally in a city where the

acceleration due to gravity is 9.81 m/s 2. The puck experiences no significant air resistance as it falls. The height of the platform above the ground is 2.00 m. What is the angle below the horizontal of the velocity of the puck just before it hits the ground
Physics
1 answer:
kozerog [31]3 years ago
7 0

Answer:

θ = 12.60°

Explanation:

In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:

\theta=tan^{-1}(\frac{v_y}{v_x})       (1)

θ: angle below he horizontal

vy: y component of the velocity just after the puck hits the ground

vx: x component of the velocity

The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:

v_x=v_o

vo: initial velocity = 28.0 m/s

The y component is calculated with the following equation:

v_y^2=v_{oy}^2+2gy         (2)

voy: vertical component of the initial velocity = 0m/s

g: gravitational acceleration = 9.8 m/s^2

y: height

You solve the equation (2) for vy and replace the values of the parameters:

v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}

Finally, you use the equation (1) to find the angle:

\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°

The angle below the horizontal is 12.60°

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a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s

a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49

Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.

To learn more about oscillations Please click on the given link:

brainly.com/question/26146375

#SPJ4

This is incomplete question Complete Question is:

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?

4 0
1 year ago
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