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Ilia_Sergeevich [38]
3 years ago
8

A golf ball is hit with an initial velocity of 15 meters per second at an angle of 35 degrees above horizontal. What is the vert

ical component of the golf ball’s initial velocity ?
Physics
1 answer:
polet [3.4K]3 years ago
6 0

Answer:

8.6 m/s

Explanation:

The components of the initial velocity of the ball can be found by using the equations:

v_x = v cos \theta\\v_y = v sin \theta

where

v_x is the horizontal component

v_y is the vertical component

v is the magnitude of the velocity

\theta is the angle above the horizontal

Substituting:

v = 15 m/s

\theta=35^{\circ}

We find:

v_x = (15)(cos 35^{\circ})=12.3 m/s\\v_y = (15)(sin 35^{\circ})=8.6 m/s

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jek_recluse [69]

Given,

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\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}

Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.

On substituting the known values,

\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}

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How much heat does it take to raise a<br> cup of water (2.34 x 10-4 m3) from<br> 15.0 °C to 75.0 °C?
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Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

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