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fredd [130]
1 year ago
13

A golfer strikes her tee shot with the following velocities, vx = 106 m/s & vy = 66.2 m/s. a. What is the velocity of flight

? b. The time of flight? c. The horizontal distance traveled?​
Physics
1 answer:
lara [203]1 year ago
3 0

Answer:

a) V ≈ 125 m/s;  b) Δt = 13.24 s;  c) ΔS ≈ 1450 m

Explanation:

a) We have just to calculate the vector resultant.

V² = 106² + 66.2²

V² = 15618.44

V ≈ 125 m/s

b) The time of flight is equal to the time to reach the maximum height summed to the time to reach the land.

In vertical:

V = V₀ + a * t

V = 66.2 - g * t

0 = 66.2 - 9.8 * t

t ≈ 6.76 s

So: Δt = 13.24 s

c) In horizontal:

V = ΔS / Δt

106 = ΔS / 13.52  ⇒  ΔS = 106 * 13.52

ΔS = 106 * 13.52

ΔS = 1433,12

ΔS ≈ 1450 m

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If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.

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As T = 2\pi\sqrt{\frac{m}{k}}, and the 2 springs have the same mass

2\pi\sqrt{\frac{m}{k_B}} = 2\pi\sqrt{\frac{m}{k_A}}

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In a hydraulic lift, if the pressure exerted on the liquid by one piston is increased by 100 N/m2 , how much additional weight c
shepuryov [24]

Answer:

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Explanation:

By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:

Hydraulic Lift - Before change

P = \frac{F}{A}

Hydraulic Lift - After change

P + \Delta P = \frac{F + \Delta F}{A}

Where:

P - Hydrostatic pressure, measured in pascals.

\Delta P - Change in hydrostatic pressure, measured in pascals.

A - Cross sectional area of the hydraulic lift, measured in square meters.

F - Hydrostatic force, measured in newtons.

\Delta F - Change in hydrostatic force, measured in newtons.

The additional weight is obtained after some algebraic handling and the replacing of all inputs:

\frac{F}{A} + \Delta P = \frac{F}{A} + \frac{\Delta F}{A}

\Delta P = \frac{\Delta F}{A}

\Delta F = A\cdot \Delta P

Given that \Delta P = 100\,Pa and A = 25\,m^{2}, the additional weight is:

\Delta F = (25\,m^{2})\cdot (100\,Pa)

\Delta F = 2500\,N

The additional mass needed for the additional weight is:

\Delta m = \frac{\Delta F}{g}

Where:

\Delta F - Additional weight, measured in newtons.

\Delta m - Additional mass, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

If \Delta F = 2500\,N and g = 9.807\,\frac{m}{s^{2}}, then:

\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }

\Delta m = 254.92\,kg

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

3 0
3 years ago
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