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xxMikexx [17]
1 year ago
6

A punted football is observed to have velocity components vhorizontal = 15 m/s to the right and vvertical = 1.25 m/s directed do

wnward at a height h = 1.3 m above the ground. Use a Cartesian coordinate system with the origin located on the ground at the position the football was punted and assume it encountered no air resistance. Determine the football's initial horizontal velocity magnitude v h in terms ofw roe in. Wermeal, g, and h.
Physics
1 answer:
enot [183]1 year ago
5 0

Answer:

v₀ₓ = 15 m / s,  v_{oy} = 5.2 m / s

v = 15.87 m / s ,   θ = 19.1

Explanation:

This is a projectile launch problem. The horizontal speed that is constant throughout the entire path is worth 15 m / s, instead the vertical speed changes in value due to the acceleration of gravity, let's look for the initial vertical speed

                      Vy² =v_{oy}² - 2 g y

                      v_{oy}² = v_{y}² + 2 g y

                       v_{oy} = √ (v_{y}² + 2 gy

Let's calculate

                    v_{oy} = √ (1.25² + 2 9.8 1.3)

                    v_{oy} = √ (27.04)

                    v_{oy} = 5.2 m / s

 The initial speed can be calculated by the initial speed

                   v = √ v₀ₓ² + v_{oy}²

                   v = RA (15² + 5.2²)

                   v = 15.87 m / s

We look for the angle with trigonometry

                 tan θ = voy / vox

                 θ = tan⁻¹ I'm going / vox

                θ = tan⁻¹ 5.2 / 15

                θ = 19.1

The answer is

              v₀ₓ = 15 m / s

              v_{oy} = 5.2 m / s

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Answer:

The answers and workings is in the Explanation section

Explanation:

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I = V/R =6/3 =2 Amps of current

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Since the second lamp is connected in series with the first, the total resistance will R₂ = 3 + 3 = 6Ω

2 resistance in series  R₂ =  6Ω

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The current is 1 Amps

Answer =  6Ω  and 1 Amps

<em>If a third identical lamp is connected in series, the total resistance is now _________Ω.  The current through all three lamps in series is now _________ Amps.  </em>

Since the third lamp is connected in series with the first and second, the total resistance will R₃ = 3 + 3 + 3 = 9Ω

total resistance of the 3 lamps R₃ =  9Ω

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Answer =  9Ω  and 0.67 Amps

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The formula for power P = I*V =120*3 = 360 Watts

power P  = 360 Watts

Answer = 360 Watts

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From P =I*V, make I the subject of the formula, I = P/V =90/120 = 0.75

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Answer =  0.75 Amps

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Current I =P/V = 75/120 = 0.625 Amps.

Answer = 0.625 Amps

<em>If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it? </em>

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The power rating of the bulb is 60W = 60/1000 = 0.06 KiloWatt

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4 0
2 years ago
__________ curves help lessen the effect of the force of the forward motion on your vehicle as it enters the curve.
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Answer:

Banked

Explanation:

Banked curves are formed when the inner edge is below the outer edge.

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5 0
2 years ago
The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
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Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
2 years ago
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Kamila [148]

Answer:

zero or 2π is maximum

Explanation:

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When the wave travels in the same direction

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      Xt = A [sin (kx-wt + φ₁) + sin (kx-wt + φ₂)]

We are going to develop trigonometric functions, let's call

     a = kx + wt

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We develop breasts of double angles

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to have a maximum of the sine function, the cosine of fi must be maximum

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the possible values ​​of each phase are

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6 0
2 years ago
How far must 5N force pull a 50g toy car if 30J of energy are transferred?​
Alborosie

Answer: 6 m

Explanation:

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7 0
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