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kozerog [31]
3 years ago
8

B. Projectile on cliff (range)

Physics
1 answer:
dimulka [17.4K]3 years ago
5 0

Answer:

x = 41.28 m

Explanation:

This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.

Let's start by using trigonometry to find the initial velocity

         cos 25 = v₀ₓ / v₀

         sin 25 = Iv_{oy} / v₀

         v₀ₓ = v₀ cos 25

         v_{oy} = v₀ sin 25

         v₀ₓ = 22 cos 25 = 19.94 m / s

         v_{oy} = 22 sin 25 = 0.0192 m / s

let's use movement on the vertical axis

         y = y₀ + v_{oy} t - ½ g t²

     

when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m

         0 = 21 + 0.0192 t - ½ 9.81 t²

         4.905 t² - 0.0192 t - 21 = 0

         t² - 0.003914 t - 4.2813 =0

we solve the quadratic equation

        t = \frac{ 0.003914\  \pm \sqrt{0.003914^2 + 4 \ 4.2813 }   }{2}

        t = \frac{0.003914 \ \pm 4.13828}{2}

        t₁ = 2.07 s

        t₂ = -2.067 s

since time must be a positive scalar quantity, the correct result is

        t = 2.07 s

now we can look up the distance traveled

         x = v₀ₓ t

         x = 19.94  2.07

         x = 41.28 m

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Maksim231197 [3]

Answer:

Tt = 70 + 135e^-0.031t

13 minutes

Explanation:

Given that :

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Temperature after 2.5 minutes = 195°

Temperature of room, Ts= 70

Using the relation :

Tt = Ts + Ce^-kt

Temperature after time, t

When freshly poured, t = 0

205 = 70 + Ce^-0k

205 = 70 + C

C = 205 - 70 = 135°

T after 2.5 minutes to find proportionality constant, k

Tt = Ts + Ce^-kt

195 = 70 + 135e^-2.5k

125 = 135e^-2.5k

125 / 135 = e^-2.5k

0.9259 = e^-2.5k

Take In of both sides :

−0.076989 = - 2.5k

k = −0.076989 / - 2.5

k = 0.031

Equation becomes :

Tt = 70 + 135e^-0.031t

t when Tt = 160

160 = 70 + 135e^-0.031k

90 = 135e^-0.031t

90/135 = e^-0.031t

0.6667 = e^-0.031t

In(0.6667) = - 0.031t

−0.405465 = - 0.031t

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8 0
2 years ago
the roque requried to turn the crank on an ice cream maker is 4.50 N.m how much work does it take to turn the crank through 300
Alexus [3.1K]

Answer:

the work required to turn the crank at the given revolutions is 8,483.4 J

Explanation:

Given;

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W = 28.278 J

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4 0
3 years ago
Consider the classic problem with holiday lights, one little bulb goes out and the whole string goes out. First consider a strin
Flura [38]

Answer:

<em>a. 0.33 Amp</em>

<em>b. 2.4 Volt</em>

<em>c. 0</em>

<em>d. 2.45 Amp</em>

<em>e. Infinite</em>

<em>f. Series is safer</em>

Explanation:

<u>Series Connection of Resistors</u>

When two or more resistors are connected in series, the current through each one of them is the same, and the voltage divides depending on the particular value of each resistance. If all the resistances are equal, then the voltage is equally divided.

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Since all the bulbs are connected in series the current is the same for all of them.

b. The voltage is equally divided, so each bulb has  120/50= 2.4 V

c. If one of the bulbs burns out and its resistance becomes infinite, then the series circuit is open and no current flows through it, neither through the rest of the bulbs. The typical case of the whole string going out.

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e. If the bulbs were connected in parallel, all of them would have the same voltage, and the total current will be equally divided among them. In that case, a short circuit in one of the bulbs will cause a parallel short, theoretically producing an infinite current and making the short circuit protection blow up.

f. The condition described above makes the strings be made of series-connected bulbs which is safer than the parallel circuit. If a single bulb shorts, the entire string goes out in a series connection, but the breaker would trigger disconnection of the house circuit if it's a parallel connection. That is why we must deal with unusable strings instead of burning cables.

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Hence:

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3 years ago
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VashaNatasha [74]

Answer:

no

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