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Mandarinka [93]
3 years ago
6

5. A family of ducks is swimming in a pond at a speed of 3 m/s when a gust of wind hits them. By the time they reach the other s

ide of the lack, the ducks are moving at a speed of 7 m/s. How long did it take for them to cross the lake if they accererated at a rate of 1 m/s2
Physics
1 answer:
ella [17]3 years ago
6 0

Answer:

The time taken by the duck to cross the lake is, t= 4 s

Explanation:

Given data,

The initial speed of the ducks, u = 3 m/s

The final speed of the ducks, v = 7 m/s

The acceleration of the duck, a = 1 m/s²

The formula for the acceleration is,

                               a = (v - u) / t

∴                               t = (v - u) / a

Substituting the given values in the above equation,

                                t = (7 - 3) / 1

                                  = 4 s

Hence, the time taken by the duck to cross the lake is, t= 4 s

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SIZIF [17.4K]

The solution to the questions are given as

  • t=40.39 \mathrm{sec}
  • \varepsilon &=(0.12v)e^{0.057t}
  • the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}

\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$

\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}

\varepsilon &=(0.12v)e^{0.057t}

(b) Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$

\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}

c)

In conclusion, the direction of the induced current will be Counterclockwise.

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2 years ago
A 0.540-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 56.0 g/s. If the sand lands in the bu
romanna [79]

Answer:

a) 12.8212 N

b) 12.642 N

Explanation:

Mass of bucket = m = 0.54 kg

Rate of filling with sand  = 56.0 g/ sec = 0.056 kg/s

Speed of sand = 3.2 m/s

g= 9.8 m/sec2

<u>Condition (a);</u>

Mass of sand = Ms = 0.75 kg

So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg

== > total weight = 1.29 × 9.8 = 12.642 N

Now impact of sand = rate of filling × velocity = 0.056 × 3.2 =  0.1792 kg. m /sec2=0.1792 N

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i-e

scale reading = 12.642 N+0.1792 N = 12.8212 N

<u>Codition (b);</u>

bucket mass + sand mass = 0.54 +0.75=1.29 kg

==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)

6 0
3 years ago
8. How is the crystal size different for extrusive and intrusive igneous rocks?
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<span>Igneous rocks which form by the crystallization of magma at a depth within the Earth are called intrusive rocks. Intrusive rocks are characterized by large crystal sizes, i.e., their visual appearance shows individual crystals interlocked together to form the rock mass.  hope that helped</span>
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Hold the paper up to a mirror so
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Answer:

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