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3 years ago

The process called _____ destroys old oceanic crust at subduction zones.

Physics

1 answer:

ZanzabumX [31]3 years ago

5 0

The process is called subduction

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An "emergency blow" is a procedure used by military submarines to quickly rise to the surface in case of trouble. It involves us

sleet_krkn [62]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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4 years ago

A softball with a mass of 0.11 kg goes moves at a speed of 12 m/s, then the ball is hit by a bat and rebounds in the opposite di

dimulka [17.4K]

=1244 and that's from my math book

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3 years ago

A person travels by car from one city to another with differem constan1 speeds between pairs of cities. She drives for 30.0 min

Montano1993 [528]

Answer:

a.52.9 km/h

b.90 km

Explanation:

We are given that

$v_1=89km/h$

$t_1=30min$

$v_2=100km/h$

$t_2=12min$

$v_3=40km/h$

$t_3=45 min$

Time spend on eating lunch and buying ga=15 min.

a.Total time=30+12+45+15=102 minute= $\frac{102}{60}=1.7 hour$

1 hour=60 minutes

Distance= $speed\times time$

$d_1=v_1\times t_1=80\times\frac{30}{60}=40km$

$d_2=100\times \frac{12}{60}=20 km$

$d_3=40\times \frac{45}{60}=30 km$

Total distance= $d_1+d_2+d_3=40+20+30=90km$

Average speed= $\frac{total\;speed}{total\;time}$

Using the formula

Average speed= $\frac{90}{1.7}=52.9Km/h$

b.Total distance between the initial and final city lies along the route=90 km

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3 years ago

If two cars are 5 m apart, and one car has a mass of 2,565 kg and the other

alekssr [168]

Answer:

Explanation:

just did

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3 years ago

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

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)$$

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

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