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2 years ago

By what means can the internal energy of a closed system increase?.

Physics

2 answers:

kirill115 [55]2 years ago

7 0

Answer:

when work is done on the system or heat comes into the system

Verizon [17]2 years ago

4 0

Answer:

if it becomes an open system or heat is added

Explanation:

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Q 28.7: A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 T. What is the maximum Hall

Zinaida [17]

Answer:the maximum Hall voltage across the strip= 0.00168 V.

Explanation:

The Hall Voltage is calculated using

Vh= B x v x w

Where

B is the magnitude of the magnetic field, 5.6 T

v is the speed/ velocity of the strip, = 25 cm/s to m/s becomes 25/100=0.25m/s

and w is the width of the strip= 1.2 mm to meters becomes 1.2 mm /1000= 0.0012m

Solving

Vh= 5.6T x 0.25m/s x 0.0012m

=0.00168T.m²/s

=0.00168Wb/s

=0.00168V

Therefore, the maximum Hall voltage across the strip=0.00168V

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2 years ago

How can meteorologists use the jet stream to predict the weather ?

barxatty [35]

<span>Jet streams act as an invisible director of the atmosphere and are largely responsible for changes in the weather across the globe.
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3 years ago

Which material is very strong and tough but shows very little elongation as it absorbs energy? A. spider silk B.rubber C.Kevlar®

Furkat [3]

C. Kevlar

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3 years ago

Read 2 more answers

At a meeting of physics teacher in Montana, the teachers were asked to calculate where a flour sack would land if dropped from a

Harlamova29_29 [7]

At a distance of 469.2 m from the original point below the airplane.

Explanation:

First of all, we have to calculate the time it takes for the sack to reach the ground.

To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

s = 300 m is the vertical displacement

u = 0 is the initial vertical velocity

t is the time

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find the it takes for the sack to reach the ground:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(300)}{9.8}}=7.82 s$

Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:

$v_x = 60 m/s$