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3 years ago

Two glass rods are rubbed together. One of the glass rods are brought near (but

Physics

1 answer:

ruslelena [56]3 years ago

5 0

Answer:

The correct answer is - do nothing.

Explanation:

If two same or identical objects, in this instance two glass rods, were rubbed together they neither attract nor repel each other as there is no transfer of electrons from one material to the other material as there is no difference in their charges.

Only two different objects are able to transfer electrons from one material to the other material if rubbed together, to produce the conditions for static electricity to be observed. Therefore there would be no charge on the glass rod so there would be no charge to balls to repel or attract.

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According to ____________ , the randomness of the universe is constantly increasing.

Darina [25.2K]

Answer:

According to " The second law of thermodynamics", the randomness of the universe is constantly increasing?

Explanation:

So answer option C. Have a great summer.

8 0

3 years ago

Roger the turtle moved 450 m in 2 hours the first day and the next day he moved in a straight line at a steady speed of 15 cm/s

Sonja [21]

Answer:

a). Roger travel total a distance of 2070m

b). The average speed of the whole tip $V_{p}$ 0.16025 $\frac{m}{s}$

Explanation:

$V= \frac{x}{t} = \frac{450m}{2 hr} * \frac{1 hr}{3600 s} = \frac{450m}{7200 s}\\ V= \frac{450m}{7200s} =0.125 \\V= 0.0625 \frac{m}{s} \\$

a).

First trip 450m

Second trip :

$V=\frac{x}{t} \\x=V*t \\x= 15 \frac{cm}{s} *3 hr\\x= 15 (\frac{cm}{s}*\frac{1m}{100 cm}) * 3 hr (\frac{3600s}{1hr})\\ x=0.15 \frac{m}{s} * 3600s\\x= 1620 m$

So final distance is

450m+1620m=2070m

b).

$V_{p} = \frac{0.15 \frac{m}{s}+0.0625 \frac{m}{s} }{2} =0.10625 \frac{m}{s}$

The average speed is $V_{p}$ 0.10625 $\frac{m}{s}$

8 0

3 years ago

A commercial jet liner takes off with an average acceleration of 3 g. How long does it take to reach the end of its runway which

ANEK [815]

Answer:

The time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.

Explanation:

Given;

average acceleration of the commercial Jet liner, a = 3g = 3 x 9.8 m/s² = 29.4 m/s²

distance traveled by the commercial Jet liner, s = 1542 m

The time taken for the commercial Jet liner to reach the end of its runway is calculated as follows;

s = ut + ¹/₂at²

where;

u is the initial velocity of the commercial Jet liner = 0

s = 0 + ¹/₂at²

s = ¹/₂at²

2s = at²

$t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2 \times 1524}{29.4} } \\\\t = 10.18 \ s$

Therefore, the time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.

6 0

3 years ago

The ___ of a gas is due to the force exerted on the walls of the

Furkat [3]

Answer:

its molecules

Explanation:

7 0

3 years ago

Falculate the speed of a wave with frequency of 10KHz and wave length of 2m

aleksklad [387]

Answer:20000m/s

Explanation:

Frequency=10khz

Frequency=10x1000

Frequency=10000hz

Wavelength=2m

Speed= frequency x wavelength

Speed=10000 x 2

Speed=20000

speed=20000m/s

8 0

4 years ago