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Arte-miy333 [17]

3 years ago

8

Ways to increase friction??

1 answer:

natima [27]3 years ago

8 0

Create a “rougher” or more adhesive point of contact,Press the two surfaces together harder., etc.

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Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

aev [14]

Answer:

Explanation:

Given

charge of first body $q_1=2.5\ mu C$

charge of second body $q_2=-3.5\ mu C$

Particle 1 is at origin and particle 2 is at $x=0.6\ m$

third Particle which charge +q must be placed left of $2.5\mu C$ because it will repel the q charge while $-3.5\mu C$ will attract it

suppose it is placed at a distance of x m

$F_{1q}=\frac{kq(2.5)}{x^2}$

$F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}$

$F_{1q}+F_{2q}=0$

$\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0$

$\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}$

$\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}$

$0.6+x=1.1832x$

$x=3.27\ m$

5 0

4 years ago

Why couldnt mendeleev organize the entire table during his research

NARA [144]

Cause he left out the noble gases out of the periodic table for one good reason, 1: He did not know them

6 0

3 years ago

Read 2 more answers

Reply quick pls

Sholpan [36]

The ball was moving for 0.8seconds
This is because 4/5 x 1 = 0.8

Hope this helps and good luckkkk :)

4 0

3 years ago

A 60cm long string of an ordinary guitar is tuned to produce the note a4 (frequency 440hz) when vibrating in its fundamental mod

pishuonlain [190]

Answer:

0.78 m

Explanation:

The relationship between wavelength and frequency of a wave is given by

$v=f \lambda$

where

v is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

For the sound wave in this problem, we have

$f=440 Hz$ is the frequency

v = 344 m/s is the speed of sound in air

Substituting into the equation and re-arranging it, we find the wavelength:

$\lambda=\frac{v}{f}=\frac{344 m/s}{440 Hz}=0.78 m$

8 0

3 years ago

Read 2 more answers

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

7 0

3 years ago