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3 years ago

11

Inexperienced physics teachers often demonstrate the use of the electroscope by touching it with a charged glass rod at a single

point. More experienced teachers typically drag the length of the rod across the top of the electroscope to increase the desired effect. Why does this help

1 answer:

jek_recluse [69]3 years ago

5 0

Answer:

This is because The glass rod of the electroscope is an insulator therefore only charge transferred to the ball is at the point of contact on the rod. Thus, When the charge rod is dragged across the top of the electroscope, by the experienced teacher the more charge is transferred to electroscope thereby producing a greater effect

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During which strokes does the piston move downward in a four-stroke internal combustion engine?

yuradex [85]

I believe that it is the third stroke

3 0

3 years ago

Read 2 more answers

A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh

irina1246 [14]

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

$K=\frac{1}{2}mv^2$

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

$U=mgh$

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

$a=\frac{v-u}{t}$

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

$u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s$

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

$\Delta v = 0 - (+9.8 m/s)=-9.8 m/s$

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

$a=\frac{v-u}{t}$

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

$v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s$

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - 0=-9.8 m/s$

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

$\Delta v = v -u$

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s$

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

$F=mg$

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

$mg = ma$

which means that the acceleration is

$a= g = -9.8 m/s^2$

and the negative sign means it points downward.

7 0

3 years ago

Light of wavelength 500 nm illuminates a round 0.50-mm diameter hole. A screen is placed 6.0 m behind the slit.. Find the width

Paul [167]

Answer:

15mm

Explanation:

We know that for circular holes first dark spot is given by

sin စ = 1.22 λ/D

Also we know that at the same time

tan စ = r/L

So

r = L tanစ = 6 x tan( arcsin(1.22x 500 x10^9/0.50 x 10^ 3))

= 0.0073 m = 7.3 mm

However since the size is twice that so 14.6 mm which is approx 15mm

5 0

3 years ago

At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field

vesna_86 [32]

Answer:

$9.7\times 10^{-5} T$

Explanation:

Velocity = $5\times 10^4i-2\times 10^4j$

r= $0.03i+0.05j$

r= $\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058$

v= $\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}$

We know that

$B=\frac{mv}{qr}$

Where q= $1.6\times 10^{-19} C$

Mass of proton= $1.67\times 10^{-27} kg$

Using the formula

$B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}$

$B=9.7\times 10^{-5} T$

3 0

3 years ago

The current theory of the structure of the

irina1246 [14]

Answer:

pt 1: $m=1.66698*10^{21} kg$

Pt 2: $KE=1212.23531 J$

Explanation:

Information Given: (p = density)

l = 5200km d = 35km p = 2700kg/ $m^{2}$

Part 1: Mass

Find volume

$V=(l)^2(d)$
$V=(4.2*10^6)^2(35*10^3)$
$V=61.74*10^{16}$

Find Mass

$m=Vp$
$m=(61.74*10^{16})(2700)$
$m=1.66698*10^{21}$

Part 2: Kinetic Energy

$v=\frac{3.8cm}{yr}*\frac{m}{100cm}*\frac{yr}{365d}*\frac{d}{24hr}*\frac{hr}{3600s}$
$v=1.20497*10^{-9}$

$KE=\frac{1}{2}mv^2$

$KE=\frac{1}{2} (1.66698*10^{21})(1.20497*10^{-9})^2$

$KE=1212.23531 J$

Part 3: Jogger Speed

set up, because I don't have the mass :(

Information given:

$KE_{jogger}$

$KE=\frac{1}{2}mv^2$
$v_{jogger} =\sqrt{\frac{2KE}{m_{jogger} } }$

Input the values

Hope it helps :)

6 0

3 years ago