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3 years ago

6

A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on a dart board. It hits at point Q

on the rim, vertically below P, 0.19 s later.
(a) What is the distance ?
(b) How far away from the dart board is the dart released?

1 answer:

Kryger [21]3 years ago

3 0

D=s(t) so it would be d=10(.19) d=.19 FOR BITH SNDWERS

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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

State the law of conservation of momentum<br>

trapecia [35]

Answer:

Law of conservation of momentum states that. For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.

Explanation:

Hope it helps

8 0

2 years ago

Read 2 more answers

A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released

GREYUIT [131]

Answer:

$6.75\mu C/m^2$

Explanation:

We are given that

Diameter,d= $1\mu m=1\time 10^{-6} m$

$1\mu m=10^{-6} m$

Radius,r= $\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m$

Density, $\rho=900kg/m^3$

Total number of electrons,n=39

Charge on electron = $1.6\times 10^{-19} C$

Total charge= $q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C$

Distance,s=2mm= $2\times 10^{-3} m$

Mass = $density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg$

Initial velocity,u=0

Final speed,v=4.5 m/s

$v^2-u^2=2as$

$(4.5)^2-0=2a(2\times 10^{-3})$

$20.25=4a\times 10^{-3}$

$a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2$

Force,F=ma

$qE=ma$

$q(\frac{\sigma}{2\epsilon_0})=ma$

$\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}$

$\epsilon_0=8.85\times 10^{-12}$

$\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2$

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3 years ago

A black hole in the universe is a An empty region in space b- A massive collapsed star c- A moon that is always turned to its da

marusya05 [52]

Answer:

b. a massive collapsed star

Explanation:

A black hole in the universe is nothing but a massive collapsed star. When the size of the star crosses a particular limit it cannot holds its mass and it collapses under it own self. This is called supernova. A black hole is actually a region in space where gravity is so strong that even light cannot escape through it. Gravity so strong because the matter has been pressed into a tiny space. hence option b is correct

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3 years ago

Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal po

zvonat [6]

Answer:

$0.293I_0$

Explanation:

When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.

Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:

$I_1=\frac{I_0}{2}$

Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:

$I_2=I_1 cos^2 \theta$

where

$\theta$ is the angle between the axes of the two polarizer

Here we have

$\theta=40^{\circ}$

So the intensity after the 2nd polarizer is

$I_2=I_1 (cos 40^{\circ})^2=0.587I_1$

And substituting the expression for I1, we find:

$I_2=0.587 (\frac{I_0}{2})=0.293I_0$

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2 years ago