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3 years ago

Where must an object be placed to form an image 30.0 cm from a diverging lens with a focal length of 43.0 cm?

1 answer:

6 0

Using lens equation;

1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)

Substituting;

1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm

Therefore, the object should be place 99.23 cm from the lens.

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Two people on ice skates push against each other. The person on the left is 50kg, the person on the right is 60kg. If the person

Zarrin [17]

I got -3.6 m/s but I had to do conservation of momentum for this question. Which involves Newtons third law but with simply that law I do not know how to complete this question. If you would like me to post my work I will though! Sorry

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3 years ago

Two long, parallel wires carry currents of different magnitudes. if the amount of current in each wire is doubled, what happens

Mrac [35]

The force per unit of length between two wires carrying current is
$\frac{F}{L}= \frac{\mu_0 I_1 I_2}{2 \pi r}$
where I1 and I2 are the currents in the two wires, while r is the distance between them.

We can see from the formula that the force is proportional to the product between I1 and I2: $F \sim I_1 I_2$
so, if we double both I1 and I2, we get a factor 4:
$F' \sim (2I_1 )(2I_2)=4 I_1 I_2 =4 F$
so, the force between the wires will be 4 times the original value.

4 0

3 years ago

2: Assume we have created a packet-switched internet. Using the TCP/IP protocol suite, we

yaroslaw [1]

Answer:

Shown below as 'The advantage is that the router will always choose the faster route to send the packet and each connection is independent.'

Explanation:

The advantage is that the router will always choose the faster route to send the packet and each connection is independent.

Hope this helps!

Kind regards

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2 years ago

The route followed by a hiker consists of three displacement vectors, X, Y and Z. Vector X is along a measured trail and is 1430

poizon [28]

Answer:

magnitude : 1635.43 m
Angle: 130°28'20'' north of east

Explanation:

First, we will find the Cartesian Representation of the $\vec{X}$ and $\vec{Y}$ vectors. We can do this, using the formula

$\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ its the magnitude of the vector and θ the angle. For $\vec{X}$ we have:

$\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )$

$\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )$

where the unit vector $\hat{i}$ points east, and $\hat{j}$ points north. Now, the $\vec{Y}$ will be:

$\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )$

Now, taking the sum:

$\vec{X} + \vec{Y} + \vec{Z} = 0$

This is

$\vec{Z} = - \vec{X} - \vec{Y}$

$(Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )$

$(Z_x , Z_y) = ( \ - 1062.70 m \ , \ 2200 m \ - \ 956.86 m \ )$

$(Z_x , Z_y) = ( \ - 1062.70 m \ , \ 1243.14 m\ )$

Now, for the magnitude, we just have to take its length:

$|\vec{Z}| = \sqrt{Z_x^2 + Z_y^2}$

$|\vec{Z}| = \sqrt{(- 1062.70 m)^2 + (1243.14 m)^2}$

$|\vec{Z}| = 1635.43 m$

For its angle, as the vector lays in the second quadrant, we can use:

$\theta = 180\° - arctan(\frac{1243.14 m}{ - 1062.70 m})$

$\theta = 180\° - arctan( -1.1720)$

$\theta = 180\° - 45\°31'40''$

$\theta = 130\°28'20''$

5 0

3 years ago

A point charge Q at the center of a sphere of radius R produces an electric flux of Î¦ coming out of the sphere. If the charge r

Dafna1 [17]

Remains the same

Explanation:

According to Gauss's law, the electric flux through a closed surface is proportional to the charge enclosed by the surface. So no matter how big or small we make the surface that encloses the charge, the electric flux remains the same because it only depends on the enclosed charge, not surface area.

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2 years ago