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3 years ago

Where in a horseshoe magnet is the magnetic field maximum??

Physics

2 answers:

Kazeer [188]3 years ago

8 0

In a Horsehoe magnet, maximum magnetic field is concentrated between the poles.

Hope this helps!

Fed [463]3 years ago

6 0

The magnetic field is maximum at the pole of the magnet ! you think at the end but no , not at the end but a little backward of the end ! so in short you can say, at pole !

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vekshin1

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A compound is found to have a molar mass of 598 g/mol. if 0.0358 g of the compound is dissolved in enough water to make 175 ml o

zalisa [80]

π=iMRT

Where, π is Osmotic pressure,
i=1 for non-electrolytes,
M is molar concentration of dissolved species (units of mol/L)
R is the ideal gas constant = 0.08206 L atm mol⁻¹K⁻¹,
T is the temperature in Kelvin(K),

Here, to calculate M convert into standard units mg tog, ml to L, c to Kelvin
M= ( $\frac{35.8}{598}$ *10⁻³ )/ 0.175 =(5.987 *10⁻⁵)mol / 0.175L = 34.21*10⁻⁵ mol/L

π=iMRT=(1)*(34.21*10⁻⁵)*(0.08206)*(298.15)=837×10⁻⁵= 8.37×10⁻³ atm
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3 0

3 years ago

A small particle has charge -2.00 uC and mass 1.50×10-4 kg. It moves from point A, where the electric potential is V(A) = 200 V,

Svet_ta [14]

Answer:

$v_b=6.13 m/s.$ .

Explanation:

Since no external force is acting on the system.

Therefore, Total energy remains constant before and after.

So, Total energy of system= energy due to potential applied+kinetic energy

$T.E=qV_a+\dfrac{1}{2}mv_a^2=qV_b+\dfrac{1}{2}mv_b^2\\\dfrac{1}{2}mv_b^2=\dfrac{1}{2}mv_a^2+q(V_a-V_b)\\$

(Here v=velocity ,V=potential ,q=charge and m=mass).

Putting values .

We get, $v_b=6.13 m/s.$ .

At point B charged particle is moving faster as compared to point A.

Hence, it is the required solution.

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4 years ago

A camera lens used for taking close-up photographs has a focal length of 22.0 mm. The farthest it can be placed from the film is

Arada [10]

Answer:

p = 6.64 cm

Explanation:

For this exercise we use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively

They tell us the focal length f = 2.2 cm and that the image as far as it can go is q = 3.29 cm, let's find the position of the object that creates this image

$\frac{1}{p} = \frac{1}{f} - \frac{1}{q}$

1 / p = 1 / 2.2 - 1/3.29

1 / p = 0.15059

p = 6.64 cm

therefore the farthest distance from the object is 6.64 c

3 0

3 years ago