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2 years ago

8

Why should a glass slide and a coverslip be held by the edges?

1 answer:

AleksAgata [21]2 years ago

8 0

THIS IS NOT THE EXACT ANSWER BUT IT MIGHT HELP
The cover slips serves two purposes: (1) it protects the microscope's objective lens from contacting the specimen, and (2) it creates an even thickness (in wet mounts) for viewing.

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There are 2 concentric cylinders. These cylinders are very long with length L. The inner cylinder has a radius R1 and is a solid

OLga [1]

To get the charge along the inner cylinder, we use Gauss Law
E = d R1/2εo
For the outer cylinder the charge can be calculated using
E = d R2^2/2εoR1
where d is the charge density
Use these two equations to get the charge in between the cylinders and the capacitance between them.

5 0

3 years ago

A -5.0 μC charge experiences a 11 i^ N electric force in a certain electric field. [Recall that i^ is a unit vector in the x-dir

Pachacha [2.7K]

Answer:

$\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

Explanation:

given,

charge = -5.0 μC

Electric force, F = 11 i^ N

force would a proton experience = ?

we know

$\vec{F} = q \vec{E}$

$11\hat{i} = -5 \times 10^{-6} \vec{E}$

$\vec{E} =-2.2 \times 10^{6}\hat{i}$

we know charge of proton is equal to 1.6 x 10⁻¹⁹ C

using formula

$\vec{F} = q \vec{E}$

$\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}$

$\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

Force experienced by the photon in the same field is equal to $\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

3 0

2 years ago

20 cm long 10 cm wide and 5 cm thick as a mass of 500 g determine the greatest pressure that can be exerted by block on the flat

uysha [10]

100000 Pascal

Explanation:

pressure= force/area

Max pressure= force/min area

so f=5

min area= 5×10^-5

5÷5*10^-5 = 100000pascal

8 0

2 years ago

A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the

barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q, is expressed as:

$KE=\frac{kQq}{r}\ \ \ \ \ \ \ ...i$

$k$ is Coulomb's constant.

#The electric field, $E$ at radius r is expressed as:

$E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii$

From i and ii, we have:

$KE=Eqr$

$r=(KE)/Eq$

#Substitute actual values in our equation:

$r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m$

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0

2 years ago

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The vertical height of an ocean wave from a crest to a trough is 10 meters. What is the amplitude of the ocean wave?

natima [27]

The amplitude of the wave would be 5 meters.

6 0

2 years ago

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