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Tpy6a [65]
3 years ago
13

A microscope that uses light and more than one lens

Physics
2 answers:
ale4655 [162]3 years ago
5 0
A microscope that uses light and more than one lens are use to view cells because the light causes no harm to the cells and is able to see and differ the different cells which the naked eye cannot see
gulaghasi [49]3 years ago
4 0
A compound optical microscope or just a optical microscope is a microscope that uses light and has more than one lens.
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A small, positively charged ball is moved close to a large, positively charged ball. Which describes how the small ball likely r
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Answer:c

Explanation:

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O How are electricity and magnetism related?
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Magnetism is the product of a moving charged particle. We can have electricity without magnetism but we can not have magnetism without electricity.An electro magnet is made so that we have a soft metal core and electricity around it. A bar magnet is a normal magnet in bar shape with permanent magnetism.
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these are the types of attractions found between molecules. their strength determines the state of the substance at room tempera
Elina [12.6K]
<h2>Answer</h2>

The physical state of the elements depends upon the <u>attraction forces </u>and their <u>kinetic energy</u>.

<h2>Explanation</h2>

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7 0
3 years ago
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Observe the given figure and find the the gravitational force between m1 and m2.​
Leno4ka [110]

Answer:

The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N

Explanation:

The details of the given masses having gravitational attractive force between them are;

m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m

The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;

F =G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}

Where;

F = The gravitational force between m₁ and m₂

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

r₂ = 0.1 m + 0.15 m = 0.25 m

Therefore, we have;

F = 6.67430 \times 10^{-11} \ N \cdot m^2/kg \times \dfrac{20 \ kg\times 50 \ kg}{(0.1 \ m+ 0.15 \ m)^{2}} \approx 1.06789 \times 10^{-6} \ N

The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N

8 0
3 years ago
Calculate the induced electric field (in V/m) in a 40-turn coil with a diameter of 11 cm that is placed in a spatially uniform m
Naddik [55]

Answer:

Explanation:

Given that,

Number of turn N = 40

Diameter of the coil d= 11cm = 0.11m

Then, radius = d/2 = 0.11/2 =0.055m

r = 0.055m

Then, the area is given as

A =πr²

A = π × 0.055²

A = 9.503 × 10^-3 m²

Magnetic Field B = 0.35T

Magnetic field reduce to zero in 0.1s, t = 0.1s

so we want to find induce electric field. To find the electric field,(E) we need to find the electric potential (V).

E.M.F is given as

ε = —N • dΦ/dt

Where magnetic flux is given as

Φ = BA

Then, ε = —N • dΦ/dt

ε = —N • dBA/dt

ε = —NBA/t

Then, its magnitude is

ε = NBA/t

Inserting the values of N, B, A and t

ε = 40×0.35×9.503×10^-3/0.1

ε = 1.33 V

Then, using the relationship between Electric field and electric potential

V = Ed

ε = E•d

E = ε/d

E = 1.33/0.11

E = 12.09 V/m

7 0
4 years ago
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