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3 years ago

How is a refracting telescope similar to a compound microscope?

Physics

2 answers:

Flura [38]3 years ago

5 0

They both have an objective lens and an eyepiece lens.

s2008m [1.1K]3 years ago

4 0

Answer:

There are two similarities between refracting telescope and compound microscope.

1. both have objective lens and eyepiece lens

2. both the lenses are convex in nature.

Explanation:

Objective lens is near the object and eyepiece is near the eye through which we can see the image.

Convex or converging lens is used to converge the rays of light coming from the object.

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HELP PLEASE NOWWWW !!!! ( 1 ) Jill and Scott both road their bikes for 30 minutes. Jill traveled 5 kilometers and Scott trave

katovenus [111]

Jill and Scott both moves for 30 minutes

now if Jill cover 5 km distance and Scott cover 10 km distance

now we know that the formula of speed is given as

$speed = \frac{distance}{time}$

now we will have

speed of Jill

$v = \frac{5}{0.5} = 10 km/h$

speed of Scott

$v = \frac{10}{0.5} = 20 km/h$

so correct answer here is

Scott had the faster speed since he rode at 20 k/h while Jill only traveled 10 km/h.

#2

distance travelled by each car is given as

$d = 400 miles$

now here it is given that

time taken by green car

$t_1 = 10 hours$

time taken by yellow car

$t_2 = 8 hours$

now we can find the speed of two cars

$speed = \frac{distance}{time}$

speed of green car

$v = \frac{400}{10} = 40 mph$

speed of yellow car

$v = \frac{400}{8} = 50 mph$

so correct answer will be

The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

6 0

3 years ago

Read 2 more answers

What is the potential difference per unit length between two infinitely long concentric cylindrical shells with inner radius 1.5

Vinil7 [7]

Answer:

165.8 V/m

Explanation:

The capacitance of a long concentric cylindrical shell of length, L and inner radius, a and outer radius, b is C = 2πε₀L/㏑(b/a)

Since the charge on the cylindrical shells, Q = CV where V = the potential difference across the capacitor(which is the potential difference between the concentric cylindrical shells)

V = Q/C

V = Q ÷ 2πε₀L/㏑(b/a)

V = Q㏑(b/a)/2πε₀L

So, the potential difference per unit length V' is

V' = V/L = Q㏑(b/a)/2πε₀

Given that a = inner radius = 1.5 cm, b = outer radius = 5.6 cm and Q = 7.0 nC = 7.0 × 10⁻⁹ C and ε₀ = 8.854 × 10⁻¹² F/m substituting the values of the variables into the equation, we have

V' = Q㏑(b/a)/2πε₀

V' = 7.0 × 10⁻⁹ C㏑(5.6 cm/1.5 cm)/(2π × 8.854 × 10⁻¹² F/m)

V' = 7.0 × 10⁻⁹ C㏑(3.733)/(55.631 × 10⁻¹² F/m)

V' = 7.0 × 10⁻⁹ C × 1.3173/(55.631 × 10⁻¹² F/m)

V' = 9.2211 × 10⁻⁹ C/(55.631 × 10⁻¹² F/m)

V' = 0.16575 × 10³ V/m

V' = 165.75 V/m

V' ≅ 165.8 V/m

6 0

3 years ago

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.46 m. The mug

Tresset [83]

Answer:

(a). The horizontal velocity is 1.46 m/s.

(b). The direction of the mug's velocity just before it hit the floor is 74.7° below the horizontal.

Explanation:

Given that,

Height of the counter = 1.46 m

Distance = 0.80 m

We need to calculate the time

Using equation of motion

$s_{y}=ut+\dfrac{1}{2}gt^2$

$s_{y}=\dfrac{1}{2}gt^2$

Put the value into the formula

$1.46=\dfrac{1}{2}\times9.8\times t^2$

$t^2=\dfrac{1.46\times 2}{9.8}$

$t=\sqrt{\dfrac{1.46\times2}{9.8}}$

$t=0.545\ sec$

Here, horizontal velocity is constant

(a). We need to calculate the velocity

Using formula of velocity

$v_{x}=\dfrac{d}{t}$

Put the value into the formula

$v_{x}=\dfrac{0.80}{0.545}$

$v_{x}=1.46\ m/s$

(b). We need to calculate the final velocity

Using equation of motion

$v_{f}^2=u^2+2as$

Put the value into the formula

$v_{f}^2=0+2\times9.8\times1.46$

$v_{f}=\sqrt{28.616}$

$v_{f}=5.34\ m/s$

The velocity is 5.34 m/s downward.

We need to calculate the direction

Using formula of direction

$\tan\theta=\dfrac{v_{y}}{v_{x}}$

$\theta=\tan^{-1}(\dfrac{5.34}{1.46})$

$\theta=74.7^{\circ}$

Hence, (a). The horizontal velocity is 1.46 m/s.

(b). The direction of the mug's velocity just before it hit the floor is 74.7° below the horizontal.

5 0

4 years ago

You need to design a 60.0-Hz ac generator that has a maximum emf of 5200 V. The generator is to contain a 180-turn coil that has

OlgaM077 [116]

Answer:

0.082 T

Explanation:

Given:

Frequency of the generator (f) = 60.0 Hz

Maximum emf of the generator (E) = 5200 V

Number of turns (N) = 180

Area per turn (A) = 0.94 m²

Magnetic field magnitude (B) = ?

We know that, the angular frequency of the generator is given as:

$\omega=2\pi f$

Plug in the value of 'f' and solve for 'ω'. This gives,

$\omega=2\pi\times 60.0=120\pi\ rad/s$

Now, the maximum emf of the generator is given by the formula:

$E=NAB\omega$

Rewriting in terms of magnetic field, 'B', we get:

$B=\frac{E}{NA \omega}$

Plug in the given values and solve for 'B'. This gives,

$B=\frac{5200\ V}{180\times 0.94\ m^2\times 120\pi\ rad/s}\\\\B=\frac{5200\ V}{63786.897}\\\\B=0.082\ T$

Therefore, the magnitude of the magnetic field in which the coil rotates is 0.082 T.

8 0

4 years ago

A ball of mass 0.165 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.940 m. W

Dmitriy789 [7]

Answer:

I = 1.525 kg.m/s

Explanation:

given,

mass of the ball = 0.165 Kg

height of drop, h = 1.25 m

ball rebound and reach to height, h' = 0.940 m

impulse = ?

using conservation of energy

Potential energy is converted into kinetic energy

$mgh = \dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8 \times 1.25}$

v = 4.95 m/s

velocity of the ball after rebound

again using conservation of energy

$mgh = \dfrac{1}{2}mv'^2$

$v'=\sqrt{2gh}$

$v'=\sqrt{2\times 9.8 \times 0.94}$

v' = 4.29 m/s

impulse is equal to change in momentum

I = m ( v' - v )

I = 0.165 x ( 4.29 - (-4.95))

I = 1.525 kg.m/s

6 0

3 years ago