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adelina 88 [10]

2 years ago

12

The largest aperture telescopes are reflectors because they require a _____ focal length.

1 answer:

Stells [14]2 years ago

5 0

The largest aperture telescopes are reflectors because they require a shorter focal length. I think that is the answer! I hope that it helps!

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A scientist travelled 80 kilometres each day in the buggy.

7nadin3 [17]

Explanation:

In one day a scientist can travel = 80 km

In 10 days. a scientist can travel = 80* 10

= 800 km.

4 0

3 years ago

Read 2 more answers

A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loo

m_a_m_a [10]

Answer:

The induced emf is $\epsilon = 0.1041 \ V$

Explanation:

From the question we are told that

The radius of the circular loop is $r = 9.50 \ cm = 0.095 \ m$

The intensity of the wave is $I = 0.0215 \ W/m^2$

The wavelength is $\lambda = 6.90\ m$

Generally the intensity is mathematically represented as

$I = \frac{ c * B^2 }{ 2 * \mu_o }$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4 \pi *10^{-7} N/A^2$

B is the magnetic field which can be mathematically represented from the equation as

$B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }$

substituting values

$B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }$

$B = 1.342 *10^{-8} \ T$

The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.095)^2$

$A = 0.0284$

The angular velocity is mathematically represented as

$w = 2 * \pi * \frac{c}{\lambda }$

substituting values

$w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }$

$w = 2.732 *10^{8} rad \ s^{-1}$

Generally the induced emf is mathematically represented as

$\epsilon = N * B * A * w * sin (wt )$

At maximum induced emf $sin (wt) = 1$

So

$\epsilon = N * B * A * w$

substituting values

$\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}$

$\epsilon = 0.1041 \ V$

7 0

2 years ago

A hill is 132 m long and makes an angle of 12.0 degrees with the horizontal. As a 54 kg jogger runs up the hill, how much work d

TEA [102]

Answer:

14523.55J

Explanation:

The work done by the jogger against gravity is given by the following equation;

$W=mgh.................(1)$

where m is the mass, g is acceleration due to gravity taken as $9.8m/s^2$ and h is the height of the hill.

Since the length of the hill is 132m and it is inclined at 12 degrees to the horizontal, the height is thus given as follows;

$h=132sin12^o\\h=27.44m$

Substituting this into equation (1) with all other necessary parameters, we obtain the following;

$W=54*9.8*27.44\\W=14523.55J$

4 0

2 years ago

An astronaut is moving in space when a big explosion occurs about 50 metres BEHIND him. How will the astronaut come to know abou

kakasveta [241]

The matter from the explosion can reach him, hitting him. He should be able to feel that.

6 0

2 years ago

Read 2 more answers

A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal

Karolina [17]

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

Initial speed of the car, u = 25 m/s

Final speed, v = 0 (it stops)

The formula for the net force is :

F = ma

a is acceleration of the car

$a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2$

Let d be the breaking distance. It can be calculated using third equation of motion as :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m$

So, the required distance covered by the car is 44.64 m.

4 0

2 years ago