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3 years ago

The light of a star that can be seen from Earth is called what? (Apparent or Absolute Magnitude)

1 answer:

3 0

A glance at the night sky above Earth shows that some stars are much brighter than others. Apparent brightness is how we see the star's brightness from Earth. The absolute brightness is what a star would look like if there was a standard distance that we were looking at all of the stars from.

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Which device uses a rotating magnetic field to produce an electric current?

Kay [80]

The correct answer is C) a motor.
In particular, we are talking about an AC motor, which produces an alternating current. In an AC motor, a coil is immersed in a rotating magnetic field. Due to the motion of the magnetic field,the angle between the direction of the field and the surface enclosed by the coil changes. As a result, the magnetic flux through the coil changes over time (the magnetic flux is given by:
 $\phi=BAcos\theta$ 
where B is the intensity of the magnetic field, A is the area enclosed by the coil and $\theta$ is the angle between the direction of B and the perpendicular to the plane of the coil). For Faraday-Newmann-Lenz law, this change in flux induces an electromotive force (emf) into the coil, according to:
 $emf=- \frac{d \Phi}{dt}$ 
where the numerator is the variation of magnetic flux and dt is the time interval. This emf in the coil produced an electrical current in the circuit.

7 0

2 years ago

Read 2 more answers

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

5 0

2 years ago

7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent

Rudiy27

Answer:

Lens at a distance = 7.5 cm

Lens at a distance = 6.86 cm (Approx)

Explanation:

Given:

Object distance u = 12 cm

a) Focal length = 20 cm

b) Focal length = 16 cm

Computation:

a. 1/v = 1/u + 1/f

1/v = 1/20 + 1/12

v = 7.5 cm

Lens at a distance = 7.5 cm

b. 1/v = 1/u + 1/f

1/v = 1/16 + 1/12

v = 6.86 cm (Approx)

Lens at a distance = 6.86 cm (Approx)

5 0

2 years ago

An elephant and a mouse would both have zero weight in gravity-free space. If they were moving toward you with the same speed, w

Dovator [93]

The elephant and the mouse having zero weight in a gravity free space will not bump into you at the same effect.

Explanation: 

When both are in a gravity free space, the weights are zero, as we know that the $\text {weight of the body}=\text {mass of the body} \times \text {acceleration due to gravity}$

$\text {here, the weight of elephant}=\text {mass of elephant } \times \text {zero gravti} y=zero$

$\text {similarly,weight of mouse}=\text {mass of mouse } \times \text {zero gravity}=zero$

But when they will acquire the speed of same magnitude, say v, their different masses will acquire different momentum, which will make the difference in effect while bumping.

$\text { momentum of elephant }=\text { mass of elephant } \times v$ $\text { momentum of mouse = mass of mouse } \times v$

And as we know $\text { mass of elephant }>\text { mass of mouse }$ Therefore, effect of impact by elephant will be more than that of mouse . An elephant breaking into you will take you back faster than a mouse in space hits you.