If you double the diameter of a telescope, you double its light-collecting area.

WILL MARK BRAINLIEST JUST PLEASE HELP

mixer [17]

Answer:

40N in either direction is the answer

6 0

2 years ago

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Anybody know these ?

pashok25 [27]

Answer:

1. a) 72 N.

2. a) 2 m/s².

Explanation:

Given the following data;

1. Mass = 90kg

Acceleration = 0.8 m/s²

To find the force;

Force = mass * acceleration

Force = 90 * 0.8

Force = 72 Newton.

2. Mass = 50kg

Force = 100N

To find the magnitude of acceleration;

Acceleration = force/mass

Acceleration = 100/50

Acceleration = 2 m/s²

5 0

2 years ago

PLEASE HELP!!!! :D When you look at the Sun through a filtered telescope, the visible portion of the Sun appears blotchy.

olya-2409 [2.1K]

<span>When an individual looks through a filtered telescope in which he or she observes the sun, the portion where it appears blotchy is likely to be called the sunspots while the layer of the sun where it shows where it occurs is called the photosphere.</span>

5 0

3 years ago

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Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .