All categories

3 years ago

Light passing through the center of a lens will carry on undeviated.

Physics

1 answer:

White raven [17]3 years ago

7 0

Answer: true

Plz mark brainliest:)

You might be interested in

The is the sac-like structure that holds the testes.

Brilliant_brown [7]

Answer:

Scrotum

Explanation:

5 0

3 years ago

Read 2 more answers

How does the energy moves from the suns core to the photosphere?

Minchanka [31]

Answer: Energy from the core travels by radiation through the radiative

zone, then by convection through the convection zone.

Explanation:

4 0

2 years ago

A triply charged ion with velocity 7.00 × 10^6 m/s moves in a path of radius 36.0 cm in a magnetic field of 0.55 T in a mass spe

saw5 [17]

Answer:

Mass of ion will be $22\times 10^{-13}kg$

Explanation:

We have given ion is triply charged that is $q=3\times 1.6\times 10^{-19}=4.8\times 10^{-19}C$

Radius r = 36 cm = 0.36 m

Velocity of the electron $v=7\times 10^6m/sec$

Magnetic field B = 0.55 T

We know that radius of the path is given by $r=\frac{mv}{qB}$

$m=\frac{rqB}{v}=\frac{0.36\times 4.8\times 10^{-19}\times 7\times 10^6}{0.55}=22\times 10^{-13}kg$

6 0

3 years ago

An electron follows a helical path in a uniform magnetic field of magnitude 0.340 T. The pitch of the path is 6.00 µm, and the m

dedylja [7]

Answer:

The electron's speed is 34007.35 m/s

Explanation:

It is given that,

Magnetic field, B = 0.34 T

Magnetic force on the electron, $F=1.85\times 10^{-15}\ N$

The electron follows a helical path. We have to find the speed of an electron. The formula for magnetic force is given by :

$F=B\times q\times v$

q = charge on an electron, $q=1.6\times 10^{-19}\ C$

v = velocity of an electron

$v=\dfrac{F}{Bq}$

$v=\dfrac{1.85\times 10^{-15}\ N}{0.34\ T\times 1.6\times 10^{-19}\ C}$

v = 34007.35 m/s

Hence, this is the required solution.

4 0

2 years ago

What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from

ElenaW [278]

When is at the end of the runway the velocity of the plane is given by the equation $vf^{2}=0+2*a*s$ where s=1800 m is the runway length. Thus
$vf^{2}=2*5*1800=18000 (m/s)^{2}$
$vf =134.164 (m/s)$

At half runway the velocity of the plane is
$v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
$
$v= \sqrt{9000}=94.87 ( \frac{m}{s})$

Therefore at midpoint of runway the percentage of takeoff velocity is
‰ $P= \frac{v}{vf}= \frac{94.87}{134.164}=0.707$

6 0

3 years ago