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2 years ago

How do lenses and mirrors compare in their interactions with light?

Physics

2 answers:

slamgirl [31]2 years ago

7 0

D. lenses focus light , mirrors do not

Alisiya [41]2 years ago

4 0

Answer:

C. lenses refract light; mirrors do not

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A bus starting from rest moves with a uniform acceleration of 0.1 m s -2 for 2 minutes. Find (a) the speed acquired, (b) the dis

KiRa [710]

Answer:

S=720m anf vf=12m/s

Explanation:

acceleration=a=0.1m/s²

time taken=t=2minutes=2×60=120seconds

vi=0m/s

vf=?

distance covered=S=?

by using second equation of motion

S=vit+1/2at²

S=0m/s×120seconds+1/2(0.1m/s²)(120s)²

S=0m/s+1/2×1440m

S=720m

and now we have to find the vf

vf=vi+at

vf=0m/s+(0.1m/s)(120s)

vf=12m/s

i hope it will help you

8 0

3 years ago

For your senior project, you are designing a Geiger tube for detecting radiation in the nuclear physics laboratory. This instrum

Paha777 [63]

Answer:

Maximum linear charge density = 84.14 nC/m

Explanation:

Looking at this question, The electric field of a line charge of infinite length is given by : Er = (1/(2πεo)) x (λ/r)

r = the distance from the center of the line of charge

λ = the linear charge density of the wire.

Now looking at the equatiom, due to the fact that Er varies inveresely with r, its maximum value will occur at the surface of the wire where r = R, the radius of the wire:

And so, Emax = (1/(2πεo)) x (λ/R)

Let's make λ the subject of the equation and we get;

λ = 2πεo(REmax)

From the question, R = 0.55/2 = 0.275cm

Also, Emax = 5.50 × 10^(6) N/C

Let's take the value of the electric constant to be εo = 8.854 x 10^(-9) C^(2) / Nm^2

R = 0.275mm = 0.000275m

Plugging these values into the equation, we get;

λ = 2π x 8.854 x 10^(-12) x 0.000275 x 5.50 × 10^(6) = 84.14 nC/m

4 0

3 years ago

Use the worked example above to help you solve this problem. An airboat with mass 4.30 102 kg, including the passenger, has an e

umka21 [38]

Answer:

a) a = 1,865 m / s² and b) t = 8.1 s

Explanation:

a) Let's use Newton's second law to find acceleration, we can work the equation in scalar form because displacement and force have the same direction

F = m .a

a = F / m

a = 8.02 10² /4.3 10²

a = 1,865 m / s²

b) We use kinematic relationships in one dimension

vf = vo + at

vf = 0 + a t

t = vf / a

t = 15.1 / 1.865

t = 8.1 s

3 0

3 years ago

you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k

alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ $v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}$

→ $v_{p}=150$ km/h , $v_{w}=50$ km/h

→ $v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11$ km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is $tan^{-1}\frac{50}{150}=18.4$ °

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0

3 years ago

A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge

Digiron [165]

Answer:

F = 2.49 x 10⁻⁹ N

Explanation:

The electrostatic force between two charged bodies is given by Colomb's Law:

$F = \frac{kq_1q_2}{r^2}\\$

where,

F = Electrostatic Force = ?

k = colomb's constant = 9 x 10⁹ N.m²/C²

q₁ = charge on proton = 1.6 x 10⁻¹⁹ C

q₂ = second charge = 1.4 C

r = distace between charges = 0.9 m

Therefore,

$F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(1.6\ x\ 10^{-19}\ C)(1.4\ C)}{(0.9\ m)^2}$

8 0

3 years ago

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