All categories

3 years ago

An object is held at a distance of 20cm from a concave lens of focal length 80cm. What is the

Physics

1 answer:

Sladkaya [172]3 years ago

5 0

<h3>Answer</h3>

it is -16

<h2>explanation </h2><h3>Here, u=−20cm,f=−80cm,v=?,h1=2cm,h2=?</h3><h3>From 1v−1u=1f,1v=1f+1u=1−80−120=−1−480=−580</h3><h3>v=−805=−16cm.</h3><h3>Thus, image is at 16 cm from the lens on the same side as the object.</h3><h3>As h2h1=vu∴h2=vuh1=−16−20×2cm=1.6cm</h3><h3>This is the size of the image.</h3>

You might be interested in

the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity

NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

$\int\limits^s_0 {a} \, ds = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\$

Remember that

$v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2$

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0

3 years ago

Which of the following statements about floating objects is always true? a. The object’s density is greater than the density of

Yuki888 [10]

It should be C. If the object is denser than the fluid, it will sink. If it isn't, it will float

6 0

3 years ago

Read 2 more answers

An electron is released from rest in a weak electric field given by vector E = -1.30 10-10 N/C Äµ. After the electron has travel

Murljashka [212]

Answer:

v = 6.45 10⁻³ m / s

Explanation:

Electric force is

F = q E

Where q is the charge and E is the electric field

Let's use Newton's second law to find acceleration

F- W = m a

a = F / m - g

a = q / m E g

Let's calculate

a = -1.6 10⁻¹⁹ / 9.1 10⁻³¹ (-1.30 10⁻¹⁰) - 9.8

a = 0.228 10² -9.8

a= 13.0 m / s²

Now we can use kinematics, knowing that the resting parts electrons

v² = v₀² + 2 a y

v =√ (0 + 2 13.0 1.6 10⁻⁶)

v = 6.45 10⁻³ m / s

4 0

3 years ago

A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what

irinina [24]

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

$\Delta L=\frac{PL}{AE}$

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

$A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2$

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

$0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg$

Mass of the climber = 69.38 kg

6 0

3 years ago

A small plastic bead has been charged to -15nC.

marishachu [46]

Answer:

Explanation:

q = - 15 n C = - 15 x 10^-9 c

(a) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on proton, F = charge on proton x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (1.67 x 10^-27)

a = 1.6 x 10^14 m/s²

(B) the direction of acceleration is towards the bead, as the force is attractive.

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on electron, F = charge on electron x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (9.1 x 10^-31)

a = 3 x 10^17 m/s²

(D) the direction of acceleration is away from the bead, as the force is repulsive.

8 0

3 years ago