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3 years ago

What is the magnification when an object is placed at 2f from the pole of the convex mirror?

Physics

1 answer:

Gekata [30.6K]3 years ago

5 0

Answer:

Linear magnification = 1/3

Explanation:

Given:

Convex mirror

Object's distance from pole = 2f

Find:

Linear magnification

Computation:

Object distance, u = −2f

So,

1/v + 1/u = 1/f

1/v + 1/(-2f) = 1/f

1/v = 1/f + 1/2f

BY taking LCM

1/v = 3 / 2f

v = 2f / 3

Magnification, M = -v / u

So,

Magnification, M = (2f / 3) / 2f

Magnification, M = 2f / 6f

Magnification, M = 2 / 6

Linear magnification = 1/3

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A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m verticall

lorasvet [3.4K]

Explanation:

Given:

t = 20 seconds

x = 3000 m

y = 450 m

a) To find the vertical component of the initial velocity $v_{0y}$ , we can use the equation

$y = v_{0y}t - \frac{1}{2}gt^2$

Solving for $v_{0y}$ ,

$v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}$

$\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}$

$\:\:\:\:\:\:\:=120.5\:\text{m/s}$

b) We can solve for the horizontal component of the velocity $v_{0x}$ as

$x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}$

$v_{0x} = 150\:\text{m/s}$

4 0

3 years ago

One end of a horizontal spring with force constant 130.0 N/m is attached to a vertical wall. A 3.00 kg block sitting on the floo

Nuetrik [128]

Answer:

a) v = 0

b) The aceleration is 1.41 $m/s^{2}$

c) The block is accelerating away from the wall.

Explanation:

First, you need to think about the effect this constant force is causing in the spring: it causes a displacement in the equilibrium point of the system, therefore we need to know where it sits now:

At equilibrium no movement is present reducing friction to 0:

$\sum{F} = 0 = F_{spring} - F_{external}$

$F_{spring} = F_{external}$

$Kx = F_{external}$

$x = \frac{F_{external}}{K}=\frac{88}{130}=0.68m=68cm$

This means that the spring can be compressed with the single force up to 68 cm, Any further compression will cause an unbalanced system and the occilation of the mass.

The spring can't be compressed by the given force to 80 cm, therefore it must have been compressed by another force and then released.

In this case, the instantanous speed is 0, since the block has just been released.

In the same instant we can stimate the free body diagram of forces by the next two equations:

$\sum_y{F}={F_N-W}=0\\\sum_x{F}={F_{spring}-F_{external}-F_{friction}}=ma$

For the y axis:

$F_N = W = mg = 3*9.8 = 29.4N$

To calculate the force of friction:

$F_{friction} = \mu_k F_N=0.4*29.4 = 11.76N$

Therefore for x axis:

${Kx-F_{external}-F_{friction}}=ma$

$a = \frac{130*0.8-88-11.76}{3} = \frac{104-88-11.76}{3}=\frac{4.24}{3}=1.41\frac{m}{s^2}$

7 0

3 years ago

Three children are riding on the edge of a merry‑go‑round that has a mass of 10^5 and a radius of 1.80 m . The merry‑go‑round is

zimovet [89]

Answer: 22.01 rpm.

Explanation:

If we assume no external torques are present, the angular momentum must be conserved.

L1 = L2

I1 . ω1 = I2 ω2 (1)

I1 = MR2 /2 + m1. R2 + m2. R2 + m3. R2

I2 = MR2 /2 + m1. R2 + m3. R2

(as the 28.0 kg child moves to the center, he has no part anymore in the rotational inertia)

As I units are SI units, it is advisable to convert the angular velocity to rad/sec, as follows:

22.0 rev/min. (2π/rev) . (1min/60 sec) = 2.3 rad/sec

Solving for ω2 in (1):

ω2 = 2.31 rad/sec = 2.31 (1 rev/2 π). (60sec/1min) = 22.01 rpm

3 0

3 years ago

A plate has an area of 1m square. It slide down an incline plane , having an angle of inclined 60° to horizontal, which a veloci

Solnce55 [7]

Answer:

$\mu =0.169704 Ns/m^2$

Explanation:

From the question we are told that

Area of plate $a=1m^2$

Angle of inclination $\theta=60 \textdegree$

Velocity $v=0.75m/s$

Thickness of oil $t=2mm$

Weight of plate $w=90N$

Generally the equation for shear force is mathematically given by

$F=mgsin45 \textdegree$

$F=90sin45 \textdegree$

$F=63.639N$

$F=\mu* \frac{v}{t}$

$F=\mu* \frac{0.75}{2*10^-^3}$

$\mu* \frac{0.75}{2*10^-^3}=63.639N$

$\mu =\frac{63.639}{375}$

Therefore viscosity is given by

$\mu =0.169704 Ns/m^2$

3 0

3 years ago

A pair of narrow slits, separated by 1.8 mm, is illuminated by a monochromatic light source. Light waves arrive at the two slits

Katyanochek1 [597]

Answer:

Answer:

750 nm

Explanation:

= separation of the slits = 1.8 mm = 0.0018 m

λ = wavelength of monochromatic light

= screen distance = 4.8 m

= position of first bright fringe =

= order = 1

Position of first bright fringe is given as

λ = 7.5 x 10⁻⁷ m

λ = 750 nm

Explanation:

5 0

3 years ago