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3 years ago

consider the mirror from the last question. an object 4cm tall stands 10cm in front of a converging mirror of focal length 5cm.

Physics

2 answers:

aleksandr82 [10.1K]3 years ago

6 0

<span>An Object 4 Cm Tall Is Placed 12 Cm From A Divergi... | Chegg.com</span>

RSB [31]3 years ago

4 0

Answer:

This question is incomplete, here is the complete question;

"Consider the mirror shown here. An object 4 cm tall stands 10 cm in front of a converging mirror of focal length 5 cm.

The image formed by this mirror is: (Select all that apply.)

real

virtual

enlarged

the same size

diminished

Inverted

erect"

Answer: Virtual, Upright and Enlarged

Explanation:

The formula for image formed by a converging mirror is given by;

$\frac{1}{f} =\frac{1}{u} +\frac{1}{v}$ ------------------- eqn 1

where f = focal length = 5 cm

u = object height = 4 cm

v = image height = ?

substitute for u and f in eqn 1 to find v.

$\frac{1}{5} =\frac{1}{4} +\frac{1}{v}$

$\frac{1}{v} =\frac{1}{5} -\frac{1}{4}$

v = -20 cm

From the value of the image formed i.e -20 cm

The negative sign means the image is behind the mirror and this means image is;

1. Virtual

2. Upright

3. Enlarged

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Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain

Luden [163]

Answer:

$P=1362\ W$

$t'=251.659\ s$ is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, $T_i=18^{\circ}C$

time taken to vapourize half a liter of water, $t=18\ min=1080\ s$

desity of water, $\rho=1\ kg.L^{-1}$

So, the givne mass of water, $m=1\ kg$

enthalpy of vaporization of water, $h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}$

specific heat of water, $c=4180\ J.kg^{-1}.K^{-1}$

Amount of heat required to raise the temperature of given water mass to 100°C:

$Q_s=m.c.\Delta T$

$Q_s=1\times 4180\times (100-18)$

$Q_s=342760\ J$

Now the amount of heat required to vaporize 0.5 kg of water:

$Q_v=m'\times h_{fg}$

where:

$m'=0.5\ kg=$ mass of water vaporized due to boiling

$Q_v=0.5\times 2256.4$

$Q_v=1.1282\times 10^{6}\ J$

Now the power rating of the boiler:

$P=\frac{Q_s+Q_v}{t}$

$P=\frac{342760+1128200}{1080}$

$P=1362\ W$

Now the time required to heat to boiling point form initial temperature:

$t'=\frac{Q_s}{P}$

$t'=\frac{342760}{1362}$

$t'=251.659\ s$

6 0

3 years ago

Explain in terms of impulse how padding reduces forces in a collision. State this in terms of a real example, such as the advant

Lena [83]

Answer:

Impulse = Average force x time of contact

Explanation:

Impulsive force is a force which is very large but applied on a body for a very small duration of time.

Impulse is given by the change in momentum of the body.

Impulse = Average force x small time interval

When padding is there, the time interval of contact is large and thus, the force exerted by the body is small.

So, when a person falls on the tile floor, there is no compression and thus, the time of contact is very small and thus the impulsive force is very large, due to which the body may damage.

So, when a person falls on the carpeted floor, there is a compression and thus, the time of contact is comparatively large and thus the impulsive force is small, due to which the body may safe.

3 0

3 years ago

two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco

Reika [66]

Answer:

$t=750s$

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

$\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt$

$x_f$ is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

$x_0_1=6km\\x_0_2=0$

We have $x_f_1=x_f_2$ . Thus, solving for t:

$x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s$

8 0

3 years ago

HELPPP PLEASEEE!!!!!

devlian [24]

Answer:

1) The speed of sound increases

2) 440 Hz

3) 29°C

4) 17°C

5) 434 Hz

6) 12 m/s

7) 17.3 m

Explanation:

1) The speed of sound increases

2) V = f×λ

f = V/λ = 343/0.78 = 439.744 ≈ 440 Hz

3) V = f×λ

512 × 0.68 = 348.16 m/s

348.16 - 331 = 17.16

T = 17.16/0.6 = 28.6 ≈ 29°C

4) Increase in speed = 350 - 340 = 10

Increase in temperature = 10/0.6 = 16.67° ≈ 17°C

5) f = V/λ = 343/0.79 = 434 Hz

6) 331 + 0.6×30 - (331 × 0.6 ×10) = 12 m/s

7) V = 331 + 0.6×25 = 346m/s

λ = 346/20 = 17.3 m

5 0

3 years ago

I need help answering 30A and B. Help a guy out?

dmitriy555 [2]

Uh so I'm no master at this subject, but all stuffs accelerate at 9.8 m/s squared. So you multiply the 9.8 and the 0.20 it's given for reasons unknown other than that's what I see in my notes... and that gives you 1.96 m/s squared.

As for B, I have no idea. I think you may multiply the 1.96 by 4. Tell me your thoughts and maybe we can work it out together

5 0

3 years ago