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3 years ago

If size of an image is 10 cm and the magnification produced by a mirror is +1, what will be the size of object?

Physics

1 answer:

olya-2409 [2.1K]3 years ago

3 0

Answer:

h = 10 cm

Explanation:

The magnification produced by a mirror is given by :

$m=\dfrac{h'}{h}$

h' is the image size and h is the object size

We have, h = 10 cm, m = +1

So,

$m=\dfrac{h'}{h}\\\\h=\dfrac{10}{1}\\\\h=10\ cm$

So, the size off the object is same as that of the size of the image i.e. 10 cm.

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Which times and distances are represented by the function? select three options. the starting distance, at 0 hours, is 300 miles

Naya [18.7K]

The times and distances are represented by the function are

at 2 hours, the traveler is 725 miles from home.
at 3 hours, the distance is constant, at 875 miles
the total distance from home after 6 hours is 1,062.5 miles.

<h3>What is distance?</h3>

The distance is the length of path travelled by a body when moving with some speed and taking time t.

Given function is

D (t) = 300t +125 for 0≤t<2.5

D (t) = 875 for 2.5 ≤t≤3.5

D(t)= 75t +612.5 for 3.5<t≤6

According to the given function, for 2 hours, the traveler will travel from home,

D (2) = 300x2 +125 = 725 miles.

For time between 2.5 to 3.5 hr, the distance is constant i.e. 875 miles.

For total time 6hr, the distance travelled will be

D(6) = 75 x 6 +612.5 = 1062.5 miles.

Thus, the correct options are picked up.

Learn more about distance.

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4 0

2 years ago

If a typical smelt weighs 225 g, what is the total mass of pcbs in a smelt in the great lakes? If a typical smelt weighs 225 g,

photoshop1234 [79]

Answer;

= 2.34 x 10^-4 g pcbs

Explanation;

-A typical smelt has 1.04 ppm of PCB is great lakes.

Therefore;

The amount of PCB in a smelt in great leak can be calculated by multiplying the mass of smelt by the pcbs present in a smelt

= 225 g * ( 1.04/100000)

= 2.34 x10^-4 g pcbs

7 0

3 years ago

An object of mass 0.5 kg is swung in uniform circular motion. The radius is 2 meters, and the force exerted is 4 N. Calculate th

Katarina [22]

Using
F= mv²/r
4 = 0.5×v² / 2
8 /0.5 = v²
v²=16
v= √16
v= 4 ms-¹

8 0

3 years ago

Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven,

andrew-mc [135]

Energy = (power) x (time)

-- <u>For the toaster:</u>

Power = 1.4 kW = 1,400 watts

Time = 5.4 minutes = 324 seconds

Energy = (1,400 W) x (324 s) = 453,600 Joules

-- <u>For the CFL bulb:</u>

Power = 11 watts

Time = 10.5 hours = 37,800 seconds

Energy = (11 W) x (37,800 s) = 415,800 Joules

-- The toaster uses energy at 127 times the rate of the CFL bulb.

-- The CFL bulb uses energy at 0.0079 times the rate of the toaster.

-- The toaster is used for 0.0086 times as long as the CFL bulb.

-- The CFL bulb is used for 116.7 times as long as the toaster.

-- The toaster uses 9.1% more energy than the CFL bulb.

-- The CFL bulb uses 8.3% less energy than the toaster.

7 0

3 years ago

Yes this nees helppppppppppppppp

kozerog [31]

Answer:

lol

Explanation:

it was funny

6 0

3 years ago