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Xelga [282]
4 years ago
12

The linear magnification produced by a spherical mirror is +1/3. Analysing this value. (i) state the type of mirror and (ii) pos

ition of the object with respect to the pole of the mirror. Draw ray diagram to justify your answer.

Physics
1 answer:
elena-s [515]4 years ago
7 0

Answer :

Explanation :

It is given that :

Magnification, m = \frac{+1}{3}

(1 ) Since, the magnification is positive it means it is a convex mirror.

(2)  The image is formed at the back of mirror and the image is virtual and erect.

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Classify each of the following hydro carbons as alkane,alkene or alkyne
timofeeve [1]

Answer:

a.)alkene

b.)alkane

d.)alkana

about c..I think there is something wrong with the question

5 0
3 years ago
How much power is used in a machine that produces 15 Joules of work in 3 seconds? Use the formula P = W/t.
Marta_Voda [28]
P=w/t


w=15
t=3
therefore, 5 watts (b)
7 0
3 years ago
If 3600 j of work is done in 3.0 s what is the power<br>0.00083W<br>1200W<br>3600W<br>11000W
Viktor [21]

Answer:

1200 W

Explanation:

Power is given by the ratio between work done and time taken:

P=\frac{W}{t}

where W is the work done and t the time taken.

In this problem, W = 3600 J and t = 3.0 s. Therefore, the power in this exercise is

P=\frac{3600 J}{3.0 s}=1200 W

5 0
4 years ago
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Which type of graph would be best for showing the percentage of people in a family with different jobs? A. Circle graph B. Scatt
musickatia [10]

Answer:

A

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6 0
4 years ago
How much heat is absorbed by a 75g iron skillet when it’s temperature rises from 5c to 20c?
alexira [117]

Answer: Q=499.5 J  

Explanation:

The heat (thermal energy) absorbed by the iron skillet can be found using the following equation:

Q=m.C.\Delta T   (1)

Where:

Q is the heat  (absorbed)

m=75 g is the mass of the element (iron in this case)

C is the specific heat capacity of the material. In the case of iron is C=0.444\frac{J}{g\°C}

\Delta T=T_{f}-T_{i}=20\°C - 5\°C= 15\°C is the variation in temperature

Knowing this, lets rewrite (1) with these values:

Q=(75 g)(0.444\frac{J}{g\°C})(15\°C)  (2)

Finally:

Q=499.5 J  

6 0
3 years ago
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