We can rearrange the mirror equation before plugging our values in.
1/p = 1/f - 1/q.
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p <-- cross multiplication
13.33cm = p
Now that we have the value of p, we can plug it into the magnification equation.
M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'
So the height of the image produced by the mirror is 9.6cm.
Answer:
λ = 3.2 x 10⁻⁷ m = 320 nm
Explanation:
The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:
v = fλ
where,
v = c = speed of the electromagnetic waves (UV rays) = speed of light
c = 3 x 10⁸ m/s
f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz
λ = wavelength of the electromagnetic waves (UV rays) = ?
Therefore, substituting the values in the relation, we get:
3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)
λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)
<u>λ = 3.2 x 10⁻⁷ m = 320 nm</u>
So, the radiation of <u>320 nm</u> wavelength is absorbed by Ozone.
It would be f=ma so
a.
f=m <span>× a</span>
Answer:
it is sooo easy u need to use magnetic panels on the rail and put the magnet on the train it works under the principal of magnetic
Explanation:
12) q = mCΔT
125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)
T = 82.0°C
13) Solving for ΔT:
ΔT = q / (mC)
a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C
b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C
c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C
d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C
e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C
f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C
14) q = mCΔT
q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)
q = 502,000 J
20) q = mCΔT
q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)
q = 742,000 J
24) q = mCΔT
q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)
q = -0.091 J
