Answer:
A) d_o = 20.7 cm
B) h_i = 1.014 m
Explanation:
A) To solve this, we will use the lens equation formula;
1/f = 1/d_o + 1/d_i
Where;
f is focal Length = 20 cm = 0.2
d_o is object distance
d_i is image distance = 6m
1/0.2 = 1/d_o + 1/6
1/d_o = 1/0.2 - 1/6
1/d_o = 4.8333
d_o = 1/4.8333
d_o = 0.207 m
d_o = 20.7 cm
B) to solve this, we will use the magnification equation;
M = h_i/h_o = d_i/d_o
Where;
h_o = 3.5 cm = 0.035 m
d_i = 6 m
d_o = 20.7 cm = 0.207 m
Thus;
h_i = (6/0.207) × 0.035
h_i = 1.014 m
I pretty sure that 3 is b and 4 is A and 5 I need a full picture
Answer:
2.73×10¯³⁴ m.
Explanation:
The following data were obtained from the question:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Wavelength (λ) =?
Next, we shall determine the energy of the ball. This can be obtained as follow:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Energy (E) =?
E = ½m²
E = ½ × 0.113 × 43²
E = 0.0565 × 1849
E = 104.4685 J
Next, we shall determine the frequency. This can be obtained as follow:
Energy (E) = 104.4685 J
Planck's constant (h) = 6.63×10¯³⁴ Js
Frequency (f) =?
E = hf
104.4685 = 6.63×10¯³⁴ × f
Divide both side by 6.63×10¯³⁴
f = 104.4685 / 6.63×10¯³⁴
f = 15.76×10³⁴ Hz
Finally, we shall determine the wavelength of the ball. This can be obtained as follow:
Velocity (v) = 43 m/s
Frequency (f) = 15.76×10³⁴ Hz
Wavelength (λ) =?
v = λf
43 = λ × 15.76×10³⁴
Divide both side by 15.76×10³⁴
λ = 43 / 15.76×10³⁴
λ = 2.73×10¯³⁴ m
Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.
Answer:
The value of resistance when power is 1100 watts =
= 50 ohms
Explanation:
Power
= 2200 Watts
Resistance
= 25 ohms
Power
= 1100 Watts
Resistance
= we have to calculate
Given that the power in an electric circuit varies inversely with the resistance
⇒ P ∝ 
⇒
= 
⇒
= 
⇒
= 50 ohms
This is the value of resistance when power is 1100 watts.
Answer:
so pressure in A must be one third the pressure in B
Explanation:
We shall apply gas law to the cylinders A and B . Since their quantity are same so their no of mole will also be same .
For cylinder A
Temperature T , volume 3V , pressure P₁ , no of mole = n
so
P₁ X 3V = n R T
For cylinder B
Temperature T , volume V , pressure P₂ ,no of mole = n
so
P₂ X V = n R T
From the two equation above
P₁ X 3V = P₂ X V

P₁ = P₂ / 3
so pressure in A must be one third the pressure in B