Refer to the diagram shown below.
i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω
Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10
Also,
R₂*i = 9.5 (2)
Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A
From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)
Answer: 0.08 Ω
Answer: 
Explanation:
Given
Magnitude of charge is 
Force experienced is 
Electric field intensity is the electrostatic force per unit charge

Thus, the electric field intensity is 
Answer:
The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Explanation:
Given that,
Wavelength = λ
For D to be small,
We need to calculate the minimum width
Using formula of minimum width


Where, D = width of slit
= wavelength
Put the value into the formula

Here,
should be maximum.
So. maximum value of
is 1
Put the value into the formula


(b). If the minimum number is 50
Then, the width is


(c). If the minimum number is 1000
Then, the width is


Hence, The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Answer:
Conduction
Explanation:
When one object touches another heat moves through it (think of a saucepan on a hot stove). Aluminium foil is a great conductor of heat, which means it is a poor insulator when it is in direct contact with something hot.
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