Answer:
30 minutes
Explanation:
Energy per time is constant, so:
E₁ / t₁ = E₂ / t₂
m₁C₁ΔT₁ / t₁ = m₂C₂ΔT₂ / t₂
(1 kg) C (70°C − 25°C) / 15 min = (1.5 kg) C (80°C − 20°C) / t
(1 kg) (45°C) / 15 min = (1.5 kg) (60°C) / t
3/min = 90 / t
t = 30 min
1.Light-collecting area
2.Angular resolution
In order to measure the resistance in the circuit, we need to know the voltage V and the current I in the circuit, this way we can calculate the resistance using the formula:

In order to calculate the current, we can use an amperemeter that must be in series with the circuit, this way it will not affect the circuit.
And in order to calculate the voltage, we can use a voltmeter that must be in parallel with the resistance, this way it will not affect the circuit.
The correct option that shows an amperemeter in series and a voltmeter in parallel is the fourth option.
Complete question:
At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.
[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]
Answer:
The time it will take the particle to pass through point (P) again is 1.639 ns.
Explanation:
F = qvB
Also;

solving this two equations together;

where;
m is the mass of electron = 9.11 x 10⁻³¹ kg
q is the charge of electron = 1.602 x 10⁻¹⁹ C
B is the strength of the magnetic field = 3.47 x 10⁻³ T
substitute these values and solve for t

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.
Answer:
I'm pretty sure it's 20m/s because 1300m divided by 65 seconds is 20 so I think it's 20m/s
Explanation: