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maria [59]
2 years ago
6

When a beam of charged particles moves through a magnetic field, what is the evidence that particles in the beam have momentum g

reater than the value mv?
Physics
1 answer:
Vsevolod [243]2 years ago
4 0

The charged particles are often deflated in a magnetic field.

<h3>What is a magnetic field?</h3>

The term charge refers to a positive or negative entity. The can be created when a charge is made to pass through a conductor in a magnetic field.

A magnetic field is created when we have a north pole and a south pole. The charged particles could be made to pass through the electric field and when that happens, we can see a pattern a shown in the image attached.

Thus, we can see that the charged particles are often deflated in a magnetic field.

Learn more about magnetic field:brainly.com/question/23096032

#SPJ1

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When a metal bonds with a nonmetal, they form a(n)
MArishka [77]
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Answer:

a) 3.242

b) 1291.178 KJ

c) 3.59 cents

Explanation:

a) First all the temperature units should be in Kelvin. -30°C + 273.15 = 243.15K and 45°C + 273.15 = 318.15K. Coefficient of performance for refrigerator is COP = TL/(TH-TL) where TL is Lower Temperature and TH is the Higher Temperature. COP = 243.15/(318.15-243.15) = 3.242

b) COP  = Cooling Effect/Work Input. Cooling Effect (or Desired Output) is 4186 kJ.

So Work Input = Cooling Effect/COP = 4186KJ/3.242= 1291.178 KJ

c) 3.6 MJ (3600kJ) is for 10 cents, using ratio for cost: energy used

\frac{10}{3600 kJ} =\frac{x}{1291.178kJ}

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4 years ago
Calculate the period of wave whose frequency is 50Hz​
miss Akunina [59]

Explanation:

time period= 1/frequency.

= 1/50 = 0.02 second.

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5 0
3 years ago
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If a car is speeding down a road at 40 miles/hour (mph), how long is the stopping distance d40 compared to the stopping distance
tia_tia [17]
Assume that the deceleration due to braking is a ft/s².

Note that
40 mph = (40/60)*88 = 58.667 ft/s
25 mph = (25/60)*88 = 36.667 ft/s

The final velocity is zero when the car stops, therefore
v² - 2ad = 0, or d = v²/(2a)
where
v = initial speed
a = deceleration
d = stopping distance.

The stopping distance, d₄₀, at 40 mph is
d₄₀ = 58.667²/(2a)
The stopping distance, d₂₅, at 25 mph is
d₂₅ = 36.667²/(2a)

Therefore
d₄₀/d₂₅ = 58.667²/(2a) ÷ 36.667²/(2a)
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Answer:
The stopping distance at 40 mph is 2.56 times the stopping distance at 25 mph.
8 0
3 years ago
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Lesechka [4]

Explanation:

Please attach the figure next to your question in order to help you.. Wish u the best and Have a great Day. Good Luck!

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