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Lisa [10]
2 years ago
13

A magnetron in a microwave oven emits electromagnetic waves with a frequency f-2450 MHz. What magnetic field strength is require

d for electrons to move in circular paths with this frequency?
Physics
1 answer:
mart [117]2 years ago
6 0

Answer:

Magnetic field, B = 0.073 T

Explanation:

It is given that,

A magnetron in a microwave oven emits electromagnetic waves with a frequency 2450 MHz, f=2450\times 10^6Hz

We need to find the magnetic field required for electrons to move in circular paths with this frequency.

The formula is given by :

B=\dfrac{m\omega}{q}

q and m are charge and mass of electron

B=\dfrac{9.1\times 10^{-31}\times 2\pi (2450\times 10^6)}{1.6\times 10^{-19}}

B = 0.073 T

Hence, this is the required solution.

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A wave is a disturbance that transmits energy from one location to another without transmitting matter with it.
GenaCL600 [577]
I think it’s true!!!
5 0
2 years ago
Given a die, would it be more likely to get a single 6 in six rolls, at least two 6s in twelve rolls, or at least one-hundred 6s
vichka [17]

Answer:

Explanation:

In first case we are interested in one time 6 in six rolls

Thus probability = number of chances required/Total chances

= 1/6

Similarly in the second case probability = 2/12 = 1/6

In the same way in last case probability = 100/600 = 1/6

The probability is the same . Thus all the cases has equal chances  

4 0
2 years ago
A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

-0.02 = 0.02cos(0 + \phi) => \phi = \pi

2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))

5 0
2 years ago
Read 2 more answers
A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float
storchak [24]

Answer:

 t ’= \frac{1450}{0.6499 + 2 v_r},  v_r = 1 m/s       t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

        v_b - v_r = d / t

By the time the boat goes down the river

        v_b + v_r = d '/ t'

let's subtract the equations

       2 v_r = d ’/ t’ - d / t

       d ’/ t’ = 2v_r + d / t

       t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }

In the exercise they tell us

         d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

         d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

      t ’= \frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \  + 2 \ v_r}

      t ’= \frac{1450}{0.6499 + 2 v_r}

In order to complete the calculation, we must assume a river speed

          v_r = 1 m / s

       

let's calculate

      t ’= \frac{ 1450}{ 0.6499 + 2 \ 1}

      t ’= 547.19 s

8 0
2 years ago
A 6.5 l sample of nitrogen at 25◦c and 1.5 atm is allowed to expand to 13.0 l. the temperature remains constant. what is the fin
ollegr [7]
Since the temperature of the gas remains constant in the process, we can use Boyle's law, which states that for a gas transformation at constant temperature, the product between the gas pressure and its volume is constant:
pV=k
which can also be rewritten as
p_1 V_1 = p_2 V_2 (1)
where the labels 1 and 2 mark the initial and final conditions of the gas.

In our problem, p_1 = 1.5 atm, V_1 =6.5 L and V_2 =13.0 L, so the final pressure of the gas can be found by re-arranging eq.(1):
p_2 = p_1  \frac{V_1}{V_2}= (1.5 atm) \frac{6.5 L}{13.0 L}=0.75 atm

Therefore the correct answer is
<span>1. 0.75 atm</span> 
8 0
2 years ago
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