Two microwave frequencies are authorized for use in microwave ovens: 895 and 2560 mhz. calculate the wavelength of each.

Answers

Answer 1

Answer: Wavelength = (speed) / (frequency)

The speeds of the two possible signals are equal, just like
all other forms of electromagnetic radiation.

Wavelength of 895 MHz = (3 x 10⁸ m/s) / (8.95 x 10⁸/s) = 0.335 m

Wavelength of 2560 MHz = (3 x 10⁸ m/s) / (2.56 x 10⁹/s) = 0.117 m

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A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation

Answers

Answer:

Explanation:

acceleration of test tube

= ω² R

= (2πn)² R

= 4π²n²R

n = no of rotation per second

= 3700 / 60

= 61.67

R = .10 m

acceleration

= 4π²n²R

= 4 x 3.14² x 61.67² x .10

= 14999 N Approx

A 372-g mass is attached to a spring and undergoes simple harmonic motion. Its maximum acceleration is 17.6 m/s2 , and its maximum speed is 1.75 m/s. a)Determine the angular frequency. b)Determine the amplitude. c)Determine the spring constant.

Answers

Answer with Explanation:

We are given that

Mass , m=372 g= $(372)/(1000)=0.372 Kg$

1 kg=1000g

Maximum acceleration, a= $17.6 m/s^2$

Maximum speed ,v=1.75 m/s

a.We know that

Maximum acceleration, a= $A\omega^2$

Maximum speed, v= $\omega A$

$17.6=A\omega^2$

$1.75=A\omega$

$(17.6)/(1.75)=(A\omega^2)/(A\omega)=\omega$

Angular frequency, $\omega=10.06 rad/s$

b.Substitute the value of angular frequency

$1.75=A(10.06)$

$A=(1.75)/(10.06)=0.17 m$

Hence, the amplitude=0.17 m

c.Spring constant,k= $m\omega^2$

Using the formula

$k=0.372* (10.06)^2$

Hence, the spring constant,k=37.6 N/m

A bicycle racer is going downhill at 11.0 m/s when, to his horror, one of his 2.25 kg wheels comes off when he is 75.0 m above the foot of the hill. We can model the wheel as a thin-walled cylinder 85.0 cm in diameter and neglect the small mass of the spokes. (a) How fast is the wheel moving when it reaches the bottom of hill if it rolled without slipping all the way down? (b) How much total kinetic energy does it have when it reaches bottom of hill?

Answers

Answer:

a.) Speed V = 29.3 m/s

b.) K.E = 1931.6 J

Explanation: Please find the attached files for the solution

Final answer:

The wheel's speed at the bottom of the hill can be found through the conservation of energy equation considering both translational and rotational kinetic energy, while the total kinetic energy at the bottom of the hill is a sum of translational and rotational kinetic energy.

Explanation:

These two questions address the physics concepts of conservation of energy, kinetic energy, and rotational motion. To answer the first question, (a) How fast is the wheel moving when it reaches the bottom of the hill if it rolled without slipping all the way down?, we need to consider the potential energy the wheel has at the top of the hill is completely converted into kinetic energy at the bottom. This includes both translational and rotational kinetic energy. Solving for the final velocity, vf, which would be the speed of the wheel, we get vf = sqrt((2*g*h)/(1+I/(m*r^2))), where g is the acceleration due to gravity, h is the height of the hill, I is the moment of inertia of the wheel, m is the mass of the wheel, and r is the radius of the wheel.

For the second question, (b) How much total kinetic energy does it have when it reaches bottom of the hill?, we use the formula for total kinetic energy at the bottom of the hill, K= 0.5*m*v^2+0.5*I*(v/r)^2. Substituting the value of v found in the first part we find the kinetic energy which we can use the formula provided in the reference information.

Learn more about Conservation of Energy and Rotational Motion here:

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Which volcanoes are formed by pyroclastic deposits? Select all that apply. A. cinder cone B. stratovolcano C. shield volcano

Answers

Answer:

Its a cinder cone cause after it all falls down to make deposits.

cinder cone and stratovolcano

A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -7000, 0> kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is <-4000, -780, 0> N.Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.6 seconds after the probe is launched to 22.9 seconds after the launch?

Answers

Answer:

The change in momentum is $\Delta p = <-1200 , -234 , 0> kg \cdot m/s$

Explanation:

From the question we are told that

The mass of the probe is $m = 170 kg$

The location of the prob at time t = 22.9 s is $A = <5600, 7200,0>$

The momentum at time t = 22.9 s is $p = < 51000, -7000, 0> kg m/s$

The net force on the probe is $F = <-4000 , -780 , 0> N$

Generally the change in momentum is mathematically represented as

$\Delta p = F * \Delta t$

The initial time is 22.6 s

The final time is 22.9 s

Substituting values

$\Delta p = <-4000 , -780 , 0> * (22.9 - 22.6)$

$\Delta p = <-4000 , -780 , 0> * (0.3)$

$\Delta p = <-1200 , -234 , 0> kg \cdot m/s$

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1025 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 55 volts?

Answers

Explanation:

Given data:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Distance d between the plates = 1 mm = 1×10⁻³m

Voltage of the battery is emf = 100 V

Resistance = 1025 ohm

Solution:

In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

$V = V_(0)(1-e^{(-t)/(RC) } )\n(V)/(V_(0) ) = 1-e(^{(-t)/(RC) }) \ne^{(-t)/(RC) } = 1- (V)/(V_(0) )$