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Airida [17]
2 years ago
5

What is the frequency of a microwave that has a wavelength of 0.050 m?

Physics
2 answers:
Elanso [62]2 years ago
5 0

Answer:

6x10^9Hz

Explanation:

we use a formula that relates the frequency and the wavelength:

f=\frac{c}{\lambda}

where f is the frequency of the wave, c is the speed of light c=3x10^8m/s, and \lambda is the wavelength.

We know that the wavelength is: \lambda=0.050m, so substituting the known values in the equation for the frequency we get:

f=\frac{3x10^8m/s}{0.05m} \\f=6x10^9s^{-1}=6x10^9Hz

The frequency is:

6x10^9Hz

Fudgin [204]2 years ago
4 0
<span>Frequency    Wavelength 1/4 Wavelength 1/20 Wavelength <span>1/100 Wavelength</span></span><span>
600 MHz          0.5 meters         12.5 cm               2.5 cm                       5.0 mm

if that helps at all</span>
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When a 5.0 kg box is hung from a spring, the spring stretches to 50 mm beyond its relaxed length. (a) In an elevator acceleratin
Shtirlitz [24]

Answer:

a) the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

Explanation:

Given that;

Gravitational acceleration g = 9.81 m/s²

Mass m = 5 kg

Extension of the spring X = 50 mm = 0.05 m

Spring constant k = ?

 we know that;

mg = kX  

5 × 9.81 = k(0.05)

k = 981 N/m

a)

Given that; Acceleration of the elevator a = 2 m/s² upwards

Extension of the spring in this situation = X1

Force exerted by the spring = F

we know that;

ma = F - mg

ma = kX1 - mg

we substitute

5 × 2 = 981 × X1 - (5 ×9.81 )

X1 = 0.06019 m

X1 = 60.19 mm

Therefore the spring  will stretch 60.19 mm with the same box attached as it accelerates upwards

B)

Acceleration of the elevator = a

The spring is relaxed i.e, it is not exerting any force on the box.

Only the weight force of the box is exerted on the box.

ma = mg

a = g

a = 9.81 m/s² downwards.

Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

6 0
2 years ago
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be obser
Alona [7]

Answer:

<h2><em>6000 counts per second</em></h2>

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

<em></em>

<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>

5 0
3 years ago
A car travels 100 m while decelerating to 8 m/s in 5 s.<br> a) What was its initial speed?
viktelen [127]

Answer:

Vi = 32 [m/s]

Explanation:

In order to solve this problem we must use the following the two following kinematics equations.

v_{f} =v_{i} - (a*t)\\

The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.

where:

Vf = final velocity = 8[m/s]

Vi = initial velocity [m/s]

a = acceleration = [m/s^2]

t = time = 5 [s]

Now replacing:

8 = Vi - 5*a

Vi = (8 + 5*a)

As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.

v_{f}^{2} = v_{i}^{2} - (2*a*d)

where:

d = distance = 100[m]

(8^2) = (8 + 5*a)^2 - (2*a*100)

64 = (64 + 80*a + 25*a^2) - 200*a

0 = 80*a - 200*a + 25*a^2

0 = - 120*a + 25*a^2

0 = 25*a(a - 4.8)

therefore:

a = 0 or a = 4.8 [m/s^2]

We choose the value of 4.8 as the acceleration value, since the zero value would not apply.

Returning to the first equation:

8 = Vi - (4.8*5)

Vi = 32 [m/s]

6 0
3 years ago
N an experiment, researchers want to determine if the __________ variable causes changes in the __________ variable.
chubhunter [2.5K]

C

independent; dependent

7 0
3 years ago
Read 2 more answers
the hydrometer with the density of liquid to be 800 kg metre per square is the volume of the submerged part of the hydrometer is
julia-pushkina [17]

Answer:

Mass = 0.04 Kg

Explanation:

Given the following data;

Density = 800 kg/m³

Volume = 5 * 10^{-5} m³

To find the mass of the object;

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the formula;

Density = \frac {mass}{volume}

Making mass the subject of formula, we have;

Mass = density * volume

Substituting the values into the formula, we have;

Mass = 800 * 5 * 10^{-5}

Mass = 0.04 Kg

8 0
3 years ago
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