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2 years ago

What is the frequency of microwaves of wavelength 3 cm?

Physics

1 answer:

tiny-mole [99]2 years ago

4 0

Answer:

10 GHz

Explanation:

Applying,

v = λf.................... Equation 1

Where v = speed of microwave, λ = wavelength, f = frequency.

make f the subject of the equation

f = v/λ................ Equation 2

Note: Microwave is an electromagnetic wave, and all electromagnetic wave have the same speed, which is 3×10⁸ m/s

From the question,

Given: λ = 3 cm = 0.03 m

Constant; v = 3×10⁸ m/s

Substitute these values into equation 2

f = 3×10⁸/(0.03)

f = 10¹⁰ Hz

f = (10¹⁰/10⁹) GHz

f = 10 GHz

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How would electron domains impact the magnetism of the substance?

Vanyuwa [196]

Magnetic domain structure is responsible for the magnetic behavior of ferromagnetic materials like iron, nickel, cobalt and their alloys, and ferrimagnetic materials like ferrite. ... Magnetic domains form in materials which have magnetic ordering; that is, their dipoles spontaneously align due to the exchange interaction.

4 0

2 years ago

John is carrying a shovelful of snow. The center of mass of the 3.00 kg of snow he is holding is 15.0 cm from the end of the sho

Andru [333]

Answer:

James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down

Explanation:

Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards

so here we can say

Upwards force = downwards Force + weight of snow

while if we find the other force which is acting downwards

then for that force we can say that net torque must be balanced

so here we have

$F_{down} L_1 = W_{snow} L_2$

so here we have

$F_{down} = \frac{L_2}{L_1} (W_{snow})$

so here we can say that upward force by which we push up is always more than the downwards force

8 0

2 years ago

A solid cylinder is released from the top of an inclined plane of height 0.81 m. From what height, in meters, on the incline sho

Jlenok [28]

Answer:

same 0.81m

Explanation:

in this problem if we assume there no resistance of any sort. and we apply the energy conservation

change in Potential energy = change in kinetic energy

mgh = 0.5mv^2

gh = 0.5v^2

the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.

so at the bottom

put h = 0.81m

9.81 * 0.81 * 2 = v^2

v=3.99 m/s

both CYLINDER and SPHERE will have same velocity at the bottom if released from the same height irrespective of shape and size

3 0

2 years ago

A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done

Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

$W = F*d$

<em>Where</em>:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

$F = W_{x} = mgsin(\theta)$

<em>Where:</em>

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²

θ: is the degree angle to the horizontal = 30°

$F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N$

Now, we can find the work:

$W = F*d = 24.53 N*20 m = 490.6 J$

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

6 0

2 years ago

Ricardo, of mass 80 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 30 kg canoe. When the canoe is at r

Deffense [45]

Answer:

m=57.65 kg

Explanation:

Given Data

Ricardo mass m₁=80 kg

Canoe mass m₂=30 kg

Canoe Length L= 3 m

Canoe moves x=40 cm

When Canoe was at rest the net total torque is zero.

Let the center of mass is at x distance from the canoe center and it will be towards the Ricardo cause. So the toque around the center of mass is given as

$m_{1}(L/2-x)=m_{2}x+m_{2}(L/2-x)$

We have to find m₂.To find the value of m₂ first we need figure out the value of.As they changed their positions the center of mass moved to other side by distance 2x.

2x=40

x=40/2

x=20 cm

Substitute in the above equation we get

$m_{x}=\frac{m_{1}(L/2-x)-m_{2}x }{L/2+x}\\m_{x}=\frac{80(\frac{3}{2}-0.2 )-30*0.2}{3/2+0.2}\\m_{x}=57.65 kg$

4 0

2 years ago