2 years ago

1. Compare and Contrast microwaves with visible light using wavelength, frequency and energy

Physics

1 answer:

coldgirl [10]2 years ago

4 0

Visible light has more energy than microwaves but it has a smaller wavelength than microwave

You might be interested in

A MEMS-based accelerometer has a mass of m = 2 grams, an equivalent spring constant of k = 5 N/m, and an equivalent damping coef

pychu [463]

Answer:

The natural frequency = 50 rad/s = 7.96 Hz

Damping ratio = 0.5

Explanation:

The natural frequency is calculated in this manner

w = √(k/m)

k = spring constant = 5 N/m

m = mass = 2 g = 0.002 kg

w = √(5/0.002) = 50 rad/s

w = 2πf

50 = 2πf

f = 50/(2π) = 7.96 Hz

Damping ratio = c/[2√(mk)] = 0.1/(2 × √(5 × 0.002)) = 0.5

5 0

3 years ago

Bart stole a watermelon and ran 5,000 feet from the cops and they chase lasted 0.1 hours how fast was Bart running in miles per

liraira [26]

I am not as sure but I think it is 9.469 miles

5 0

2 years ago

what is the magnitude of g at a height above earth surface where free fall acceleration equal 6.5m/s2

ZanzabumX [31]

The magnitude of gravity is expressed in terms of its acceleration. So the magnitude of ' g ' at that altitude is exactly 6.5 m/s^2.

8 0

3 years ago

An object experiences an impulse, moves and attains a momentum of 200 kg·m/s. If its mass is 50 kg, what is its velocity?

Nady [450]

Answer:

4 m/s

Explanation:

Momentum is defined as:

$p=mv$

where

m is the mass of the object

v is its velocity

For the object in this problem, we know:

p = 200 kg m/s is the momentum

m = 50 kg is the mass

Solving for the velocity, we find:

$v=\frac{p}{m}=\frac{200}{50}=4 m/s$

8 0

2 years ago

A physics professor is pushed up a ramp inclined upward at 30.0° above the horizontal as she sits in her desk chair, which slide

11111nata11111 [884]

Answer:

V = 3.17 m/s

Explanation:

Given

Mass of the professor m = 85.0 kg

Angle of the ramp θ = 30.0°

Length travelled L = 2.50 m

Force applied F = 600 N

Initial Speed u = 2.00 m/s

Solution

Work = Change in kinetic energy

$F_{net}d = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}\\\frac{2F_{net}d }{m} = v^{2} -u^{2}\\ v^{2} =\frac{2F_{net}d }{m} +u^{2}\\ v^{2} =\frac{2(600cos30 - 85\times 9.8 \times sin30) \times 2.5 }{85} +2.00^{2}\\ v^{2} = 10.066\\v = 3..17m/s$

7 0

3 years ago