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xenn [34]
3 years ago
12

Cellular phones use _____?

Physics
2 answers:
mina [271]3 years ago
6 0

3. D-Microwaves

1. C-electromagnetic spectrum

2. B-radio waves

^^(for those doing the three question physical science test)

dangina [55]3 years ago
3 0
Let's see: frequency of cellular phone waves (GSM phones) is (800-1900 MHz). If we look at the table of the electromagnetic spectrum, we can see that this range is contained within the frequencies of the microwaves, which include waves in the range 300 MHz-300 GHz.

So, summarizing, the correct answer is "microwaves".
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A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point
seropon [69]

since centripetal acceleration is always towards the center of the circle

so at the given position where speed and acceleration is given the center coordinate will be towards the center of circle

also we know that

a_c = \frac{v^2}{r}

11.6 = \frac{2.5^2}{R}

R = 0.54 m

so the coordinates of the center will be

(x, y) = (3.20 , (3.5 + 0.54))

(x, y) = (3.20, 4.04)

so the coordinate is (3.20 m, 4.04 m)

6 0
3 years ago
What happens to light waves from a star as the star moves toward Earth?
Musya8 [376]
This shifts the star’s spectral lines toward the blue end of the spectrum. If the star is moving away from us, it’s waves are effectively stretched out when they reach earth, increasing their wavelength. This shifts the star’s spectral lines toward the red end of the spectrum.
3 0
4 years ago
Read 2 more answers
A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.64 10-2 kg/s. the density of the gasoline is 735
Irina18 [472]
Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m

Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²

Let v =  speed of pumping the gasoline, m/s
Then the mass flow rate is 
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s

The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s

Answer:  2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
7 0
4 years ago
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A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a
GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}

T = 7.83 X10⁻⁷ s

6 0
3 years ago
Please i need your help w/h/w you make the following measurements if an object:42kg, and 22m​
Rufina [12.5K]

Answer:

1.9 kg/m^3

Explanation:

The density of an object is given by

d=\frac{m}{V}

where

m is the mass of the object

V is its volume

In this problem,

m = 42 kg

V = 22 m^3

Substituting into the equation, we find the object's density:

d=\frac{42 kg}{22 m^3}=1.9 kg/m^3

8 0
3 years ago
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