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Valentin [98]
3 years ago
9

The type of waves that transmit cell phone messages are most likely

Physics
2 answers:
alukav5142 [94]3 years ago
8 0
They're high frequency radio waves. They're in the category of UHF ... Not quite high enough in frequency to be called microwaves.
Softa [21]3 years ago
6 0
Cell phones use RF waves from nearby cell towers. RF waves fall in between FM radio waves and microwaves
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Consider a mass initially moving at 7.50 m/s. How does it take to move 3.5 km (Be sure to convert to meters) if it accelerates a
Tomtit [17]
The mass is moving by uniformly accelerated motion, with initial velocity v_i=7.50 m/s and acceleration a=0.55 m/s^2. Its position at time t is given by the following law:
x(t)=v_i t + \frac{1}{2}at^2
where we take the initial position x_i=0 since we are only interested in the distance traveled by the mass.

If we put x(t)=d=3.5 km=3500 m into the equation, the corresponding time t is the time it takes for the mass to travel this distance:
\frac{1}{2}at^2+v_it-d=0
4.9t^2+7.5t-3500 =0
And the two solutions for the equation are:
t=-25.5 s --> negative, we can discard it
t=24.0 s --> this is the solution to our problem
5 0
4 years ago
If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits___, ____, ____
nalin [4]

So, the complete sentence is If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits<u> 84, 1, </u>and <u>5</u>, and all planets will orbit the sun successfully for the best conditions.

When the mass of the sun is larger, Earth moves around the sun at a faster pace and When the mass of the sun is smaller, Earth moves around the sun at a slower pace.

When Earth is closer to the sun, its orbit becomes faster and When Earth is farther from the sun, its orbit becomes slower.

When Earth is closer to the sun, there will be a hotter climate. A little movement that takes one closer to the sun could lead to a huge impact, as the sun is very hot.

So, it can be concluded that If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits<u> 84, 1, </u>and <u>5</u>, and all planets will orbit the sun successfully for the best conditions.

Learn more about Sun here:

brainly.com/question/15837114

#SPJ10

6 0
2 years ago
A quantity of a gas has an absolute pressure of 400 kPa and an absolute temperature of 110 degrees kelvin. When the temperature
KiRa [710]

Answer:

A) 854.46 kPa

Explanation:

P₁ = initial pressure of the gas = 400 kPa

P₂ = final pressure of the gas = ?

T₁ = initial temperature of the gas = 110 K

T₂ = final temperature of the gas = 235 K

Using the equation

\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}

Inserting the values

\frac{400}{110}=\frac{P_{2}}{235}

P₂ = 854.46 kPa

3 0
3 years ago
A steel sphere sits on top of an aluminum ring. The steel sphere (a= 1.1 x 10^-5/degrees celsius) has a diameter of 4.000 cm at
Mars2501 [29]

Answer:

C

Explanation:

To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):

d = d₀ + d₀αT

for the sphere, we were given

D₀ = 4.000 cm

α = 1.1 x 10⁻⁵/degrees celsius

we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T             EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

α = 2.4 x 10⁻⁵/degrees celsius

we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T       EQN 2

Since @ the temperature T at which the sphere fall through the ring, d=D

Eqn 1 = Eqn 2

4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms

0.006=5.18x10⁻⁵T

T=115.7K

4 0
3 years ago
A laser beam of wavelength 600 nm is incident on two slits that are separated by 0.02 mm. What is the separation between adjacen
Liula [17]

Answer:

option D

Explanation:

given,

wavelength = 600 nm

width of separation = 0.02 mm

L = 5 m

for mth order maxima

d \times \dfrac{y_m}{L}=m\lambda

for (m+1)th order maxima

d \times \dfrac{y_{m+1}}{L}=(m+1)\lambda

now,

y_m=\dfrac{mL\lambda}{d}      and

y_{m+1}=\dfrac{(m+1)L\lambda}{d}

hence,

\Delta y = y_{m+1} - y_m

\Delta y =\dfrac{L\lambda}{d}

\Delta y =\dfrac{5 \times 600 \times 10^{-9}}{0.02 \times 10^{-3}}

\Delta y =0.15\ m

\Delta y =15\ cm

hence, the correct answer is option D

4 0
4 years ago
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