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Furkat [3]
3 years ago
14

In comparison to radio waves, visible light has:

Physics
1 answer:
Alexxx [7]3 years ago
8 0

Answer:

Visible light has a shorter wavelength than radio waves

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MRU(movimiento rectilineo uniforme) un movil viaja en linea recta con una velocidad media de 12metros /segundos durante 9segundo
liubo4ka [24]

Answer:

a) 141.6m

b) 8.4m/s

Explanation:

a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:

x_1=v_1t_1=(12m/s)(9s)=108m\\\\x_2=v_2t_2=(4.8m/s)(7s)=33.6m\\\\x_T=x_1+x_2=108m+33.6m=141.6m

hence, the total distance is 141.6m

b) the mean velocity of the total trajectory is given by:

v_m=\frac{v_1+v_2}{2}=\frac{12m/s+4.8m/s}{2}=8.4\frac{m}{s}

hence, the mean velocity is 8.4 m/s

8 0
3 years ago
A mover loads a 100 kg box into the back of a moving truck by
hammer [34]

Explanation:

Mechanical Advantage (MA)

MA=d1d2=FoutFin ; d1 is the distance of effort, d2 is the distance the object is moved

7 0
3 years ago
A ball with an initial velocity of zero is dropped from a table onto the ground. After 2 s, the
GenaCL600 [577]
  • Initial velocity=u=0m/s
  • Acceleration=a=10m/s^2
  • Time=t=2s

Final velocity=v

\\ \sf\longmapsto v=u+at

\\ \sf\longmapsto v=0+10(2)

\\ \sf\longmapsto v=20m/s

5 0
3 years ago
Read 2 more answers
Is sand a dependent or independent variable
anastassius [24]

Answer:

dependent variable

Explanation: Because sand changes when water touches it therefore it would be considered a dependent variable

6 0
3 years ago
Read 2 more answers
A 2.2 m long simple pendulum oscillates with a period of 4.8 s on the surface of unknown planet. What is the surface gravity of
Vsevolod [243]

Answer:

g=3.76\ m/s^2

Explanation:

Given that,

The length of a simple pendulum, l = 2.2 m

The time period of oscillations, T = 4.8 s

We need to find the surface gravity of the planet. The time period of the planet is given by the relation as follows :

T=2\pi \sqrt{\dfrac{l}{g}} \\\\T^2=4\pi ^2\times \dfrac{l}{g}\\\\g=\dfrac{4\pi ^2 l}{T^2}

Put all the values,

g=\dfrac{4\pi ^2 \times 2.2}{(4.8)^2}\\\\=3.76\ m/s^2

So, the value of the surface gravity of the planet is equal to 3.76\ m/s^2.

7 0
3 years ago
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