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cupoosta [38]
3 years ago
11

¿Por qué los mensajes llegan en forma inmediata de un celular a otro?

Physics
1 answer:
KATRIN_1 [288]3 years ago
7 0

Porque el internet ayuda a la rapidez del mensaje.

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The greater the mass of an object
lesya [120]

Answer:

The greater the velocity, the greater the Force needs to be, and the greater the fiction is

Explanation:

I don't know what you are working on so here are a few responses

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3 years ago
Why do people cover their flowers on a cool night. Think about radiation absorption conduction and temperature
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When the temperature lowers the plants can freeze and die. One way to prevent that is to cover the plants so they dont freeze 

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3 years ago
Yellow-green light has a wavelength of 560 nm. What is its frequency?
Natasha2012 [34]
5.4 x 1014Hz
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3 years ago
Read 2 more answers
Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th
alex41 [277]

Answer: 321 J

Explanation:

Given

Mass of the box m=3\ kg

Force applied is F=25\ N

Displacement of the box is s=15\ m

Velocity acquired by the box is v=6\ m/s

acceleration associated with it is a=\dfrac{F}{m}

\Rightarrow a=\dfrac{25}{3}\ m/s^2

Work done by force is W=F\cdot s

W=25\times 15\\W=375\ J

change in kinetic energy is \Delta K

\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J

Therefore, the magnitude of work done by friction is 321\ J

3 0
3 years ago
A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.
Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

\Delta K = \Delta E_{p} = q\Delta V

<u>Where</u>:

K: is the kinetic energy

E_{p}: is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV                                    

\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J

Now, the kinetic energy of the particle as it moves through point B is:

\Delta K = K_{f} - K_{i}

K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!      

8 0
3 years ago
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