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miv72 [106K]
3 years ago
15

The technological challenges of installing cellphones across the globe

Physics
1 answer:
Leona [35]3 years ago
3 0

Answer:

I don't know sorry For this question

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What is the acceleration of a dog that runs from 3 m/s to 6 m/s over a distance of 90m?
KonstantinChe [14]

Answer:

solution given:

acceleration (a)=?

initial velocity (u)=3m/s

final velocity (v)=6m/s

distance (s)=90m

we have

v²=u²+2as

substituting value

6²=3²+2*a*90

36=9+180a

36-9=180a

a=25/180

<u>a=0.1388m/s²</u>

6 0
3 years ago
How much thermal energy is needed to melt 1.25 kg of water at its melting point? Use Q = masslaten heat of fusion.
Amanda [17]

Answer:

Latent heatnof fusion = 417.5 J

Explanation:

Specific latent heat of fusion of water is 334kJ.kg-1.

The heat required to melt water when it's ice I called latent heat because there is no temperature change, the only change observed is change in physical structure.

The amount of heat required to change 1 kg of solid to its liquid state (at its melting point) at atmospheric pressure is called Latent heat of Fusion.

Latent heat = ML

Latent heat= 1.25 kg * 334kJ.kg-1

Latent heat = 1.25*334 *(J/kg)*kg

Latent heat = 417.5 J

8 0
3 years ago
An electron and a second particle both move in circles perpendicular to a uniformmagnetic field. The mass of the second particle
Katarina [22]

Answer:

The change on the second particle is 2.93\times 10^{-16}\ C.

Explanation:

The period of revolution of the particle in the magnetic field is given by the formula as follows :

T=\dfrac{2\pi m}{Bq}

It is given that the magnetic field is uniform. The mass of the second particle is the same as that of a proton but thecharge of this particle is different from that of a proton.

m_s=m_p

If both particles take the same amount of time to go once around their respective circles. So,

T_e=T_s\\\\\dfrac{2\pi m_e}{Bq_e}=\dfrac{2\pi m_s}{Bq_s}\\\\\dfrac{m_e}{q_e}=\dfrac{m_p}{q_s}\\\\q_s=\dfrac{m_pq_e}{m_e}\\\\q_s=\dfrac{1.67\times 10^{-27}\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}\\\\q_s=2.93\times 10^{-16}\ C

So, the change on the second particle is 2.93\times 10^{-16}\ C.

7 0
3 years ago
PLEASE HELP I WILL GIVE MAX POINTS
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you can collect water and shine a light though it and record your finings

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3 years ago
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Which of the following is a physical property of wood
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The density, hard, strong, and rough.
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