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Ira Lisetskai [31]
3 years ago
9

You describe a friend’s position by including distance, direction, and what other term?

Physics
2 answers:
garik1379 [7]3 years ago
5 0

Answer:

Acceleration

Explanation:

Kipish [7]3 years ago
3 0

Answer: Acceleration

Explanation:

You might be interested in
1. A body has a velocity of 72 km/hr. Find its value in m/s.
CaHeK987 [17]

72 Km/hr

= 72000 m/ 60×60 s

= 72000 m/ 3600 s

= 20 m/s

Answer is 20 m/s.

Hope it helps! Please do comment

8 0
3 years ago
Deduce the relation between kwh and joule define 1 watt​
dimulka [17.4K]

Explanation:

We know that kWh is a commercial unit of energy.

Also, 1 Joule = 1 watt × 1 seconds

1\ kWh=1kW \times 1\ h

Since ,1 h = 3600 s and 1kW = 1000 W

1\ kWh=1000\ W \times 3600\ s\\\\1\ kWh = 36\times 10^5\ W-s\\\\1\ kWh=36\times 10^5\ J

So, the relation is : 1\ kWh=36\times 10^5\ J

1 watt is defined as the power of an appliance when the energy is transferred at a rate of 1 Joule per second.

5 0
3 years ago
Name 2 things centripetal force acts on.
ira [324]

The centripetal force acts upon an object moving in a circle at constant speed.  The centripetal force acts perpendicular to the direction of motion , the speed of object will remain constant.

3 0
3 years ago
What phase are daughter cells in as a result of mitosis
NemiM [27]
They are in interphase
5 0
4 years ago
(a) Calculate the number of free electrons per cubic meter for gold assuming that there are 1.5 free electrons per gold atom. Th
KonstantinChe [14]

Answer:

Part A:

n=8.85*10^{28}m^{-3}

Part B:

Electron Mobility=3.03*10^{-3} m^2/V

Explanation:

Part A:

To calculate the number of free electrons n we use the following formula::

n=1.5N-Au

Where N-Au is number of gold atoms per cubic meter

N-Au=\frac{Density*Avogadro Number}{atomic weight}

Density = 19.32g/cm^3

Avogadro Number=6.02*10^{23} atoms/mol

Atomic weight=196.97g/mol

So:

n=1.5*\frac{Density*Avogadro Number}{atomic weight}

n=1.5*\frac{19.32*6.02*10^{23}}{196.97}

n=8.85*10^{28}m^{-3}

Part B:

Electron Mobility=\frac{Elec-conductivity}{n * charge on electron}

n is calculated above which is 8.85*10^{28}m^{-3}

Charge on electron=1.602*10^{-19}

Elec- Conductivity= 4.3*10^{7}

Electron Mobility=\frac{4.3*10^{7}}{ 8.85*10^{28} * 1.602*10^{-19}}

Electron Mobility=3.03*10^{-3} m^2/V

4 0
4 years ago
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