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LUCKY_DIMON [66]
3 years ago
15

I'LL GIVE ONE OF YOU BRAINLIEST ANSWER. PLEASE HELP ME

Physics
1 answer:
True [87]3 years ago
8 0
A) The magnitude of the electric field = (electric force)/(charge of particle)

Electric force = (charge of particle) (electric field magnitude)
= (2 C)(2.35 N/C) = 4.7 N

(electric force and electric field are always in the same direction)
so, force = 4.7 N south

b) Electric field magnitude = (electric force)/(particle charge)
= (6*10^-3 N)/(-4 C) = -0.0015 N/C = -1.5*10^-3 N/C east

For part d, I think I'll need to see part c first. 
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An astronaut's pack weighs 17.5 N when she is on earth, but only 3.24 N when she is on the surface of an asteroid. What is the a
Nimfa-mama [501]

Answer:

a= 1.81 m/s²

Explanation:

Ratio of the weights on both surfaces can be calculated as,

\frac{W_{a} }{W_{E} }=\frac{ma}{mg}

Here;

   Wa is the weight on the asteroid,       W_{E} is weight on earth,

         m is mass of pack,          a is acceleration due to an asteroid and

         g is acceleration due to gravity.

         Rearrange above equation for a,

           a=g (\frac{Wa}{W_{E} })

Substitute 3.24 N for Wa,       17.5 N for  W_{E},

          9.81 m/s² for g in the above equation,

          a= 9.81(\frac{3.24}{17.5})

          a= 1.81 m/s²

7 0
4 years ago
When the shuttle bus comes to a sudden stop to avoid hitting a dog, it decelerates uniformly at 4.5 m/s2 as it slows from 9.3 m/
saw5 [17]
V=u+at
V= speed 0 m/s
u = initial speed 9.3 m/s
a = acceleration - 4.5 (negative)
0=9.3-4.5t
t=9.3/4.5= ? :) seconds

3 0
3 years ago
A 3.80-m-long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building
Virty [35]

Answer:

Explanation:

The question is ;

3.80 m-long, 500. kg steel beam extends horizontally from the point

where it has been bolted to the framework of a new building under

construction. A 67.0 kg construction worker stands at the far end of

the beam. What is the magnitude of the torque about the point where the beam is bolted into place?

Solution:

Here two torques are in action

a) torque T1 due to weight of worker at the edge of the beam - at center of the beam

b) torque T2 due to weight of the uniform beam - at the point where the beam is bolted

So we first calculate the torque produced due to weight of the worker;

We can see that;

Distance of worker from the center of the beam = 1.9m

Mass of the worker = 67 kg

Value of g= 9.8 m/sec2

T1=force × distance from point of rotation

Here force is weight of the worker which is = mass × g=67×9.8=656.6 N

So the torque is

T1= 656.6×1.9=12225.892 Nm

or

T1 = 12225.892 Nm

Now torque by the beam itself ;

Length of the beam from its center point to the bolt point = 1.9 m

Weight force of beam acting at center point= mass× g= 500×9.8=4900 N

Torque T2 at bolt point by the beam weight = weight of beam × length of beam from its center to the bolt point = 4900×1.9=9310 Nm

T2=9310Nm

So total torque = T1+T2= 12225.892+9310=21565.892 Nm

8 0
3 years ago
PLZ HELP ME FAST! How does the placement of the metal coil closer or further away from the magnet affect the motor’s strength? W
love history [14]

A very long solenoid has a magnetic field inside which depends on the current in the wire and the number of turns per unit length of the solenoid. ... Adding more turns in the coil can also increase the total resistance of the wire.

3 0
3 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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