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Ulleksa [173]
3 years ago
9

Two 4.0 kg masses are 1.0 m apart on a frictionless table. Each has 1.0 μC of charge.What is the magnitude of the electric force

on one of the masses?
Physics
1 answer:
xeze [42]3 years ago
7 0

Coulomb's law:

  Force = (<span>8.99×10⁹ N m² / C²<span>) · (charge₁) · (charge₂) / distance²

            = (</span></span><span>8.99×10⁹ N m² / C²<span>) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C)  / (1.0 m)²

            = (8.99×10⁹ x 1×10⁻¹² / 1.0)  N

            =      8.99×10⁻³  N

            =        0.00899 N repelling.

Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.

-- '4.0 kg masses';  don't need it. 
   Mass has no effect on the electric force between them.

-- 'frictionless table';  don't need it.
   Friction has no effect on the force between them,
only on how they move in response to the force.
</span></span>
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what is the orbital speed for a satellite 3.5 x 10^8m from the center of mars? Mars mass is 6.4 x 10^23 kg
adell [148]

Answer:

v = 349.23 m/s

Explanation:

It is required to find the orbital speed for a satellite 3.5\times 10^8\ m from the center of mass.

Mass of Mars, M=6.4\times 10^{23}\ kg

The orbital speed for a satellite is given by the formula as follows :

v=\sqrt{\dfrac{GM}{r}} \\\\v=\sqrt{\dfrac{6.67\times 10^{-11}\times 6.4\times 10^{23}}{3.5\times 10^8}} \\\\v=349.23\ m/s

So, the orbital speed for a satellite is 349.23 m/s.

6 0
3 years ago
A force of 3600 N is exerted on a piston that has an area of 0.030 m2. What force is exerted on a second piston that has an area
Airida [17]
For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
p_1 = p_2
which becomes
\frac{F_1}{A_1}= \frac{F_2}{A_2}
where
F_1=3600 N is the force on the first piston
A_1=0.030 m^2 is the area of the first piston
F_2 is the force on the second piston
A_2=0.015 m^2 is the area of the second piston

If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
F_2=F_1  \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N
7 0
3 years ago
Read 2 more answers
A long solenoid consists of 1700 turns and has a length of 0.75 m.The current in the wire is 0.48 A. What is the magnitude of th
Travka [436]

Answer:

1.37 ×10^-3 T

Explanation:

From;

B= μnI

μ = 4π x 10-7 N/A2

n= number of turns /length of wire = 1700/0.75 = 2266.67

I= 0.48 A

Hence;

B= 4π x 10^-7 × 2266.67 ×0.48

B= 1.37 ×10^-3 T

4 0
3 years ago
A mountain-climber friend with a mass of 74 kg ponders the idea of attaching a helium-filled balloon to himself to effectively r
bazaltina [42]

Answer:

V=16.65 m^3

Explanation:

The volume of the balloon can be find compared the force in each cases so:

reduce 25% from 74kg

R=\frac{25}{100}*74kg=18.5kg

So the net force uproad on the balloon is

F_b=18.5kg*g

Now the density of the both gases air and helium are different however the volume is the same change offcorss the mass so:

P_h=\frac{m}{V}=0.179 kg/m^3

P_A=1.29 kg/m^3

F_b=F_A-F_H

F_b=m_a*g-m_h*g

m=P/V

18.5kg*g=(1.29kg/m^3-0.179kg/m^3*)V*g

V=\frac{18.5kg}{(1.29-0.179)kg/m^3}

V=16.65 m^3

4 0
3 years ago
Fill in the blanks for the following:
storchak [24]

Answer:

<em>a. 4.21 moles</em>

<em>b. 478.6 m/s</em>

<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm-mol^{-1}K^{-1}

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = <em>4.21 moles</em>

The equation for root mean square velocity is

Vrms = \sqrt{\frac{3RT}{M} }

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = \sqrt{\frac{3*8.314*293}{0.0319} }= <em>478.6 m/s</em>

<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>

\frac{Voxy}{Vnit} = \sqrt{\frac{Mnit}{Moxy} }

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

\frac{478.6}{Vnit} = \sqrt{\frac{14.0}{31.9} }

\frac{478.6}{Vnit} = 0.66

Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>

<em>the root mean square velocity of the oxygen gas is </em>

<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

6 0
3 years ago
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