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MrRa [10]
4 years ago
9

Two charged bees land simultaneously on flowers that are separated by a finite distance. For a few moments, the charged bees res

t on the flowers. The charged bees both generate an electric field, and while the charged bees are resting on the flowers, the net electric field at some distance between them is zero. (a) Do the bees have the same or opposite signs of charge?
Same, the electric fields point in opposite directions and therefore cancel at some midpoint.
Same, the electric fields multiply together to equal zero.
Opposite, the electric fields point in the same direction summing to zero.
Opposite, the net electric field due to the two bees points in a direction perpendicular to the direction from one bee to the other.
Physics
1 answer:
damaskus [11]4 years ago
3 0

Answer:

Same, the electric fields point in opposite directions and therefore cancel at some midpoint.

Explanation:

The Electric field net at some point between them is zero, only if they point in opposite direction (they cancel to the each other). In order the electric fields  have opposite direction, at some point between the bees , the bees must have the same sign of electric charge

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Two hypothetical planets of masses m1 and m2 and radii r1 and r 2 , respectively, are nearly at rest when they are an infinite d
Leto [7]

Answer:

(a) v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }

v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }

(b) Kinetic Energy of planet with mass m₁, is KE₁ =  1.068×10³² J

Kinetic Energy of planet with mass m₂, KE₂ =  2.6696×10³¹ J

Explanation:

Here we have when their distance is d apart

F_{1} = F_{2} =G\frac{m_{1}m_{2}}{d^{2}}

Energy is given by

Energy \,of \,attraction = -G\frac{m_{1}m_{2}}{d}}+\frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2

Conservation of linear momentum gives

m₁·v₁ = m₂·v₂

From which

v₂ =  m₁·v₁/m₂

At equilibrium, we have;

G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2       which gives

2G{m_{1}m_{2}}= d m_{1} v^2_1+  dm_{2} (\frac{m_1}{m_2}v_1)^2= dv^2_1(m_1+(\frac{m_1}{m_2} )^2)

multiplying both sides by m₂/m₁, we have

2Gm^2_{2}}= dv^2_1 m_2+dm_1v^2_1 =dv^2_1( m_2+m_1)

Such that v₁ = \sqrt{\frac{2Gm^2_2}{d(m_1+m_2)} }

v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }

Similarly, with v₁ =  m₂·v₂/m₁, we have

G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2\Rightarrow  2G{m_{1}m_{2}}= dm_{1} (\frac{m_2}{m_1}v_1)^2 +d m_{2} v^2_2= dv^2_2(m_2+(\frac{m_2}{m_1} )^2)

From which we have;

2G{m^2_{1}}= dm_{2} v_2^2 +d m_{1} v^2_2 and

v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }

The relative velocity = v₁ + v₂ =v_1+v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} } + m_2\sqrt{\frac{2G}{d(m_1+m_2)} } = (m_1+m_2)\sqrt{\frac{2G}{d(m_1+m_2)} }

v₁ + v₂ = (m_1+m_2)\sqrt{\frac{2G}{d(m_1+m_2)} }

(b) The kinetic energy KE = \frac{1}{2}mv^2

KE_1= \frac{1}{2} m_{1} v^2_1 \, \, \, KE_2= \frac{1}{2} m_{2} v^2_2

Just before they collide, d = r₁ + r₂ = 3×10⁶+5×10⁶ = 8×10⁶ m

v_1 = 8\times10^{24}\sqrt{\frac{2\times6.67408 \times 10^{-11}} {8\times10^6(2.00\times10^{24}+8.00\times10^{24})} } = 10333.696 m/s

v_2 = 2\times10^{24}\sqrt{\frac{2\times6.67408 \times 10^{-11}} {8\times10^6(2.00\times10^{24}+8.00\times10^{24})} } =2583.424 m/s

KE₁ = 0.5×2.0×10²⁴× 10333.696² =  1.068×10³² J

KE₂ = 0.5×8.0×10²⁴× 2583.424² =  2.6696×10³¹ J.

7 0
3 years ago
The gravitational force on a baseball is 1.4 N. What is its mass? Round to two decimal places.
valina [46]

Answer:

0.14 kg

Explanation:

W = mg

1.4 N = m (9.8 m/s²)

m = 0.14 kg

8 0
3 years ago
A table of mass 10 kg is lifted so that the gravitational potential energy of the table increases by 1470 J. How high is the tab
Kryger [21]
15 meters due to the fact that gravitational potential energy equals mass times gravity times height.
4 0
3 years ago
Read 2 more answers
8. Why would ChiN want its own currency to be undervalued relative to the U.S. dollar? How does ChiN maintain an undervalued cur
tekilochka [14]

Answer:

The Chinese government wish to keep the currency undervalued because: A weaker exchange rate makes exports more competitive and increases demand for Chinese exports. Chinese economic growth is dependent on exports, so the value of the currency plays a key role in boosting growth.

Explanation:

5 0
2 years ago
The speed of a wave on a violin A string is 288 m/s and on the G string is 128 m/s. The force exerted on the ends of the string
Katyanochek1 [597]

Answer:

\dfrac{\mu_A}{\mu_G}=0.197

Explanation:

given,

Speed of a wave on violin A = 288 m/s

Speed on the G string = 128 m/s

Force at the end of string G  = 110 N

Force at the end of string A = 350 N

the ratio of mass per unit length of the strings (A/G). = ?

speed for string A

 v_A = \sqrt{\dfrac{F_A}{\mu_A}}.......(1)

speed for string G

 v_G = \sqrt{\dfrac{F_G}{\mu_G}}........(2)

Assuming force is same in both the string

now,

dividing equation (2)/(1)

\dfrac{v_G}{v_A}=\dfrac{\sqrt{\dfrac{F_G}{\mu_G}}}{\sqrt{\dfrac{F_A}{\mu_A}}}

\dfrac{v_G}{v_A}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}

\dfrac{128}{288}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}

\dfrac{\mu_A}{\mu_G}=0.197

5 0
3 years ago
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