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Julli [10]
3 years ago
11

You have a large metal sphere that starts out isolated and has no net charge. Then you bring a positively charged rod near it wi

thout touching.
Will there be a force between the sphere and rod?

Draw a qualitative diagram showing the location of the charges in the metal sphere once the rod is brought near.

Physics
1 answer:
liq [111]3 years ago
6 0

Answer:

Yes

Explanation:

There will be a force because the large metal sphere is polarized when the rod is placed close to it, this happens because the metal sphere has free electrons and they will be attracted by the positively charged rod, the sphere will remain neutral as a whole (the same number of negative and positive charge), but a portion of the sphere will become more positive and the other more negative, creating this force.

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0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary for
Paladinen [302]

Assuming the accleration applied was constant, we have

v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)

\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)

\implies a=200\,\dfrac{\mathrm m}{\mathrm s^2}

Then the force applied to the ball is given by

F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)

\implies F=90\,\dfrac{\mathrm{kg}\,\mathrm m}{\mathrm s^2}=90.\,\mathrm N

8 0
3 years ago
A string that is 3.6 m long is tied between two posts and plucked. The string produces a wave that has a frequency of 320 Hz and
marin [14]

To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as

\lambda = \frac{v}{f}

Where,

v = Velocity

f = Frequency,

Our values are given as

L = 3.6m

v= 192m/s

f= 320Hz

Replacing we have that

\lambda = \frac{192}{320}

\lambda = 0.6m

The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,

N = \frac{L}{\lambda}

N = \frac{3.6}{0.6}

N = 6

Therefore the number of wavelengths of the wave fit on the string is 6.

5 0
3 years ago
A firecracker in a coconut blows the coconut into three pieces. two pieces of equal mass fly off south and west, perpendicular t
netineya [11]
Here we have mass that moves at ceratin speed. This means that we have momentum. The law that must be observed is law of conservation of momentum. It states that momentum before certain event is equal to a momentum after that event. Here we have three masses so we can write this as:
m_{1}  v_{1i} + m_{2}  v_{2i} + m_{3}  v_{3i} = m_{1}  v_{1f} + m_{2}  v_{2f} + m_{3}  v_{3f}
Before the firecracker blows a coconut does not move, so left side is equal to 0:
0 = m_{1} v_{1f} + m_{2} v_{2f} + m_{3} v_{3f}

We know that m1=m2=m and m3=2m. Also we are asked to find v3f so we can rewrite formula:
v_{3f} = -  \frac{m_{1}  v_{1f}  + m_{2} v_{2f} }{ m_{3} }

We must take in consideration that two parts with same mass do not move in same direction. The center of mass of these two parts moves between them at angle of 45° with respect to both south and west. The speed of a center of mass is:
v_{f} = \sqrt{ v_{1f}^{2}+ v_{2f}^{2} } \\ \\ v_{f} = 33.9m/s
This speed we can insert into formula for v3f:
v_{3f} = - \frac{m*33.9+m*33.9 }{ 2m } \\  \\ v_{3f} = - \frac{2m*33.9 }{ 2m }  \\  \\ v_{3f} = - 33.9m/s

We can see that part of a coconut with biggest mass has same speed as center of mass of two other parts. Negative sign shows that direction is opposite to direction of two pats. Biggest part moves towards north-east.
8 0
3 years ago
Read 2 more answers
6^5/2^3= what is the answer please​
kogti [31]

Answer:

972

Explanation:

6^5/2^3

=7776/8

=972

6 0
3 years ago
Read 2 more answers
Solve each problem. Be sure to show your work and give a final answer with units rounded to the given number of significant figu
Alex Ar [27]

Answer:

1. the voltage will be 2.35×12.5 = 29.4V

2. the resistance would be 9.0/6.2= 1.45ohms

3. in series they will add up thus 4+8+12= 24ohms

4. in parallel it will be 2.18ohms

8 0
3 years ago
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