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lianna [129]
3 years ago
12

A positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away

from the rod we can conclude: A) the object is positively charged B) the object is negatively charged C) the object is an insulator D) the object is a conductor E) none of the above Two small charged objects repel each other with a force F when separated by a distance d. If the charge on each object is reduced to one-fourth of its original value and the distance between them is reduced to d/2 the force becomes: A) F/16 B) F/8 C) F/4 D) F/2 E) F
Physics
1 answer:
Trava [24]3 years ago
3 0

Answer

A)   Positively charged two insulating rod are brought closed to an object  they repel each other. It means the Object is positively charged. Because similar charge repel each other.

   The correct answer is Option A.

B) we know force between to charges is calculated using Formula

         F = \dfrac{kQ_1Q_2}{r^2}.......(1)

   form the given condition in the question

          F' = \dfrac{k\dfrac{Q_1}{4}\dfrac{Q_2}{4}}{(\dfrac{r}{2})^2}

          F' = \dfrac{k\dfrac{Q_1}{4}\dfrac{Q_2}{4}}{(\dfrac{r^2}{4})}

          F' = \dfrac{4}{16}\dfrac{kQ_1Q_2}{r^2}

from equation (1)

          F' = \dfrac{F}{4}

hence, the correct answer is Option C.

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A car of mass 1200 kg traveling westward at 30. m/s is slowed to a stop in a distance of 50. m by the car’s brakes. What was the
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How many atoms are in 2Mg3(PO4)2?
gogolik [260]

Answer:

13 atoms

Explanation:

Look at the subscripts.

“Mg₃” means that there are 3 magnesium atoms; “PO₄” means that there is 1 phosphorus atom and 4 oxygen atoms. But note that the PO₄ group has a subscript too: 2. This multiplies the atom count by 2, so you have 2 phosphorus atoms and 8 oxygen atoms.

3 + 2 + 8 = 13

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8 0
3 years ago
A potential difference of 300 volts is applied to a 2.0-µf capacitor and an 8.0-µf capacitor connected in series. (a) What are t
zubka84 [21]

Answer:

a) Q1=Q2=480μC   V1=240V   V2=60V

b) Q1=96μC   Q2=384μC   V1=V2=48V

c) Q1=Q2=0C    V1=V2=0V

Explanation:

Let C1 = 2μC  and  C2=8μC

For part (a) of this problem, we know that charge in a series circuit, is the same in C1 and C2. Having this in mind, we can calculate equivalent capacitance first:

C=\frac{1}{\frac{1}{C1} +\frac{1}{C2} } = 1.6\mu C

Q_{T} = V_{T}*C_{T} = 480\mu C

V1 = \frac{Q1}{C1}=240V

V2 = \frac{Q2}{C2}=60V

For part (b), the capacitors are in parallel now. In this condition, the voltage is the same for both capacitors:

V1=V2   So,  \frac{Q1}{C1} = \frac{Q2}{C2}

Total charge is the same calculated for part (a), so:

\frac{Qt - Q2}{C1} = \frac{Q2}{C2}   Solving for Q2:

Q2 = 384μC    Q1 = 96μC.

Therefore:

V1=V2=48V

For part (c), both capacitors would discharge, since their total voltage of 300V would by applied to a wire (R=0Ω). There would flow a huge amount of current for a short period of time, and capacitors would be completely discharged. Q1=Q2=0C  V1=V2=0V

3 0
3 years ago
A 2 200-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.20 m before coming into contact
Kisachek [45]

Answer:

669042 N upward

Explanation:

W_g = Work done by gravity on pile driver

W_b = Work done by beam

u = Initial velocity

v = Final velocity

m = Mass of pile driver = 2200 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height the pile driver falls = 4.2 m

d = Bend in beam = 14 cm

Total work done = Change in Kinetic energy

W_g+W_b=\frac{1}{2}m(v^2-u^2)\\\Rightarrow mg(h+d)+Fd=0\\\Rightarrow F=-\frac{mg(h+d)}{d}\\\Rightarrow F=-\frac{2200\times 9.81(4.2+0.14)}{0.14}\\\Rightarrow F=-669042\ N

The force the beam exerts on the pile driver while the pile driver is brought to rest is 669042 N upward

8 0
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