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eimsori [14]
3 years ago
8

In the figure a block slides along a track from one level to a higher level, by moving through an

Physics
1 answer:
Scrat [10]3 years ago
7 0

Answer:

(B) 1.17 m.

Explanation:

At the highest point, the conservation of energy equation would becomes;

K.E (initial ) = K.E + P.E;

½ m(V)^2 = ½ m(V)^2 + mgh

(m*6^2)/2 = (m * 9.81 * 1.1) + (m*V^2)/2

36/2 = (9.81 * 1.1) + v^2/2

Solving the equation;

v=√14.44 m/s  

= 3.8 m/s

a = −0.6g

= -0.6*9.81

= −5.88 m/s^2

Using equation of motion,

V^2 = U^2 + 2aS

S = V^2/2a

= 14.44/12

= 1.19 m.

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Answer:

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Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC

In words, the value of q₃ must be 5.3nC.

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If there are any confusions in the instructions, the chances of making a mistake rises. So, it is very important to clarify the confusions found in the instructions before choosing things needed for the experiment and starting the experiment.

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