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Alex_Xolod [135]
3 years ago
9

Answered: A 4 kg mass is attached to a horizontal spring with the spring constant of 600 N/m and rests on a frictionless surface

on the ground. The spring is compressed 0.5 m past its equilibrium. What is the initial energy of the system.
Answer: 75 joules
Physics
1 answer:
butalik [34]3 years ago
6 0
Glad it’s figured out
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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

6 0
3 years ago
A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho
fiasKO [112]

Answer:

v_f=8.17\frac{m}{s}

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}

The box is initially at rest, so v_i=0. Solving for v_f:

v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}

5 0
3 years ago
Find the total electric charge of 1.7 kg of electrons. me=9.11×10−31kg, e=1.60×10−19C.
Gelneren [198K]

Answer:

2.99\cdot 10^{11}C

Explanation:

The mass of one electron is

m_e = 9.11\cdot 10^{-31}kg

So the number of electrons contained in M=1.7 kg of mass is

N=\frac{M}{m_e}=\frac{1.7 kg}{9.11\cdot 10^{-31}kg}=1.87\cdot 10^{30}

The charge of one electron is

e=1.60\cdot 10^{-19} C

So, the total charge of these electrons is equal to the charge of one electron times the number of electrons:

Q=Ne=(1.87\cdot 10^{30})(1.6\cdot 10^{-19}C)=2.99\cdot 10^{11}C

8 0
3 years ago
The amount of oxygen in the reactants is 4 atoms. In the products, the oxygen is distributed to water (H2O) and O2 gas. Which co
Reil [10]

Answer:

2 in front of water and 1 in front of oxygen

Explanation:

This question is describing balancing a chemical reaction. A balanced chemical reaction has the same number of atoms of each elements on both the reactant and product side. According to the question, the reactants contains 4 atoms of oxygen. The reactants give rise to water (H20) and O2 in the products side.

This reaction is most likely the decomposition of hydrogen peroxide (H2O2) as follows:

H2O2 (l) ----> H2O (l) + O2(g)

Based on the description, H2O2 will be 2H2O2 as it is said to contain four atoms of oxygen. This means that, in order to have a balanced equation, we must place coefficient 2 in front of water and coefficient 1 in front of oxygen. That is;

2H2O2 (l) ----> 2H2O (l) + O2(g)

6 0
3 years ago
Read 2 more answers
What does the REVOLUTION of Earth around the Sun bring us and how long does it take?
olasank [31]

Answer:

takes 365 days and it bring us seasons such as, spring,winter and fall.

7 0
3 years ago
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