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marta [7]
2 years ago
5

A 2 µC charge q1 and a 2 µC charge q2 are 0.3 m from the x-axis. A 4 µC charge q3 is 0.4 m from the y-axis. The distances d13 an

d d23 are 0.5 m. Find the magnitude and direction of the resulting vector R. Round your answer to the nearest tenth. The magnitude is N. The direction is °.
Physics
2 answers:
emmainna [20.7K]2 years ago
7 0
Here's the answer ladies and gents
Magnitude : 0.5n
Direction  : 0 degrees.
Marina CMI [18]2 years ago
5 0

A 2 µC charge q1 and a 2 µC charge q2 are 0.3 m from the x-axis. A 4 µC charge q3 is 0.4 m from the y-axis. The distances d13 and d23 are 0.5 m. Find the magnitude and direction of the resulting vector R. Round your answer to the nearest tenth.

The magnitude is  0.5 N.

The direction is °.0

ON EDGENUTITY

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A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy
SashulF [63]

Answer:

5.444\times 10^{-4}

Explanation:

The momentum of the neutron before and after the decay  is the same since there's no external force.

P_{sys}=const\\\\P=mv\\\\K=0.5mv^2

#The neutron is initially at rest, so after the decay:

P_A+P_B=0\\\\P_A=-P_B

#After decay, the proton has +ve direction  with a velocity v_Awhile the electron moves in a negative direction with a velocity v_B

Therefore:

P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B

Let the energy released during the decay be Q:

Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}

Hence,Kp/Ktot is 5.444x10^(-4)

4 1
2 years ago
An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th
kolbaska11 [484]

Answer:

mg=200.4 N.

Explanation:

This problem can be solved using Newton's law of universal gravitation: F=G\frac{m_{1}m_{2}}{r^{2}},

where F is the gravitational force between two masses m_{1} and m_{2}, r is the distance between the masses (their center of mass), and G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2}) is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting w_{s} be the weight on the surface:

w_{s}=mg,

mg=8*10^{2},

m=(8*10^{2})/g,

since we now that g=9.8m/s^{2} we get that the mass is

m=81.6kg.

Now we can use Newton's law of universal gravitation

F=G\frac{Mm}{r^{2}},  

where m is the mass of the astronaut and M is the mass of the earth. From Newton's second law we know that

F=ma,

in this case the acceleration is the gravity so

F=mg, (<u>becarefull, gravity at this point is no longer</u> 9.8m/s^{2} <u>because we are not in the surface anymore</u>)

and this get us to

mg=G\frac{Mm}{r^{2}}, where mg is his new weight.

We need to remember that the mass of the earth is M=5.972*10^{24}kg and its radius is 6.37*10^{6}m.

The total distance between the astronaut and the earth is

r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6} meters.

Now we can compute his weigh:

mg=G\frac{Mm}{r^{2}},

mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}},

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5 0
2 years ago
If a 110 kg go-cart traveling at a velocity of 13.41 m/s has a collision with an impulse of 615 Nxs, what is the
mafiozo [28]

Answer:

5.59 m/s

Explanation:

We are given;

Mass = 110 kg

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Force = 615 N

Time(t) = 1 s

Now, the formula for force is;

Force = mass x acceleration

Thus;

615 = 110 × acceleration

\Acceleration(a) = 615/110 = 5.591 m/s²

Now, using Newton's first law of motion, we can find acceleration (a). Thus;

v = u + at

v = 13.41 + (5.591 × 1)

v ≈ 19 m/s

So,the change in velocity is;

Final velocity(v) - Initial velocity(u) = 19 - 13.41 = 5.59 m/s

6 0
2 years ago
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Amiraneli [1.4K]

Answer:

1.034m/s

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V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

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Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

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8 0
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The primary of a step-up transformer is connected across the terminals of a standard wall socket, and resistor 1 with a resistan
alexira [117]
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3 0
2 years ago
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