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viva [34]
3 years ago
5

What is the order of magnitude of the gravitational force between two 1.0 kilogram charges that are positioned 1.0 meter apart?

(1) 11 (2) 2 (3) 1 (4) -11
Physics
1 answer:
wariber [46]3 years ago
5 0
I have the sender immediately notify me about this one and only the one you are interested please send us your requirements with you on this site for a few months ago but have not heard back yet over to you as soon and will have to get a new one is the same as last time I have to do it is not an intended solely those are the best way for me to get a good day.
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label the three planes and axis of movement on the figures. define each plane and axis of movement and axis of movement and give
TiliK225 [7]

Answer: it = kobe

Explanation

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2 years ago
Help pls i need this right now
pantera1 [17]

Answer:

The x-component of F_{3} is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O (1)

Where:

\vec F_{1}, \vec F_{2}, \vec F_{3} - External forces exerted on the ring, measured in newtons.

\vec O - Vector zero, measured in newtons.

If we know that \vec F_{1} = (70.711,70.711)\,[N], \vec F_{2} = (-126.859, 46.173)\,[N], F_{3} = (F_{3,x},F_{3,y}) and \vec O = (0,0)\,[N], then we construct the following system of linear equations:

\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N (2)

\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N (3)

The solution of this system is:

F_{3,x} = 56.148\,N, F_{3,y} = -116.884\,N

The x-component of F_{3} is 56.148 newtons.

5 0
2 years ago
A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t
sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, \rho=2.44\times 10^{-8}\ \Omega-m

Resistance in terms of resistivity is given by :

R=\dfrac{\rho l}{A}

Also, V = IR

So,

\dfrac{V}{I}=\dfrac{\rho l}{A}

A is area of wire,

\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}, r is radius, r = d/2 (diameter=d)

\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m

Out of four option, near option is (C) 17 μm.

6 0
2 years ago
قوة الجذب المركزي تكون في اتجاه
pochemuha

Answer:

تكون دائمًا متعامدة مع سرعة الجسم وتكون دائمًا في اتجاه مركز انحناء المسار

Explanation:

6 0
2 years ago
Read 2 more answers
A man does 500 J of work pushing a car a distance of 2 m. How much force does he apply? Assume there is no friction.
Dmitry [639]

The correct answer is A. 250N

Work is a product of force and distance.

That is, work done=force×distance

Therefore substituting for the values in the question:

500J=force×2m

force= 500Nm/2m=250N

another unit for work done is Nm as force as the SI unit of force is newtons and distance in meters

6 0
2 years ago
Read 2 more answers
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