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Anton [14]
3 years ago
9

Just how strong is the electric force? suppose you had two small boxes, each containing 1.0 g of protons. (a) if one were placed

on the moon by an astronaut and the other were left on the earth, and if they were connected by a very light (and very long!) string, what would be the tension in the string? express your answer in newtons and in pounds. do you need to take into account the gravitational forces of the earth and moon on the protons? why? (b) what gravitational force would each box of protons exert on the other box?
Physics
1 answer:
vlabodo [156]3 years ago
3 0
Let's use Newton's Law of Second Motion: F=ma. When no other direct force is acting on the system, the acceleration is due to the gravity. The modified equation becomes: F = mg. So, yes, you need to take into account the gravitational accelerations in the moon and on Earth. 

g,moon = 1.622 m/s²
g,Earth = 9.81 m/s²

The net force is the tension of the string:

F,Earth - F,moon = Tension
Tension = (1/1000 kg)(9.81 m/s²) - (1/1000 kg)(1.622 m/s²)
Tension = 8.188×10⁻³ N

To convert, 1 pound force is equal to 4.45 Newtons:

Tension = 8.188×10⁻³ N * 1 lbf/4.45 N
Tension = 1.84×10⁻³ lbf
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Which would most likely cause specular reflection? O a pathway with rough rocks a shiny, smooth leaf a small patch of soil a rou
andrew11 [14]

Answer:

a shiny smooth leaf

Explanation:

A shiny smooth leaf will cause specular reflection. Other choices will cause diffused reflection from the surface.

A specular reflection is similar to how a mirror or smooth surface reflects. The incident light is given off as a single ordered reflection from the surface of a body.

For this to occur, the surface incident must be smooth and without rough patterns on it.

A path way with rough rocks, small patch of soil and rough logs will give off diffused reflection

8 0
3 years ago
Read 2 more answers
18. Compared to its weight on Earth, a 5 kg object on the moon will weigh A. the same amount. B. less. C. more.
s2008m [1.1K]

Answer:

B. less

Explanation:

acceleration due to gravity on Earth, g = 9.8 m/s²

acceleration due to gravity on Moon, g = 1.6 m/s²

Given mass of the object as, m = 5 kg

Weight of an object is given as, W = mg

                                                         

Weight of the object on Earth, W = 5 x 9.8 = 49 N

Weight of the object on Moon, W = 5 x 1.6 = 8 N

Therefore, the object weighs less on the moon compared to its weight on Earth.

The correct option is "B. less"

8 0
3 years ago
Object a travels in the +x-direction before hitting a stationary object
Leto [7]
The object’s resultant angle of motion with the +x-axis after the collision is 47°

<span>From object A:
 
1) x-momentum is 5.7 × 10^4 kilogram meters/second,
2) y-momentum is 6.2 × 10^4 kilogram meters/second.
 
Now, we know, tan</span>Ф = \frac{y}{x}

⇒tanФ = \frac{6.2 × 10^4 }{5.7 × 10^4}

⇒tanФ = 1.088

⇒ Ф = tan^{-1} 1.088 
         =  47.4 ≈ 47

8 0
3 years ago
Read 2 more answers
swings a 5.5 kg cup of water in a vertical circle of radius 1.9 m. (a) What minimum speed must the cup have in this demo if the
Tanzania [10]

Answer:

4.32

Explanation:

The centripetal acceleration of any object is given as

A(cr) = v²/r, where

A(c) = the centripetal acceleration

v = the linear acceleration

r = the given radius, 1.9 m

Since we are not given directly the centripetal acceleration, we'd be using the value of acceleration due to gravity, 9.8. This means that

9.8 = v²/1.9

v² = 1.9 * 9.8

v² = 18.62

v = √18.62

v = 4.32 m/s

This means that, the minimum speed the cup must have so as not to get wet or any spill is 4.32 m/s

6 0
3 years ago
[7 pts] 1. Suppose that a car starts from rest, its engine providing an acceleration of 4 ft/s2 while air resistance provides 0.
kumpel [21]

Answer:

a) Initial Value Problem

dv/dt = 4 - 0.1v

v(0) = 0

b) solution to the IVP

v(t) = 40(1 - e^(-t/10))

c) Limiting velocity

Vo = 40 ft/s

Position of the car after 12 hours

X = 14,390 ft

Explanations:

The complete explanations of each of the sections contained in the question are in the files attached to this solution.

4 0
3 years ago
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