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Vsevolod [243]
2 years ago
15

Two negative charges that are both -0.3C push each other apart with a force of 19.2 N. How far apart are the two charges?

Physics
1 answer:
Paha777 [63]2 years ago
8 0
<span>Using Coulomb's law: k*(-0.3)*(-0.3)/(d^2)=19.2 D is the distance between the two negative charges</span>
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A cylinder with a moment of inertia I (about its axis of symmetry), mass m, and radius r has a massless string wrapped around it
Eduardwww [97]
Part b is equal to F in standards of society and it’s quality of math during the 1900s
(That was a bit of Social Studies lol)
4 0
2 years ago
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi
OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

8 0
2 years ago
A dog runs 3 miles and 4 miles west in 6 hours. What’s the dogs total distance and displacement ?
Inga [223]

Complete question:

A dog runs 3 miles east and 4 miles west in 6 hours. What’s the dogs total distance and displacement ?

Answer:

The total distance covered by the dog is  7 miles

The displacement of the dog is 1 mile west

Explanation:

Given;

initial position of the dog = 3 miles east

final position of the dog = 4 miles west

time of motion, t = 6 hours

The total distance covered by the dog is given as;

Total distance = 3 miles + 4 miles = 7 miles

The displacement of the dog is given as;

displacement = final position of the dog  - initial position of the dog

displacement =  4 miles west - 3 miles east = 1 mile west

6 0
2 years ago
To ensure the greatest nutritional benefits, dietary guidelines suggest which of the following?
oee [108]
The answer would be D. Eating nutrient-dense foods
3 0
2 years ago
Read 2 more answers
Greg walks on a straight road from his home to a convenience store 3.0 km away with a speed of 6.0 km/h. On reaching the store h
VladimirAG [237]

This question was apprently selected from the "Sneaky Questions" category.

The store is 3 km from his home, and he walks there with a speed of 6 km/hr.  So it takes him (3 km) / (6 km/hr)  =  1/2 hour to get to the store.

That's 30 minutes.  So the whole part-(a.) of the question refers to only that part of the trip, and we don't care what happens once he reaches the store.  

a). Over the first 30 minutes of his travel, Greg walks 3.0 km on a straight road, and he ends up 3.0 km away from where he started.

Average speed = (distance/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>

Average velocity = (displacement/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>

There's probably some more questions in part-(b.) where you'd need to use Greg's return trip to find the answers, but johnaddy210 is only asking us for part-(a.).

8 0
2 years ago
Read 2 more answers
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