Assuming motion is on a straight path what would result in two positive components of a vector

Two point charges each carrying a charge of + 4.5 E - 6 C are located 4.5 meters away from each other. How strong is the electro

BARSIC [14]

Answer:

0.009 N, repulsive

Explanation:

The electrostatic force between two electric charges is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, we have

$q_1 =q_2 = +4.5\cdot 10^{-6}C$ are the two charges

r = 4.5 m is their separation

Substituting into the equation, we find

$F=(9\cdot 10^9 Nm^2 C^{-2})\frac{(+4.5\cdot 10^{-6} C)(4.5\cdot 10^{-6} C)}{(4.5 m)^2}=0.009 N$

Moreover, the force is repulsive. In fact, the following rules apply:

- When two charges have same sign, they repel each other

- When two charges have opposite signs, they attract each other

7 0

2 years ago

Can someone help me with physics;((

vladimir2022 [97]

Answer: vf= 51 m/s and d= 112 m

Explanation: solution attached

4 0

3 years ago

A small sandbag is dropped from rest from a hovering hot-air balloon. (assume the positive direction is upward.) (a) after 1.5 s

marishachu [46]

solution:

We know v0 = 0, a = 9.8, t = 4.0. We need to solve for v

so,

we use the equation:

v = v0 + at

v = 0 + 9.8*4.0

v = 39.2 m/s

Now we just need to solve for d, so we use the equation:

d = v0t + 1/2*a*t^2

d = 0*4.0 + 1/2*9.8*4.0^2

d = 78.4 m

3 0

2 years ago

4 This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive

zubka84 [21]

Answer:at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B2------15km/h------->Dock

|

B1 at 20km/h

|

V

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.

3 0

3 years ago

A certain ideal gas has molar heat capacity at constant volume CV. A sample of this gas initially occupies a volume V0 at pressu

ANTONII [103]

Answer:

Explanation:

The processes are described on the image attached below. The isobaric process consists of an horizontal line, the adiabatic expansion is described by a polytropic curve:

$P_{2} \cdot V_{2}^{\gamma} = P_{3} \cdot V_{3}^{\gamma}$