231 91 Pa has_

Constants Part A wo charged dust particles exert a force of 7.5x102 N n each other What will be the force if they are moved so t

alexandr1967 [171]

Answer:

New force, $F'=48\times 10^3\ N$

Explanation:

It is given that,

Force acting between two charged particles, $F=7.5\times 10^2\ N$

We need to find the force if they are moved so they are only one-eighth as far apart.

The force between two charged particles separated at a distance of r is given by :

$F=k\dfrac{q_1q_2}{r^2}$ ............(1)

If the charges are one-eighth as far apart then, r' =(1/8)r and new force is given by :

$F'=k\dfrac{q_1q_2}{(\dfrac{r}{8})^2}$ ..........(2)

Dividing equation (1) and (2) :

$\dfrac{F}{F'}=\dfrac{1}{64}$

$F'=7.5\times 10^2\ N\times 64$

F' = 48000 N

or

$F'=48\times 10^3\ N$

Hence, this is the required solution.

8 0

3 years ago

Which is a sub-atomic particle?

Natalka [10]

A particle that is smaller than an atom or a cluster of particles.

4 0

3 years ago

To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i

ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

$m_1v_1+m_2v_2=(m_1+m_2)v$

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

$m_1(-v_1)+m_2v_2=(m1+m2)(-v2)$

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

$-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s$

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

$v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s$

hence, the time is t=0.42 s

4 0

3 years ago

It takes 15 min to drive 6.0 mi in a straight line to the local hospital. It takes 10 min to go the last 3.0 mi, 2.0 min to go t

Gala2k [10]

Answer:

36.87 km/h

Explanation:

Convert all the units in SI system

1 mile = 1609.34 m

d1 = 6 mi = 9656.04 m

t1 = 15 min = 15 x 60 = 900 s

d2 = 3 mi = 4828.02 m

t2 = 10 min = 10 x 60 = 600 s

d3 = 1 mi = 1609.34 m

t3 = 2 min = 2 x 60 = 120 s

d4 = 0.5 mi = 804.67 m

t4 = 0.5 min = 0.5 x 60 = 30 s

Total distance, d = d1 + d2 + d3 + d4

d = 9656.04 + 4828.02 + 1609.34 + 804.67 = 16898.07 m = 16.898 km

total time, t = t1 + t2 + t3 + t4

t = 900 + 600 + 120 + 30 = 1650 s = 0.4583 h

The ratio of the total distance covered to the total time taken is called average speed.

Average speed = 16.898 / 0.4583 = 36.87 km/h

6 0

3 years ago

B) Calculate the net work required to bring a 1300 kg car moving at 30 m/s to rest.

prohojiy [21]

I Lolo my name is keshav and

5 0

3 years ago

231 91 Pa has__neutrons