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3 years ago

What 3 things can an object with unbalanced forces do?

1 answer:

3 0

Forces are considered balanced when all of the combined forces lead to no change in the motion of the object. For example, when a book is sitting on a table, the force of gravity is pushing downward and the normal force is pushing upward with exactly the same amount of force. Since they are equal and opposite forces, the book does not move.

Unbalanced forces exist when there are unequal forces acting upon the object, which leads to a change in the state of motion. Unbalanced forces can lead to a change in direction, a change in speed, or both a change in direction and in speed.

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Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

3 years ago

A man can jog 10 miles in 90 minutes. What’s his speed in mph?

Yuki888 [10]

Answer:

hi there!

the correct answer to this question is: 6.67 mph

Explanation:

you convert minutes to hours

10 miles * 60 mins / 90 mins

7 0

3 years ago

Calculate the force needed to bring a 1,019-kg car to rest from a speed of 29 m/s in a distance of 118 m

son4ous [18]

Acceleration = ▵v/▵t
Time = d/v
Fisrt calculate time : ( 118/29 ) = 4 seconds
Then calculate acceleration
A = 29/4 = 7.25 m/s²
Now the force.
Force = mass * acceleration.
F= 1,019 * 7.25
F= 7,387 N

8 0

3 years ago

What are the strengths and limitations of the doppler and transit methods? What kind of planets are easiest to detect with each

Arturiano [62]

$\huge\mathfrak\red{✔Answer:-}$