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Leya [2.2K]
2 years ago
10

What it mean head on collision?​

Physics
2 answers:
olchik [2.2K]2 years ago
8 0
Head on collision means to go through a systems of for example extra curricular activities and getting your head injured with something or someone else.
murzikaleks [220]2 years ago
6 0
I don’t understand the question
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Who were we in the space/arms race with?<br> In the movie *Hidden figures*
daser333 [38]

Explanation:

The U.S. launched its first man into space in May 1961.

4 0
3 years ago
2) A 0.4kg ball moves in horizontal circle of radius 3 m at speed of 100m/s. What
sweet-ann [11.9K]

Answer:

F = 1300 N

Explanation:

F = mv²/R = 0.4(100²)/3 = 1333.3333...

6 0
2 years ago
Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |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
What is an example of a sting nuclear force ?
slavikrds [6]

Answer:

A.) stability of an atom

Explanation:

5 0
3 years ago
What are some guidelines for using self-assessments?
lakkis [162]
Decide which components you wish to assess yourself. There is no set order in which you do this. you could, for an example assess one or two elements and skip the rest.
4 0
3 years ago
Read 2 more answers
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