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Novosadov [1.4K]
2 years ago
12

Please help with this diagram as soon as possible.

Physics
1 answer:
melomori [17]2 years ago
3 0
This is the answer .....

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Two positive charges of 6.0 x 10^-6 C are separated by 0.50 m. Calculate the force of electrical attraction
SpyIntel [72]

Answer:

Force between two charges is given as 1.296 N

Explanation:

As we know that the two charges will attract or repel each other when they are placed near to each other

Here we know that electrostatic force is given as

F = \frac{kq_1 q_2}{r^2}

here we have

q_1 = q_2 = 6 \times 10^{-6} C

distance between two charges is given as

r = 0.50 m

so we have

F = \frac{(9\times 10^9)(6 \times 10^{-6})^2}{0.50^2}

F = 1.296 N

3 0
3 years ago
Will give brainiest to best answer
Gala2k [10]

Answer:

t

Explanation:

6 0
2 years ago
the human nervous system can propagate nerve impulses at about 10 squared metered per second. estimate the time it takes to trav
rusak2 [61]
Use the formular d = v x t
d = 2m
v= 100m/s

t= d / v
 = 2 / 100
 = 0.02sec
7 0
3 years ago
Read 2 more answers
Consider the electric force between a pair of charged particles a certain
Crazy boy [7]

Answer:

Doubled

Explanation:

F = (kq1q2) / r^2

F and q (Either q1 or q2) are directly proportional, so double the charge would also double the electruc force between the charges.

5 0
3 years ago
A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr
Leviafan [203]

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration a=g=9.8 m/s^2 directed downward, initial vertical position d=750 m, and initial vertical velocity v_0 = 0. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

v_f^2 -v_i^2 =2ad

substituting, we find

v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

v_f=\frac{121.2 m/s}{2}=60.6 m/s

In order to do that, we use again the same SUVAT equation substituting v_f with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m

Which means that the heigth of the packet was

h=750 m-187.4 m=562.6 m

3 0
3 years ago
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