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bazaltina [42]
2 years ago
13

What is the galaxy that our solar system is located in?

Physics
1 answer:
jeka57 [31]2 years ago
7 0

Answer:

I think the answer is Milky Way

I hope this Will help you

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What is 2m+n =k for M
kotykmax [81]
Answer : M= k-n/2
Subtract n from each side
2m=n-k
Divide out 2 from both sides to get M alone
m= k-n/2
5 0
3 years ago
A 4,000-kg car traveling at 20m/s hits a wall with a force of 80,000 N and comes to a stop. What was the impact time?
Nookie1986 [14]
Rate of change of momentum = impact force
(m*v-m*u)/t = F
4000*20/t = 80000 (note: v is zero as it stopped)
<span>soo, t = 1 sec</span>
7 0
3 years ago
a man is 9 m behind a dorr of train when it starts moving with a=2ms^-1. how far the man have to run and after what time will he
Naya [18.7K]
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8 0
2 years ago
Relate a real life phenomenon with each branch of physics
anastassius [24]

Answer:

Branches of physics with real life examples

In measuring and understanding nuclear fission (a real life phenomenon), all branches of theoretical and experimental physics have to be employed. Physics branches needed in it are, radiation detection and measurement, nuclear physics, statistical physics, thermodynamics, and almost all others.

Explanation:

4 0
3 years ago
A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the fol
PIT_PIT [208]

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a)  at 4.75 cm

b)  at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula E = \dfrac{kq }{d}

E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}

E = 1461.95 N/C

c) The electric field E is calculated as:

E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}

E = 239.76 N/C

7 0
3 years ago
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