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3 years ago

Why do you think space is good environment to demonstrate Newton's third law of motion?

Physics

1 answer:

igomit [66]3 years ago

3 0

Answer:

Yes

Explanation:

Outer space is an excellent environment to demonstrate the law of Action-Reaction, since there is no friction with boundaries, and also one is away from the Earth gravitational pull, then such force can be neglected.

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Calculate the charge of a copper particle of mass 300 mg, if 0.01% of the electron are removed from it

Alex73 [517]

Answer:

the charge will be neutral.

Explanation:According a book, metals want to "lose" electrons to achieve noble gas configuration. But at the same time, copper ions in a galvanic element are going to gain electrons from a zinc electrode. Why is it so? Is it because one of them HAS to gain/lose electrons, and zinc wins out in terms of achieving noble gas configuration?

6 0

3 years ago

Find the energy released in the fission reaction¹₀n + ²³⁵₉₂U → ⁹⁸₄₀Zr + ¹³⁵₅₂Te +3(¹₀n) The atomic masses of the fission product

ikadub [295]

Fission reaction is given

¹₀n + ²³⁵₉₂U → ⁹⁸₄₀Zr + ¹³⁵₅₂Te +3(¹₀n)

The atomic masses of the fission products are 97.912735 u for ⁹⁸₄₀Zr and 134.916450 u for ¹³⁵₅₂Te. Hence, the energy released in fission reaction is 191.715 MeV.

How to find the energy released in the given fission reaction?

We know, that the atomic mass of the elements are as follows:

⁹⁸₄₀Zr - 97.9120u

¹³⁵₅₂Te - 134.9087u

²³⁵₉₂U - 235.0483923u
n = 1.008665u

In order to find the mass difference, we will calculate the initial mass and the mass of products.

The equation for the reaction:

n + ²³⁵₉₂U → ⁹⁸₄₀Zr + ¹³⁵₅₂Te +3n

Mass of initial reagents is 1.008665u + 235.0483923u = 233.052588u

Product of the reagents is 97.9120u + 134.9087u + 3(1.008665u) = 235.846773u

Now, using the formula of

$E=(\triangledown m)c^2$

$E=(0.205815u)\frac{931.494MeV/c^2}{u} c^2=191.715MeV$

Hence, the energy released in fission reaction is 191.715 MeV.

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4 0

2 years ago

How many significant figures are in 50.5?

Art [367]

3 as a single number is considered a sf

7 0

3 years ago

In the absence of air, a penny and a feather dropped from the same height will

Roman55 [17]

When they are dropped at the same height, taking their mass into account, the feather will touch the ground last.

7 0

4 years ago

Read 2 more answers

A car initially traveling at 15 m/s North accelerates to 25 m/s North in 4 seconds. The magnitude of the average acceleration is

GarryVolchara [31]

The average acceleration of the car as it accelerates from 15 m/s to 25 m/s will be 2.5 m/s².

<h3>What is Average acceleration?</h3>

Average acceleration is the average rate of change of velocity with respect to time. Mathematically -

a = Δv/Δt

Given that a car is traveling at 15 m/s in north direction and in 4 seconds it accelerates to 25 m/s in the north direction only. Therefore, we can write -

initial velocity [u] = 15 m/s

final velocity [v] = 25 m/s

time taken [Δt] = 4 s

The magnitude of the average acceleration can be calculated as follows-

a = Δv/Δt

Δv = v - u = 25 - 15

Δv = 10 m/s

Therefore -

a = 10/4

a = 2.5 m/s².

Therefore, the average acceleration of the car will be 2.5 m/s².

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1 year ago