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Varvara68 [4.7K]
3 years ago
7

Question 1 of 10

Physics
1 answer:
Marianna [84]3 years ago
6 0

Answer:

Option D. ²²²₉₀Th

Explanation:

Let the unknown be ⁿₘZ. Thus, the equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

Next, we shall determine n, m and Z. This can be obtained as follow:

For n:

226 = 4 + n

Collect like terms

226 – 4 = n

222 = n

n = 222

For m:

92 = 2 + m

Collect like terms

92 – 2 = m

90 = m

m = 90

For Z:

ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th

Therefore, the complete equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th

Thus, the unknown is ²²²₉₀Th

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A car is moving at 14 m/s. After 30 s, its speed inncreased to 20 m/s. What is the acceleration over time
Juli2301 [7.4K]

Answer:

0.2 m/s^2

Explanation:

initial speed 14m/s

final speed 20m/s

acceleration:

(20m/s - 14m /s) /30s = (6m/s)/30s = 0.2 m/s^2

7 0
2 years ago
If you are driving 128.4 km/h along a straight road and you look down for 3.0s, how far do you travel during this inattentive pe
ser-zykov [4K]

Answer:

107 m

Explanation:

Convert km/h to m/s:

128.4 km/h × (1000 m / km) × (1 h / 3600 s) = 35.67 m/s

Distance = rate × time

d = 35.67 m/s × 3.0 s

d = 107 m

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3 years ago
What is the difference in the speed of the generator with a small magnet and a generator with a large magnet?
Delvig [45]
With a small magnet with a generator it will be taken up quickly because how small it is while with a big generator it would take more force for it for the generator to attach because the larger the magnet that heavier it will be because it is attached to the North Pole magnet
8 0
2 years ago
Read 2 more answers
When drawing a Bohr model for an element that has 16 electrons, how many electrons would be placed in the third energy level?
lubasha [3.4K]
There would be 6 electrons placed on the third energy level.
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2 years ago
2×3.14√(1.0m/(9.8〖ms〗^(-1) )=)
il63 [147K]

This is the period in a simple harmonic motion which is 2 seconds in this question.

<h3>What is Period ?</h3>

The period of an oscillatory object can be defined as the total time taken  by a vibrating body to make one complete revolution about a reference point.

We are given the below question

2×3.14√(1.0m/(9.8〖ms〗^(2) )= T

This question can as well be expressed as

2π√(L/g) which is equal to period T.

In a nut shell, Period T = 2×3.14√(1.0m/9.8)

T = 6.28√0.102

T = 6.28 × 0.32

T = 2.006 s

Therefore, the period T of the oscillation is 2 seconds approximately.

Learn more about Period here: brainly.com/question/12588483

#SPJ1

8 0
11 months ago
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