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2 years ago

Which of the following is an example of a force?

Physics

1 answer:

Naya [18.7K]2 years ago

7 0

I think its d . i hope this helps

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In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright frin

atroni [7]

Answer:

Y = 0.0254 m = 25.4 mm

Explanation:

The formula for the fringe spacing in Young's Double-slit experiment is given by the following formula:

$Y = \frac{\lambda L}{d}$

where,

Y = fringe spacing = 0.0176 m

λ = wavelength = 425 nm = 4.25 x 10⁻⁷ m

L = Distance between screen and slits

d = slit separation

Therefore,

$\frac{0.0176\ m}{4.25\ x\ 10^{-7}\ m} = \frac{L}{d}\\\\\frac{L}{d} = 41411.76$

Now, for:

λ = 614 nm = 6.14 x 10⁻⁷ m

$Y = (6.14\ x\ 10^{-7}\ m)(41411.76)\\$

8 0

3 years ago

What are the three categories of adult development?

andreev551 [17]

Physical development , Cognitive Development , psychosocial development I think!

5 0

3 years ago

When you push a 1.90-kg book resting on a tabletop, you have to exert a force of 2.10 N to start the book sliding. Once it is sl

zimovet [89]

Static frictional force = ƒs = (Cs) • (Fɴ)
2.26 = (Cs) • m • g
2.26 = (Cs) • (1.85) • (9.8)
Cs = 0.125
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
1.49 = (Cκ) • m • g
1.49 = (Cκ) • (1.85) • (9.8)
Cκ = 0.0822

3 0

3 years ago

What is the function of layer of air trapped under the hovercraft

aev [14]

Answer:b

Explanation:

3 0

3 years ago

You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi

Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0

3 years ago