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2 years ago

Help! I can't remember how to calculate this.

Physics

1 answer:

Mars2501 [29]2 years ago

5 0

You need to draw two lines from the head of the arrow: one parallel to the axis (which will be bent) and the other one passing through the center of the converging lens <span>(which will continue straight) </span>and see where they intersect.

If the arrow is at an infinite distance, the image will be point-like in F.
If the arrow is at a distance greater than 2F, the image will be upside down between F and 2F.
If the arrow is at a distance of 2F, the image will be upside down in 2F.
If the arrow is at a distance between F and 2F, the image will be upside down beyond 2F.
If the arrow is at a distance F, all rays proceed parallel to each other and no image is formed.
If the arrow is at a distance smaller than F, the image will be virtual.

Therefore, the correct matches are 1B, 2C, 3D, and 4A.

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Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an

8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 $m/s^2$ where as radial the component of acceleration is 8.77 $m/s^2$

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

$r=e^u$

So the transverse component of acceleration are given as

$a_{\theta}=(ru''+2r'u')\\$

Here

$r=e^u\\r=e^{\pi/4}\\r=2.1932 m$

$r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m$

$a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\$

The transverse component of acceleration is 26.32 $m/s^2$

The radial component is given as

$a_r=r''-r\theta'^2$

Here

$r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m$

$a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2$

The radial component of acceleration is 8.77 $m/s^2$

6 0

2 years ago

Which force is greater the earth’s pull on the moon or the moon’s pull on the earth

Vitek1552 [10]

Answer:

The earth's pull on the moon

Explanation:

Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth.

7 0

2 years ago

The peak intensity of radiation from a star named sigma is 2 x 10^6 mmkay. What is the average surface temperature of Sigma roun

Igoryamba

Answer:

1.45 K

I had the same question and i got it right.

8 0

3 years ago

Read 2 more answers

Determine the Mutual Inductance per unit length between two long solenoids, one inside the other, whose radii are r1 and r2 (r2

Triss [41]

Answer:

M' = μ₀n₁n₂πr₂²

Explanation:

Since r₂ < r₁ the mutual inductance M = N₂Ф₂₁/i₁ where N₂ = number of turns of solenoid 2 = n₂l where n₂ = number of turns per unit length of solenoid 2 and l = length of solenoid, Ф₂₁ = flux in solenoid 2 due to magnetic field in solenoid 1 = B₁A₂ where B₁ = magnetic field due to solenoid 1 = μ₀n₁i₁ where μ₀ = permeability of free space, n₁ = number of turns per unit length of solenoid 1 and i₁ = current in solenoid 1. A₂ = area of solenoid 2 = πr₂² where r₂ = radius of solenoid 2.

So, M = N₂Ф₂₁/i₁

substituting the values of the variables into the equation, we have

M = N₂Ф₂₁/i₁

M = N₂B₁A₂/i₁

M = n₂lμ₀n₁i₁πr₂²/i₁

M = lμ₀n₁n₂πr₂²

So, the mutual inductance per unit length is M' = M/l = μ₀n₁n₂πr₂²

M' = μ₀n₁n₂πr₂²

3 0

2 years ago

A very good insulator has an R-value of 29. The material's heat transfer coefficient is

Fittoniya [83]

Answer:

b. 0.034

Explanation:

The heat transfer coefficient of a material (U-value) is equal to the reciprocal of its R-value, therefore:

$U = \frac{1}{R}$

where

R is the R-value of the material

For the insulator in this problem,

R = 29

Substituting into the equation, we find the heat transfer coefficient:

$U=\frac{1}{29}=0.034$

4 0

2 years ago