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3 years ago

Four engineers propose designs for door locks that are triggered by a sample of

Physics

1 answer:

sveta [45]3 years ago

7 0

Question:

A. Engineer A: "When the sample alpha decays, it will give off alpha particles. These will

trigger a photosensor (a device that senses forms of light) that will open the lock."

B. Engineer B: "When the sample alpha decays, it only releases large amounts of heat.

This will trigger a thermometer that will open the lock."

C. Engineer C: "When the sample alpha decays, it will lose mass, This will trigger a scale

that will open the lock."

D. Engineer D: "When the sample alpha decays, it will give off negatively charged alpha

particles. Because alpha particles are electrons, they will complete a circuit that will

open the lock."

Answer:

The correct option is;

A. Engineer A: "When the sample alpha decays, it will give off alpha particles. These will trigger a photosensor ( a device that senses forms of light) that will open the lock"

Explanation:

Here, we note that in alpha decay (α-decay) there is an emission of an alpha particle or helium nucleus with the transformation of the parent element nucleus into that of a different element, having reduced mass and atomic numbers

Alpha particles are a form of ionizing radiation with a mass of 6.64 × 10⁻²⁷ kg and therefore they can be sensed by a photo sensor but will be slowly sensed by a scale.

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A circular rod has a radius of curvature R = 9.09 cm and a uniformly distributed positive charge Q = 6.49 pC and subtends an ang

Digiron [165]

Answer:

E = 1.19 N/C

Explanation:

Let's first determine the length of the arc which can be given as:

L= Rθ

where:

L = length of the arc

R = radius of curvature

θ = angle in radius

L = (9.09×10⁻²m)(2.59)

L = (0.0909)(2.59)

L = 0.235431 m

Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]$

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]$

$E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]$

Since $\lambda = \frac{Q}{L}$

where;

L = length

Q = charge

λ = density of the charge;

then substituting $\frac{Q}{L}$ for λ, we have :

$E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]$

$E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}$

substituting our given parameter; we have:

$E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}$

E = 1.1889 N/C

E = 1.19 N/C

∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C

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3 years ago

body is suspended from the ceiling with two wires that make an angle of 40° with the ceiling. The weight of the body is 150N. Wh

steposvetlana [31]

<span>The wires are suspended diagonally, meaning the tension in each is directed diagonally as well (along the wire). We can, however, talk about the horizontal and vertical components of the tension force. The horizontal tension force in each wire is Tcosθ, while the vertical tension force is Tsinθ. The horizontal tension forces balance one another, because the wires are pulling in opposite directions. This means Tcosθ=Tcosθ, which is obvious but not very helpful. If you look at the vertical tension forces on the other hand, you can see that they must balance the weight of the body for the system to be in equilibrium. In other words: Tsinθ+Tsinθ=mg You're given enough information in the question to solve for T using this equation!</span>

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3 years ago

Will give brainliest

Nostrana [21]

Answer:

I think 18 not entirely sure though

Explanation:

Group 18 are the noble group so 16 + 2 = 18

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The pressure exerted by a column of liquid is equal to the product of the height of the column times the gravitational constant

almond37 [142]

Answer:

The column of methanol is = 12.27 m high

Explanation:

The pressure of mercury (p) = ghd,................... equation 1

Where g = acceleration due to gravity, h= height, d = density.

d = 13.6 g/ml convert to kg/m³ = 13.6 g/ml(1000 (kg/m³)/(g/ml))

d =13600 kg/m³., h= 713mm = 713/1000 = 0.713 m., g= 9.80 m/s²

Substituting these values into equation 1,

P = 13600 × 0.713 × 9.8 = 95028.64 N/m².

If the pressure of the mercury supports the pressure exerted by the methanol

∴ pressure that supports the mercury = pressure of the methanol

p = ghd

making h the subject of formula,

h = p/gd ........................... equation 2

Where p = 95028.64 N/m², g = 9.8 m/s², d = 0.79 g/ml = (0.79 × 1000) kg/m³

d = 790 kg/m³.

Substituting these values into equation 2

h = 95028.64/(9.8×790)

h = 95028.64/7742

h = 12.27 m.

The column of methanol is = 12.27 m high

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3 years ago

A man swings a bucket on a 120-cm rope with a tangential speed of 4.5 m/s. If the magnitude of the centripetal force is 25 N, wh

Paraphin [41]

Answer:

the buckets mass is 346g if im not mistaken

Explanation:

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3 years ago