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Natalka [10]
3 years ago
9

Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first bl

ock is pulled with a force of 100 N. What is the tension in the string connecting block 100 to block 99? What is the tension in the string connecting block 50 to block 51?
Physics
1 answer:
Ivanshal [37]3 years ago
4 0

Answer:

The tension in the string connecting block 50 to block 51 is 50 N.

Explanation:

Given that,

Number of block = 100

Force = 100 N

let m be the mass of each block.

We need to calculate the net force acting on the 100th block

Using second law of newton

F=ma

100=100m\times a

ma=1\ N

We need to calculate the tension in the string between blocks 99 and 100

Using formula of force

F_{100-99}=ma

F_{100-99}=1

We need to calculate the total number of masses attached to the string

Using formula for mass

m'=(100-50)m

m'=50m

We need to calculate the tension in the string connecting block 50 to block 51

Using formula of tension

F_{50}=m'a

Put the value into the formula

F_{50}=50m\times a

F_{50}=50\times1

F_{50}=50\ N

Hence, The tension in the string connecting block 50 to block 51 is 50 N.

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Fed [463]

Answer:

\underline{ \boxed{F_m =  3.35 { \times 10}^{ - 2} N}}

Explanation:

since \: in \: this \: case \to \: \underline{ v }\: is \perp \: to \:\underline{ q} \\ then \: the \: magnetic \: force \: (F_m) \: is \:  max = qvB \sin(90)  \\ F_m = qvB  \times (1) = qvB \\ hence \to \\ F_m = qvB  \\ where \\  \: q = 0.75 C : v = 235m {s}^{ - 1}  : B = 1.9 { \times 10}^{ - 4} T  \\ therefore \to \\ F_m = 0.75 \times 235 \times 1.9 { \times 10}^{ - 4} \\ F_m = 176.25 \times 1.9 { \times 10}^{ - 4} \\ F_m = 0.0334875  = 3.35 { \times 10}^{ - 2} N \\ \boxed{F_m =  3.35 { \times 10}^{ - 2} N}

3 0
3 years ago
A ray diagram for a refracted light ray is shown.
Rina8888 [55]

The dashed line represents the<u> normal </u>to the boundary

Explanation:

Refraction is a phenomenon that occurs when a ray of light crosses the interface between two mediums.

When this occurs, the ray of light changes direction and speed.

The new direction of the ray of light is given by Snell's law:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1 is the index of refraction of the 1st medium

n_2 is the index of refraction of thre 2nd medium

\theta_1 is the angle of incidence, which is the angle between the direction of the incident ray and the normal to the boundary

\theta_2 is the angle of refraction, which is the angle between the direction of the refracted ray and the normal to the boundary

The index of refraction of a medium is the ratio between the speed of light in a vacuum (c) and the speed of light in that medium (v):

n=\frac{c}{v}

The figure in the question is missing, however you can find it in attachment.

We see that the dashed line is perpendicualr to the boundary between the two mediums, so it represents the normal.

So, the correct answer is

Normal

We note that in this case the ray of light bends towards the normal, this means that \theta_2, therefore the second medium has a larger index of refraction (n_2>n_1).

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

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7 0
3 years ago
Why we don’t feel atmospheric pressure?
Veseljchak [2.6K]

Answer:

The reason we can't feel it is that the air within our bodies (in our lungs and stomachs, for example) is exerting the same pressure outwards, so there's no pressure difference and no need for us to exert any effort.

8 0
3 years ago
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3 0
3 years ago
In the sum →A+→B=→C, vector →A has a magnitude of 12.0 m and is angled 40.0° counterclockwise from the +x direction, and vector
icang [17]

Answer:

Explanation: Ok, first caracterize the two vectors that we know.

A = ax + ay = (12*cos(40°)*i + 12*sin(40°)*j) m

now, see that C is angled 20° from -x, -x is angled 180° counterclockwise from +x, so C is angled 200° counterclockwise from +x

C = cx + cy = (15*cos(200°)*i + 15*sin(200°)*j) m

where i and j refers to the versors associated to te x axis and the y axis respectively.

in a sum of vectors, we must decompose in components, so: ax + bx  = cx and ay + by = cy. From this two equations we can obtain B.

bx= (15*cos(200°) - 12*cos(40°)) m = -23.288 m

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Lets call a to the angle between -x and B, from trigonometry we know that tg(a) = by/bx, that means a = arctg(12.843/23.288) = 28.8°

So the total angle will be 180° + 28.8° = 208.8°.

For the magnitude of B, lets call it B', we can use the angle that we just obtained.

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So the magnitude of B is 26.58 m.

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3 years ago
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