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3 years ago

Which logic function(s) provides a high output in response to only seven low inputs?

Physics

1 answer:

inysia [295]3 years ago

3 0

Answer:

figuer it out we are all on our ownExplanation:

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After creating a prototype, which of the following would not be and appropriate next step in the design process of a new technol

Dafna11 [192]

The answer would be to research the need. This should have been done before the project began.

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4 years ago

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What could be the possible answer to the question ?<br><br>thankyou ~

Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$

F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

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2 years ago

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Weigt A box sitting still on the ground by itself has a weight of 700N, what is the mass? (gravity's acceleration is 9.80 m/s2)

Alex

Force=700N
Acceleration=9.8m/s^2

Using newtons law

$\\ \rm\longmapsto F=ma$

$\\ \rm\longmapsto m=\dfrac{F}{a}$

$\\ \rm\longmapsto m=\dfrac{700}{9.8}$

$\\ \rm\longmapsto m=71.4kg$

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3 years ago

An object is hung from a spring balance attached to the ceiling of an elevator cab. The balance reads 65 N when the cab is stand

lana66690 [7]

Answer:

(A) Reading will be 65 N

(B) Net force on the elevator will be 49.076 N

Explanation:

We have given the balance force = 65 N

Acceleration due to gravity $g=9.8m/sec^2$

We know that W=mg

So $65=m\times 9.8$

m = 6.632 kg

(a) In first case as the as the speed is constant so the force on the elevator will be 65 N

(B) In second case as the elevator is decelerating at a rate of $2.4m/sec^2$

So net acceleration = 9.8-2.4= $7.4m/sec^2$

So net force on elevator will be = m× net acceleration = 6.632×7.4 = 49.076 N

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3 years ago

A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of n

Oduvanchick [21]

Answer:

15.0 m/s^2

Explanation:

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3 years ago

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