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lara [203]
3 years ago
14

12. The diameter of a circle is 2.42m. Calculate itsarea in proper significant figure​

Physics
1 answer:
Sever21 [200]3 years ago
7 0

Answer:

A = 4.6 [m²]

Explanation:

The area of a circle can be calculated by means of the following equation.

A=\frac{\pi }{4} *D^{2}

where:

A = area [m²]

D = diameter = 2.42 [m]

Now replacing:

A=\frac{\pi }{4} *(2.42)^{2} \\A = 4.6 [m^{2} ]

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If a team of workers comes to a consensus, what probably happened in the meeting?
Feliz [49]

Answer:

A

Explanation:

A consensus is when you come to an agreement

7 0
3 years ago
A force of 30n is applied to a screwdriver to pry the lid off of a can of paint. The screwdriver applies 75n of force to the lid
gizmo_the_mogwai [7]
The mechanical advantage of the screwdriver that is being described above is equal to 75N. This means that for every 30N that is applied on the screwdriver, this simple machine would in turn apply 75N of force to the lid of the can. 
3 0
3 years ago
You are performing a knee extension exercise. You hold a 20kg weight at full knee extension. The weight is 0.4m from your knee j
dmitriy555 [2]

Answer:

The moment is -78.4 N-m (clockwise).

Explanation:

Given:

Mass of the object (m) = 20 kg

Distance of the object from the knee joint (d) = 0.4 m

Weight of leg is not considered.

Acceleration due to gravity (g) = 9.8 m/s²

Now, weight of the object is equal to the product of its mass and acceleration due to gravity. So,

Weight = Mass × Acceleration due to gravity

            = mg=20\times 9.8 =196\ N

We know that, moment of a force about a point is defined as the product of force applied and the perpendicular distance between the point and the line of application of force.

Moment of the given weight about the knee joint is given as:

Moment about knee joint = Weight × Distance from knee joint to weight

Moment about knee joint = 196 × 0.4 = 78.4 Nm

Now, from the diagram below, we can observe that, the weight acts vertically down and thus the sense of rotation about the knee joint at point O is clockwise. So, moment is negative.

Therefore, the moment is -78.4 N-m (clockwise).

7 0
3 years ago
At a certain instant, the earth, the moon, and a stationary 1160 kg spacecraft lie at the vertices of an equilateral triangle wh
Afina-wow [57]

Answer:

W = 1.22 \times 10^9 J

Explanation:

Initial potential energy of the given spacecraft is given as

U = -\frac{GM_e m}{r} - \frac{GM_m m}{r}

so we have

U = - \frac{Gm}{r}(M_e + M_m)

so we have

M_e = 5.98 \times 10^{24} kg

M_m = 7.35 \times 10^{22} kg

m = 1160 kg

r = 3.84 \times 10^8 m

U = - \frac{(6.67 \times 10^{-11})(1160)}{3.84 \times 10^8}(5.98 \times 10^{24} + 7.35 \times 10^{22})

U = -1.22 \times 10^9 J

now total work done to move it to infinite is given

W = 0 - U

W = 1.22 \times 10^9 J

6 0
3 years ago
Consider the 65 N light fixture supported as in the figure. Find the tension in the supporting wires.
ASHA 777 [7]

By using Lami's theorem formula, the tension in the supporting wires is 48.6 Newtons

TENSION

  • Tension is also a force having Newton as S.I unit.
  • The tension in the wire will be the same.

This question can be solved by using either vector diagram or by using  Lami's theorem.

The sum of two given angles  = 42 + 42 = 84 degrees

The third angle = 180 - 84 = 96 degrees.

Below is the Lami's theorem formula

\frac{T}{sin\alpha } = \frac{T}{sin\beta } = \frac{W}{sinY}

Where

\alpha  = \beta = 42 + 90 = 132 degrees

Y = 96 degrees

W = 65 N

By using the formula, we have

\frac{T}{sin\alpha } =  \frac{W}{sinY}

T/sin 132 = 65/sin96

Cross multiply

T = 0.743 x 65.57

T = 48.56 N

Therefore, the tension in the supporting wires is 48.6 Newtons approximately.

Learn more about Tension here: brainly.com/question/24994188

3 0
3 years ago
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