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Julli [10]
1 year ago
5

A uniform 40-N board supports two children weighing 500 N and 350 N. If the support is at the center of the board and the 500-N

child is 1.5 m from its center, How far is the 350-N child from the center?
Physics
1 answer:
serg [7]1 year ago
5 0

Answer:

b= 2.14 m

Explanation:

Given that

Weight of the board ,wt = 40 N

Wight of the first children , wt₁=500 N

Weight of the second children ,wt₂ = 350 N

The distance of the 500 N child from center ,a= 1.5 m

lets take distance of the 350 N child from center = b m

Now by taking the moment about the center of the board

We know that moment = Force x Perpendicular distance from the force

wt₁ x a = wt₂ x b

500 x 1.5 = 350 x b

b= 2.14 m

Therefore the distance of the 350 N weight child from the center is 2.14 m.

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Why are fossil fuels considered nonrenewable resources if they are still forming beneath the surface today?
AleksandrR [38]

B is the answer because it takes millions of years to form these fossil fuels and everyday we use way more than we can find we may have a surplus for now but we may run out sooner than some think

7 0
2 years ago
An object begins at position x = 0 and moves one-dimensionally along the x-axis witļi a velocity v
Liula [17]

Answer:

The answer is "between 20 s and 30 s".

Explanation:

Calculating the value of positive displacement:

\ (x_{+ve}) = \frac{1}{2} \times 15 \times  20 \\\\

          = \frac{1}{2} \times 300 \\\\=  150 \\\\

Calculating the value of negative displacement upon the time t:

(x_{-ve}) = \frac{1}{2} \times 5 \times 20- 20(t-20) \\\\

          = \frac{1}{2} \times 100- 20t+ 400 \\\\= 50- 20t+ 400 \\\\

\to X= X_{+ve} + X_{-ve} \\\\

\to  150 - 50 -20t+400 =0\\\\\to 100 -20t+400 =0 \\\\\to 500 -20t =0\\\\\to 20t =500 \\\\\to t=\frac{500}{20}\\\\\to t=\frac{50}{2}\\\\\to t= 25

That's why its value lie in "between 20 s and 30 s".

6 0
2 years ago
What mass needs to be attached to a spring with a force constant of 7N/m in order to make a simple harmonic oscillator oscillate
Alex_Xolod [135]

Answer:

Mass will be 4.437 kg

Explanation:

We have given force constant k = 7 N/m

Time period of oscillation T = 5 sec

So angular frequency \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{5}=1.256rad/sec

We know that angular frequency is given by

\omega =\sqrt{\frac{k}{m}}

1.256 =\sqrt{\frac{7}{m}}

Squaring both side

1.577 =\frac{7}{m}

m = 4.437 kg

6 0
2 years ago
Dao makes a table to identify the variables used in the equations for centripetal acceleration. A 2 column 5 rows. The first col
Zanzabum

Answer:

Column X. Tangential Speed

Column Y. radius  

Explanation:

The equation for centripetal acceleration is

           a_{c} = v² / r

Where v is the tangential velocity of the body and the radius of curvature.

To analyze this equation you must place the tangential velocity in one column and in the other the turning radius

Let's check the answers

Column X. Tangential Speed

Column Y. radius  

This is the correct answer.

5 0
2 years ago
Read 2 more answers
A circular coil 17.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this locati
sergeinik [125]

Answer:

The torque in the coil is  4.9 × 10⁻⁵ N.m  

Explanation:

T = NIABsinθ

Where;

T is the  torque on the coil

N is the number of loops = 9

I is the current = 7.8 A

A is the area of the circular coil = ?

B is the Earth's magnetic field = 5.5 × 10⁻⁵ T

θ is the angle of inclination = 90 - 56 = 34°

Area of the circular coil is calculated as follows;

A = \frac{\pi d^2}{4} \\\\A = \frac{\pi 0.17^2}{4} =0.0227 m^2

T = 9 × 7.8 × 0.0227 × 5.5×10⁻⁵ × sin34°

T = 4.9 × 10⁻⁵ N.m

Therefore, the torque in the coil is  4.9 × 10⁻⁵ N.m

5 0
2 years ago
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