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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?

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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Answer:

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Explanation:

First of all, we need to find the volume of the cylinder.

The volume of the cylinder is given by:

$V=\pi r^2 h$

where:

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$h=51.0 mm=0.051 m$ is the height

Substituting, we find

$V=\pi (0.021 m)^2 (0.051 m)=7.1 \cdot 10^{-5} m^3$

And the density is given by

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2 years ago

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Romashka-Z-Leto [24]

Answer:

v=8m/s

Explanation:

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We can calculate the speed of the cars after the collision by using

$W_f=(0.7)(1650kg+1900kg)(9.8\frac{m}{s^2})=24353J\\24353J=\frac{1}{2}(1650kg+1900kg)v^2\\v=\sqrt{\frac{24353J}{1775kg}}=3.70\frac{m}{s}$

Now , we can compute the speed of the second car by taking into account the conservation of the momentum

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Hope this helps!!

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2 years ago

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Light from a monochromatic source shines through a double slit onto a screen 5.00 m away. The slits are 0.180 mm apart. The dark

Nitella [24]

Answer:

Wavelength of incident light, $\lambda = 612 nm$

Given:

Distance between slit and screen, x = 5.00 m

slit width, d = 0.180 mm

width of the fringe, $\beta = 1.70 cm = 0.017 m$

Solution:

To calculate the wavelength of the incident light, $\lambda$ :

$\beta = \frac{x\lambda }{d}$

$\lambda = \frac{\beta d}{x}$

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2 years ago

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Answer:

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Explanation:

Given that, the speed of stolen car is,

$v_{s} =108km/hr\\v_{s} =108\times \frac{5}{18}m/s\\ v_{s} =30m/s$

As policeman start chasing the stolen car after 60 seconds.

Now suppose the speed of policeman car is, $v_{p}$

The policeman catches the stolen car at a distance of,

$S=60km\\S=60000m$

Now the distance covered by the policeman in time t is $v_{p}\times t$

And the distane cover by the thief in stolen car in time(t+60s) is $v_{s}\times (t+60sec)$ .

And these distances are equal and they are equal to 60000 m.

Therefore,

$v_{p}\times t=v_{s}\times (t+60sec)=60000m$

Therfore,

$v_{s}\times (t+60sec)=60000m\\30m/s\times (t+60sec)=60000m\\(t+60s)=2000s\\t=1940s$

Now use this value to solve for minimum speed of policeman's car.

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T = 300 N

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1 year ago