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3 years ago

A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press

Physics

1 answer:

Ghella [55]3 years ago

8 0

Answer:

$\mathbf{F_1=4.41*10^4\ N}$

$\mathbf{F_2 = 1.176*10^5 \ N}$

Explanation:

The missing image of the figure slide is attached in below.

However, from the model, it is obvious that it is in equilibrium.

As a result, the relation of the force and the torque is said to be zero.

i.e.

$\sum F = 0$ and $\sum \tau = 0$

From the image, expressing the forces through the y-axis, we have:

$F_1+F_2 = W_B + W_P \\ \\ \implies 9.8(1500+15000) \\ \\ \implies \mathtt{1.617\times 10^5 \ N}$

Also, let the force $F_1$ be the pivot and computing the torque to determine $F_2$ :

Then:

$F_1(0)+F_2(20.0) = 10.0W_B + 15.0W_P$

$F_2 = \dfrac{((10*1500)+(15*15000))*9.8}{20.0}$

$F_2 = 117600 \ N$

$\mathbf{F_2 = 1.176*10^5 \ N}$

For the force equation:

$F_1+F_2=1.617*10^5 \ N;$

where:

$F_2 = 1.176*10^5 \ N$

Then:

$F_1+1.176*10^5 \ N=1.617*10^5 \ N$

$F_1=1.617*10^5 \ N-1.176*10^5 \ N$

$F_1=44100\ N$

$\mathbf{F_1=4.41*10^4\ N}$

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Answer:

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3 years ago

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5. J. J. Thomson

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3 years ago

What are some important factors to consider when choosing a warm-up before your workout?

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<u>Answer:</u>

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3 years ago

Read 2 more answers

A jet aircraft is traveling at 262 m/s in hor-

NeTakaya

Solution :

Speed of the air craft, $S_a$ = 262 m/s

Fuel burns at the rate of, $S_b$ = 3.92 kg/s

Rate at which the engine takes in air, $S_{air}$ = 85.9 kg/s

Speed of the exhaust gas that are ejected relative to the aircraft, $S_{exh}$ =921 m/s

Therefore, the upward thrust of the jet engine is given by

$F=S_{air}(S_{exh}-S_a)+(S_b \times S_{exh})$

F = 85.9(921 - 262) + (3.92 x 921)

= 4862635.79 + 3610.32

= $4.8 \times 10^6 \ N$

Therefore thrust of the jet engine is $4.8 \times 10^6 \ N$ .

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2 years ago

Ballistic data obtained on a firing range show that aerodynamic drag reduces the speed of a .44 magnum revolver bullet from 250

m_a_m_a [10]

Answer:

0.363999909622

Explanation:

F = Force

m = Mass = 15.6 g

C = Drag coefficient

ρ = Density of air = 1.21 kg/m³

A = Surface area = $\dfrac{\pi}{4}d^2$

v = Terminal velocity = $v=210\ m/s$

s = Displacement = 150 m

$a=\dfrac{v^2-u^2}{2s}$

Force is given by

F = ma

$F=\dfrac{1}{2}\rho CAv^2\\\Rightarrow ma=\dfrac{1}{2}\rho CAv^2\\\Rightarrow m\dfrac{v^2-u^2}{2s}=\dfrac{1}{2}\rho CAv^2\\\Rightarrow C=2\times m\dfrac{v^2-u^2}{2s}\times\dfrac{1}{\rho Av^2}\\\Rightarrow C=2\times15.6\times 10^{-3}\dfrac{210^2-250^2}{2\times 150}\times\dfrac{1}{1.21\times\dfrac{\pi}{4}\times (11.2\times 10^{-3})^2(210)^2}\\\Rightarrow C=-0.363999909622$

The drag coefficient is 0.363999909622 (ignoring negative sign)

4 0

3 years ago