Work is being done in which of these situations? all motions are at a constant velocity

A rectangular beam 200 mm deep and 300 mm wide is simply supported over a span of 8 m. What uniformly distributed load per metre

Alexandra [31]

The uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.

<h3>What is bending stress?</h3>

When an object is subjected to a heavy load at a specific spot, it often experiences bending stress, which causes the object to bend and tire.

The given data in the problem is;

Bending stress, σ = 120 N/mm2

Moment of inertia, I = 8.5 × 106 mm⁴

Depth of beam, y = d/2 = 200/2 = 100 mm

Length of beam, L = 8 m = 8000 mm

Width of beam, W = 300 mm

The maximum bending moment of the beam with UDL;

$\rm W = \frac{wL^2}{8}$

From the bending equation;

$\rm M = \frac{\sigma I}{y_{max}} \\\\ M = \frac{120 \ N / mm^2 \times 8.5 \times 10^6 }{100 \ mm } \\\\ M = 10.2 \times 10^6 \ N - mm$

The maximum bending moment of the beam with UDL;

$\rm M = \frac{wL^2}{8} \\\\ 10.2 \times 10^6 = \frac{w \times 8^2}{8} \\\\ w = 1.275 \times 10^6 \ N/mm$

Hence the uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.

To learn more about the bending stress, refer;

brainly.com/question/24227487

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5 0

2 years ago

A 35 kg boy is riding a 65 kg go-cart. He pushes on the gas pedal, causing the cart to accelerate at 5 m/s2. Use the equation F

trasher [3.6K]

Answer:

$\boxed {\boxed {\sf 500 \ Newtons }}$

Explanation:

The equation for force is given:

$F=m*a$

First, we must find the total mass, which is the sum of the boy's mass and the go-cart's mass.

total mass= boy's mass + go cart's mass

The boy's mass is 35 kilograms and the go cart's is 65 kilograms.

total mass= 35 kg+ 65 kg=100 kg

Now we know the total mass and the acceleration.

$m= 100 \ kg \\a= 5 \ m/s^2$

Substitute the values into the formula.

$F=100 \ kg * 5 \ m/s^2$

Multiply.

$F= 500 \ kg*m/s^2$

1 kilograms meter per square second is equal to 1 Newton.
Our answer of 500 kg*m/s² is equal to 500 Newtons.

$F= 500 \ N$

The force exerted by the go cart engine is <u>500 Newtons.</u>

4 0

3 years ago

Read 2 more answers

A buoy is a floating device that can have many purposes, but often as a locator for ships. Collin constructs a hollow metal buoy

Yuri [45]

Answer:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water

B )angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

c ) h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

Explanation:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water this is because the substances with lesser density floats when placed in a substance with a higher density

B ) when the Buoy is at rest ( t = 0 ) and is then pushed down calculate angular frequency ( w ) of small oscillations in terms of the given variables

volume of cone = hA /3.

h = height of cone, A = Base Area.

therefore the total volume of the Buoy above water level ( at rest )

= ( $\pi r^{2} * \frac{h}{3}$ ) * 2 = $\frac{\pi r^{2} h }{6}$

The part of the buoy immersed in water = x

then the net upward force will be

fb ( force of Buoy ) = fg ( force of gravity )

note force = Mass * Acceleration

force of buoy = $\frac{\alpha }{2} *$ $\frac{\pi r^{2} }{h}$ * a = ( force of gravity ) $\alpha * T\frac{r^{2} }{4} * x * g$

therefore a = $\frac{3g}{h} * x$

angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

C ) height of the each cone

$\frac{2\pi }{w} = 1$ therefore w = 2 $\pi$

back to the angular frequency : $\frac{3g}{h} = 4\pi ^{2}$

therefore h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

4 0

3 years ago

Help me quick!!! please!!

Rudik [331]

Answer:

39.240 W

Explanation:

Let's start by calculating the work done by the engine. We can assume that it is the same work done by the weight of the object to bring it from 40m to the surface: as much energy it takes to bring it up, the same ammount it takes to bring it down. Said work is $w= \vec F\cdot \vec{h} = mg h = 1000 \times 9.81\times 40 = 392.400 J$