All categories

3 years ago

What should not be used to support a scaffold because they could collapse, sending the structure to the ground?

Physics

1 answer:

uranmaximum [27]3 years ago

5 0

Answer:

Cinder blocks

Explanation:

From safety training guides, whenever there is inadequate support, poor

construction, or a movement in the components of the scaffold such as the coupler as well as the base upon which the scaffold structure is built, there will be a great danger of collapse.

In light of this safety rule, Cinder blocks such as sandcrete blocks, bricks e.t.c shouldn't be used to support scaffold because they have a high tendency of being crushed under load

You might be interested in

_________ forces exist between objects that are not in contact with each other

valina [46]

Answer:

Gravity as well as electrostatic and magnetic attraction and repulsion provide real life examples of forces being exerted by one object on another without them being in contact with each other. Many children are aware of magnetism and have played with fridge magnets.

8 0

3 years ago

A 2.50 gram rectangular object has measurements of 22.0 mm, 13.5 mm, and 12.5 mm. what is the objectâs density in units of g/ml?

navik [9.2K]

<span>To calculate the density, we must first calculate the volume of the object. For a rectangle, the volume is equal to V=l*w*h. In this case, V=22*13.5*12.5=3712.5 mm^3. Next we have to convert the mm^3 to mL using the conversion factor of 1 mm^3=.001 mL. To get mL, we set up the expression 3712.5 mm^3 * (.001mL/1mm^3) = 3.7125 mL. Finally, to get the density, we know density is mass over volume, so we set the expression p=m/V=2.5g/3.7125mL=.6734 which rounded to 3 significant figures is .673. The answer will be .673</span>

8 0

3 years ago

One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra

sveticcg [70]

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration ( $a_{r}$ ), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

$a_{r} = 4\pi^{2}\cdot f^{2}\cdot R$ (1)

Where:

$f$ - Frequency, measured in hertz.

$R$ - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension ( $T$ ), measured in newtons, through the string, then we derive the following model:

$\Sigma F = T = m\cdot a_{r}$ (2)

Where $m$ is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

$T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R$ (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

$\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right)$ (4)

$\frac{T_{2}}{T_{1}} = 4\cdot 2$

$\frac{T_{2}}{T_{1}} = 8$

If both the radius and frequency are doubled, then the tension is increased 8 times.

5 0

3 years ago

A carnot heat engine receives 600 kj of heat from a source of unknown temperature and rejects 175 kj of it to a sink at 20°c. de

oee [108]

(b) 71%

The thermal efficiency of a Carnot heat engine is given by:

$\eta = \frac{W}{Q_{in}}$

where

W is the useful work done by the engine

$Q_{in}$ is the heat in input to the machine

In this problem, we have:

$Q_{in}=600 kJ$ is the heat absorbed

$W=600 kJ-175 kJ=425 kJ$ is the work done (175 kJ is the heat released to the sink, therefore the work done is equal to the difference between the heat in input and the heat released)