Ask question

Ask question

All categories

3 years ago

11

Which of the following caught in or caught between hazard poses the greatest risk when working in excavations and trenches? A. B

eing pinned B. Cave-ins C. Equipment falling into trenches D. Severing of underground utilities

2 answers:

RSB [31]3 years ago

6 0

I believe the correct answer is b.) cave-ins

Vlada [557]3 years ago

4 0

A being pinned if a big bolder falls on you

You might be interested in

A biker first accelerates from 0.0 m/s to 6.0 m/s in 6 s, then continues at this speed for 5 s. What is the total distance trave

Svetradugi [14.3K]

Answer:

48m

Explanation:

Given the following data;

Initial velocity = 0m/s

Final velocity = 6m/s

Time, t = 6 secs

Time, T2 = 5 secs

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Substituting into the equation;

$a = \frac{6 - 0}{6}$

$a = \frac{6}{6}$

Acceleration, a = 1m/s²

<u>To find the distance covered in the first phase;</u>

<em>Solving for distance, we would use the second equation of motion;</em>

$S = ut + \frac {1}{2}at^{2}$

<em>Substituting the values into the equation;</em>

$S = 0(6) + \frac {1}{2}*1*(6)^{2}$

$S = 0 + \frac {1}{2}*1*36$

$S = 0.5 *36$

Distance, S1 = 18m

<u>For the second phase, time T2 = 5 secs;</u>

<em>Mathematically, speed is given by the equation;</em>

$Speed = \frac{distance}{time}$

<em>Making distance the subject of formula, we have;</em>

$Distance, S = speed * time$

<em>Substituting into the above equation;</em>

$Distance, S = 6 * 5$

Distance, S2 = 30m

Total distance = S1 + S2 = 18m + 30m = 48m

Total distance = 48m

<em>Therefore, the total distance traveled by the biker is 48m.</em>

4 0

3 years ago

What happens at a convergent boundary?

LenaWriter [7]

Convergent boundaries form earthquakes, which forms mountains and islands.

8 0

3 years ago

Read 2 more answers

A river flows due east at 12 km/h. A boat crosses the 720 m wide river by maintaining a constant velocity of 72 mi/h due north r

zhenek [66]

Answer:

The boat will be 74 .17 meters downstream by the time it reaches the shore.

Explanation:

Consider the vector diagrams for velocity and distance shown below.

converting 72 miles per hour to km/hr

we have 72 miles per hour 72 × 1.60934 = 115.83 km/hr

The velocity vectors form a right angled triangle, and can be solved using simple trigonometric laws

$tan \theta = \frac{12}{115.873}$

$\theta = tan^{-1}( \frac{12}{115.873})=5.9126$

This is the vector angle with which the ship drifts away with respect to its northward direction.

<em>From the sketch of the displacement vectors, we can use trigonometric ratios to determine the distance the boat moves downstream.</em>

Let x be the distance the boat moves downstream.d

$sin(5.9126)=\frac{x}{720}$

$x= 720\times 5.9126$

$x=74.17m$

∴The boat will be 74 .17 meters downstream by the time it reaches the shore.

6 0

3 years ago

Four capacitors with capacitance 3.0 pF, 2.0 pF, 5.0 pF and X pF are connected in series to each other. If the equivalent capaci

bagirrra123 [75]

Answer:

<h2>A. 6pF</h2>

Explanation:

If unknown capacitance C1, C2, C3 and C4 are connected in series to one another, their equivalent capacitance of the circuit will be expressed as shown

$\frac{1}{C_t} = \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3} +\frac{1}{C_4} \\$

Given the capacitance's 3.0 pF, 2.0 pF, 5.0 pF and X pF connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, then to get X, we will apply the formula above;

$\frac{1}{0.83} = \frac{1}{3.0} +\frac{1}{2.0} +\frac{1}{5.0} +\frac{1}{C_4} \\\\\\1.205 = 0.333+0.5+0.2+\frac{1}{C_4} \\\\1.205 = 1.033 + \frac{1}{C_4} \\\\\frac{1}{C_4} = 1.205-1.033\\\\\frac{1}{C_4} = 0.172\\\\C_4 = \frac{1}{0.172}\\ \\C_4 = 5.8pF\\\\$

C₄ ≈ 6pF

Hence the value of the X capacitor is approximately 6pF

8 0

3 years ago

In the 25 ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int

Aleonysh [2.5K]

Answer:

a) <em>8.33 x 10^-6 Pa</em>

b) <em>8.23 x 10^-11 atm</em>

c) <em>1.67 x 10^-5 Pa</em>

d) <em>1.65 x 10^-10 atm</em>

<em></em>

Explanation:

Intensity of the light $I$ = 2500 W/m^2

speed of light $c$ <u> </u>= 3 x 10^8 m/s

a) we know that the pressure for for a totally absorbing surface is given as

$P_{abs}$ = $I/c$ = 2500/(3 x 10^8) = <em>8.33 x 10^-6 Pa</em>

b) 1 atm = 101325 Pa

$P_{abs}$ = (8.33 x 10^-6)/101325 = <em>8.23 x 10^-11 atm</em>

c) for a totally reflecting surface

$P_{ref}$ = $2I/c$ = twice the value for totally absorbing

$P_{ref}$ = 2 x 8.33 x 10^-6 = <em>1.67 x 10^-5 Pa</em>

d) 1 atm = 101325 Pa

$P_{ref}$ = 2 x 8.23 x 10^-11 = <em>1.65 x 10^-10 atm</em>

8 0

3 years ago